{"id":4098,"date":"2022-04-14T18:15:53","date_gmt":"2022-04-14T18:15:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-double-integrals-over-rectangular-regions\/"},"modified":"2022-11-09T16:40:42","modified_gmt":"2022-11-09T16:40:42","slug":"skills-review-for-double-integrals-over-rectangular-regions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-double-integrals-over-rectangular-regions\/","title":{"raw":"Skills Review for Double and Triple Integrals","rendered":"Skills Review for Double and Triple Integrals"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the general antiderivative of a given function<\/li>\r\n \t<li>Evaluate trigonometric functions at specific angle measures<\/li>\r\n \t<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\r\n \t<li>Solve rational equations by clearing denominators<\/li>\r\n \t<li>Solve for a variable in a square root equation<\/li>\r\n \t<li>Convert between radical and exponent notations<\/li>\r\n \t<li>Solve for a variable in a complex radical or rational exponent equation<\/li>\r\n \t<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the sections that discuss double and triple integrals, we will learn how to integrate functions with more than one independent variable over various types of regions in various coordinate systems. Here we will review basic integration techniques, evaluating trigonometric functions, evaluating definite integrals using the Second Fundamental Theorem of Calculus, manipulating equations, and how to find limits at infinity.\r\n<h2>Indefinite Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043393369\">Given a function [latex]f[\/latex], the <strong>indefinite integral<\/strong> of [latex]f[\/latex], denoted<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(x) dx[\/latex],<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\nis the most general antiderivative of [latex]f[\/latex]. If [latex]F[\/latex] is an antiderivative of [latex]f[\/latex], then\r\n<div id=\"fs-id1165043119692\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(x) dx=F(x)+C[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165043096049\">The expression [latex]f(x)[\/latex] is called the <em>integrand<\/em> and the variable [latex]x[\/latex] is the <em>variable of integration<\/em>.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043041347\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Power Rule for Integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165042514785\">For [latex]n \\ne \u22121[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165043250161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int x^n dx=\\dfrac{x^{n+1}}{n+1}+C[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043385541\">Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/appendix-b-table-of-derivatives\/\" target=\"_blank\" rel=\"noopener\">Appendix B: Table of Derivatives<\/a>.<\/p>\r\n\r\n<table summary=\"This is a table with two columns and fourteen rows, titled \u201cIntegration Formulas.\u201d The first row is a header row, and labels column one \u201cDifferentiation Formula\u201d and column two \u201cIndefinite Integral.\u201d The second row reads d\/dx (k) = 0, the integral of kdx = the integral of kx^0dx = kx + C. The third row reads d\/dx(x^n) = nx^(x-1), the integral of x^ndn = (x^n+1)\/(n+1) + C for n is not equal to negative 1. The fourth row reads d\/dx(ln(the absolute value of x))=1\/x, the integral of (1\/x)dx = ln(the absolute value of x) + C. The fifth row reads d\/dx(e^x) = e^x, the integral of e^xdx = e^x + C. The sixth row reads d\/dx(sinx) = cosx, the integral of cosxdx = sinx + C. The seventh row reads d\/dx(cosx) = negative sinx, the integral of sinxdx = negative cosx + C. The eighth row reads d\/dx(tanx) = sec squared x, the integral of sec squared xdx = tanx + C. The ninth row reads d\/dx(cscx) = negative cscxcotx, the integral of cscxcotxdx = negative cscx + C. The tenth row reads d\/dx(secx) = secxtanx, the integral of secxtanxdx = secx + C. The eleventh row reads d\/dx(cotx) = negative csc squared x, the integral of csc squared xdx = negative cot x + C. The twelfth row reads d\/dx(sin^-1(x)) = 1\/the square root of (1 \u2013 x^2), the integral of 1\/(the square root of (x^2 \u2013 1) = sin^-1(x) + C. The thirteenth row reads d\/dx (tan^-1(x)) = 1\/(1 + x^2), the integral of 1\/(1 + x^2)dx = tan^-1(x) + C. The fourteenth row reads d\/dx(sec^-1(the absolute value of x)) = 1\/x(the square root of x^2 \u2013 1), the integral of 1\/x(the square root of x^2 \u2013 1)dx = sec^-1(the absolute value of x) + C.\"><caption>Integration Formulas<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Differentiation Formula<\/th>\r\n<th>Indefinite Integral<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(k)=0[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int kdx=\\displaystyle\\int kx^0 dx=kx+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(x^n)=nx^{n-1}[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int x^n dx=\\frac{x^{n+1}}{n+1}+C[\/latex] for [latex]n\\ne \u22121[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\ln |x|)=\\frac{1}{x}[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\frac{1}{x}dx=\\ln |x|+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(e^x)=e^x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int e^x dx=e^x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\cos x dx= \\sin x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\cos x)=\u2212 \\sin x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\tan x)= \\sec^2 x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\sec^2 x dx= \\tan x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\csc x)=\u2212\\csc x \\cot x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\csc x \\cot x dx=\u2212\\csc x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\sec x)= \\sec x \\tan x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\sec x \\tan x dx= \\sec x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\cot x)=\u2212\\csc^2 x[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\csc^2 x dx=\u2212\\cot x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}( \\sin^{-1} x)=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-x^2}} dx= \\sin^{-1} x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\tan^{-1} x)=\\frac{1}{1+x^2}[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\frac{1}{1+x^2} dx= \\tan^{-1} x+C[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{d}{dx}(\\sec^{-1} |x|)=\\frac{1}{x\\sqrt{x^2-1}}[\/latex]<\/td>\r\n<td>[latex]\\displaystyle\\int \\frac{1}{x\\sqrt{x^2-1}} dx= \\sec^{-1} |x|+C[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165043425485\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Properties of Indefinite Integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043395041\">Let [latex]F[\/latex] and [latex]G[\/latex] be antiderivatives of [latex]f[\/latex] and [latex]g[\/latex], respectively, and let [latex]k[\/latex] be any real number.<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1165043393659\"><strong>Sums and Differences<\/strong><\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (f(x) \\pm g(x)) dx=F(x) \\pm G(x)+C[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042328714\"><strong>Constant Multiples<\/strong><\/p>\r\n\r\n<div id=\"fs-id1165042328717\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int kf(x) dx=kF(x)+C[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165043248811\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating Indefinite Integrals<\/h3>\r\nEvaluate each of the following indefinite integrals:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\frac{x^2+4\\sqrt[3]{x}}{x} dx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\frac{4}{1+x^2} dx[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\int \\tan x \\cos x dx[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165042705917\" class=\"exercise\">[reveal-answer q=\"fs-id1165042552215\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042552215\"]\r\n<ol id=\"fs-id1165042552215\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain\r\n<div id=\"fs-id1165042552227\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx=\\displaystyle\\int 5x^3 dx-\\displaystyle\\int 7x^2 dx+\\displaystyle\\int 3x dx+\\displaystyle\\int 4 dx[\/latex]<\/div>\r\nFrom the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives\r\n<div id=\"fs-id1165043312575\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 5x^3 dx-\\displaystyle\\int 7x^2 dx+\\displaystyle\\int 3x dx+\\displaystyle\\int 4 dx=5\\displaystyle\\int x^3 dx-7\\displaystyle\\int x^2 dx+3\\displaystyle\\int x dx+4\\displaystyle\\int 1 dx[\/latex]<\/div>\r\nUsing the power rule for integrals, we conclude that\r\n<div id=\"fs-id1165042407363\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx=\\frac{5}{4}x^4-\\frac{7}{3}x^3+\\frac{3}{2}x^2+4x+C[\/latex]<\/div><\/li>\r\n \t<li>Rewrite the integrand as\r\n<div id=\"fs-id1165042371846\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{x^2+4\\sqrt[3]{x}}{x}=\\frac{x^2}{x}+\\frac{4\\sqrt[3]{x}}{x}[\/latex]<\/div>\r\nThen, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have\r\n<div id=\"fs-id1165043427498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int (x+\\frac{4}{x^{2\/3}}) dx &amp; =\\displaystyle\\int x dx+4\\displaystyle\\int x^{-2\/3} dx \\\\ &amp; =\\frac{1}{2}x^2+4\\frac{1}{(\\frac{-2}{3})+1}x^{(-2\/3)+1}+C \\\\ &amp; =\\frac{1}{2}x^2+12x^{1\/3}+C \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Using the properties of indefinite integrals, write the integral as\r\n<div id=\"fs-id1165043348665\" class=\"equation unnumbered\">[latex]4\\displaystyle\\int \\frac{1}{1+x^2} dx[\/latex].<\/div>\r\nThen, use the fact that [latex] \\tan^{-1} (x)[\/latex] is an antiderivative of [latex]\\frac{1}{1+x^2}[\/latex] to conclude that\r\n<div id=\"fs-id1165042374764\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{4}{1+x^2} dx=4 \\tan^{-1} (x)+C[\/latex]<\/div><\/li>\r\n \t<li>Rewrite the integrand as\r\n<div class=\"equation unnumbered\">[latex] \\tan x \\cos x=\\frac{ \\sin x}{ \\cos x} \\cos x= \\sin x[\/latex].<\/div>\r\nTherefore,\r\n<div id=\"fs-id1165043317182\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\tan x \\cos x dx=\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\displaystyle\\int (4x^3-5x^2+x-7) dx[\/latex]\r\n\r\n[reveal-answer q=\"4078823\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"4078823\"]\r\n\r\nIntegrate each term in the integrand separately, making use of the power rule.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043259694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043259694\"]\r\n\r\n[latex]x^4-\\frac{5}{3}x^3+\\frac{1}{2}x^2-7x+C[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]210143[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Evaluate Trigonometric Functions<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong>\r\nThe unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.\r\n\r\nRemember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function's value at a given angle.\r\n\r\n<img class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Evaluating Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\r\nIf [latex] \\theta[\/latex] is an angle measure, then, using the unit circle,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan \\theta=\\frac{sin \\theta}{cos \\theta}\\\\ \\sec \\theta=\\frac{1}{cos \\theta}\\\\ \\csc t=\\frac{1}{sin \\theta}\\\\ \\cot \\theta=\\frac{cos \\theta}{sin \\theta}\\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Unit Circle to Find the Value of Trigonometric Functions<\/h3>\r\nFind [latex]\\sin \\theta,\\cos \\theta,\\tan \\theta,\\sec \\theta,\\csc \\theta[\/latex], and [latex]\\cot \\theta[\/latex] when [latex] \\theta=\\frac{\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"151548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151548\"]\r\n\r\n[latex]\\begin{align}&amp;\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &amp;\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &amp;\\tan \\frac{\\pi }{3}=\\sqrt{3} \\text{ (From}\\frac{sin \\theta}{cos \\theta})\\\\ &amp;\\sec \\frac{\\pi }{3}=2 \\text{ (From}\\frac{1}{cos \\theta})\\\\ &amp;\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\text{ (From}\\frac{1}{sin \\theta} \\text{ - do not forget to rationalize})\\\\ &amp;\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}(\\text{ (From}\\frac{cos \\theta}{sin \\theta} \\text{ - do not forget to rationalize})\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173354[\/ohm_question]\r\n\r\n<\/div>\r\nBelow are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.\r\n<table id=\"Table_05_03_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Tangent<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Secant<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>2<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosecant<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>2<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cotangent<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\r\n<strong><em>(also in Module 3, Skills Review for Arc Length and Curvature)<\/em><\/strong>\r\n<p id=\"fs-id1170572622222\">The <strong>Fundamental Theorem of Calculus, Part 2<\/strong> (also known as the <strong>evaluation theorem<\/strong>) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\r\n\r\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\r\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\r\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\r\n\r\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\r\n<div id=\"fs-id1170571678913\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1170572173704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572173704\"]\r\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\r\n\r\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\r\n\r\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill &amp; =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ &amp; =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ &amp; =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ &amp; =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ &amp; =\\frac{16}{3}-16\\hfill \\\\ &amp; =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\r\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572130389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572130389\"]\r\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\r\n\r\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\r\n\r\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\r\n\r\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill &amp; ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ &amp; =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ &amp; =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ &amp; =18-6-\\frac{2}{3}+2\\hfill \\\\ &amp; =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]\r\n\r\n[reveal-answer q=\"3887125\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3887125\"]\r\n\r\nUse the power rule.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571639103\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571639103\"]\r\n\r\n[latex]\\frac{7}{24}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]211336[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Manipulate Equations<\/h2>\r\nIt will sometimes be necessary to change the bounds of an integral which will even require manipulating the given equation.\r\n<h3>Manipulate Rational Equations<\/h3>\r\nEquations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex] \\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).\r\n\r\nOne of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.\r\n\r\nSolve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.\r\n\r\nMultiply both sides of the equation by\u00a0[latex]4[\/latex], the common denominator of the fractional coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving a Rational Equation For a Specific Variable<\/h3>\r\nSolve the equation [latex] \\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].\r\n[reveal-answer q=\"425621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425621\"]\r\n\r\nFind the least common denominator of\u00a0[latex]8[\/latex] and\u00a0[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.\r\n\r\nMultiply both sides of the equation by the common denominator,\u00a0[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving a Rational Equation For a Specific Variable<\/h3>\r\nSolve the equation [latex] x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].\r\n\r\n[reveal-answer q=\"331190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331190\"]\r\n\r\nClear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Manipulate Radical Equations<\/h3>\r\n<strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3y+18}=x &amp; \\\\ \\sqrt{x+3}=y-3 &amp; \\\\ \\sqrt{x+5}-\\sqrt{y - 3}=2\\end{array}[\/latex]<\/div>\r\nRadical equations are manipulated by eliminating each radical, one at a time until you have solved for the indicated variable.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Radical Equations<\/h3>\r\nAn equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a radical equation, solve it<\/h3>\r\n<ol>\r\n \t<li>Isolate the radical expression containing your variable of interest on one side of the equal sign. Put all remaining terms on the other side.<\/li>\r\n \t<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\r\n \t<li>Solve the resulting equation for the variable of interest.<\/li>\r\n \t<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation with One Radical<\/h3>\r\nSolve [latex]\\sqrt{15 - 2y}=x[\/latex] for [latex]y[\/latex].\r\n\r\n[reveal-answer q=\"503795\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"503795\"]\r\n\r\nThe radical is already isolated on the left side of the equal sign, so proceed to square both sides.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2y}=x &amp; \\\\ {\\left(\\sqrt{15 - 2y}\\right)}^{2}={\\left(x\\right)}^{2} &amp; \\\\ 15 - 2y={x}^{2}\\end{array}[\/latex]<\/div>\r\nNow isolate [latex]y[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll} 15 - 2y={x}^{2} \\\\ -2y={x}^{2}-15 \\\\ y=-\\frac{1}{2}{x}^{2}+\\frac{15}{2}\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the radical equation [latex]\\sqrt{y+3}=3x - 1[\/latex]\u00a0for [latex]y[\/latex].\r\n\r\n[reveal-answer q=\"719648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"719648\"]\r\n\r\n[latex]y=9{x}^{2}-6x-2[\/latex] Note: When you square the binomial on the right side of the equation you get [latex]9{x}^{2}-6x+1[\/latex]. Then, take away [latex]3[\/latex] from both sides.[\/hidden-answer]\r\n\r\n<\/div>\r\nRational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex] and [latex]{8}^{\\frac{2}{3}}[\/latex] is another way of writing [latex]\\left(\\sqrt[3]{8}\\right)^2[\/latex].\r\n<p style=\"padding-left: 30px;\">We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex].<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Rational Exponents<\/h3>\r\nA rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:\r\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\r\n<\/div>\r\n<h2>Take Limits at Infinity<\/h2>\r\n<div id=\"fs-id1165043145047\" class=\"bc-section section\">\r\n<p id=\"fs-id1165043107285\">Recall that [latex]\\underset{x \\to a}{\\lim}f(x)=L[\/latex] means [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently close to [latex]a[\/latex]. We can extend this idea to limits at infinity. For example, consider the function [latex]f(x)=2+\\frac{1}{x}[\/latex]. As can be seen graphically below, as the values of [latex]x[\/latex] get larger, the values of [latex]f(x)[\/latex] approach 2. We say the limit as [latex]x[\/latex] approaches [latex]\\infty [\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=2[\/latex]. Similarly, for [latex]x&lt;0[\/latex], as the values [latex]|x|[\/latex] get larger, the values of [latex]f(x)[\/latex] approaches 2. We say the limit as [latex]x[\/latex] approaches [latex]\u2212\\infty [\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to a}{\\lim}f(x)=2[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"717\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211025\/CNX_Calc_Figure_04_06_019.jpg\" alt=\"The function f(x) 2 + 1\/x is graphed. The function starts negative near y = 2 but then decreases to \u2212\u221e near x = 0. The function then decreases from \u221e near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.\" width=\"717\" height=\"423\" \/> The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].[\/caption]\r\n<p id=\"fs-id1165042936244\">More generally, for any function [latex]f[\/latex], we say the limit as [latex]x \\to \\infty [\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \\infty}{\\lim}f(x)=L[\/latex]. Similarly, we say the limit as [latex]x\\to \u2212\\infty [\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x&lt;0[\/latex] and [latex]|x|[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=L[\/latex]. We now look at the definition of a function having a limit at infinity.<\/p>\r\n\r\n<div id=\"fs-id1165042331960\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1.2em; font-weight: 600; text-align: center; text-transform: uppercase; background-color: #eeeeee;\">Definition<\/span><\/p>\r\n\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042970725\">(Informal) If the values of [latex]f(x)[\/latex] become arbitrarily close to [latex]L[\/latex] as [latex]x[\/latex] becomes sufficiently large, we say the function [latex]f[\/latex] has a limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165042986551\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042374662\">If the values of [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] for [latex]x&lt;0[\/latex] as [latex]|x|[\/latex] becomes sufficiently large, we say that the function [latex]f[\/latex] has a limit at negative infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165043105208\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -\\infty }{\\lim}f(x)=L[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043157752\">If the values [latex]f(x)[\/latex] are getting arbitrarily close to some finite value [latex]L[\/latex] as [latex]x\\to \\infty [\/latex] or [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] approaches the line [latex]y=L[\/latex]. In that case, the line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] (Figure 2). For example, for the function [latex]f(x)=\\frac{1}{x}[\/latex], since [latex]\\underset{x\\to \\infty }{\\lim}f(x)=0[\/latex], the line [latex]y=0[\/latex] is a horizontal asymptote of [latex]f(x)=\\frac{1}{x}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043262534\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042973921\">If [latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex] or [latex]\\underset{x \\to \u2212\\infty}{\\lim}f(x)=L[\/latex], we say the line [latex]y=L[\/latex] is a <strong>horizontal asymptote<\/strong> of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"766\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211028\/CNX_Calc_Figure_04_06_020.jpg\" alt=\"The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.\" width=\"766\" height=\"273\" \/> Figure 2. (a) As [latex]x\\to \\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]L[\/latex]. The line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex]. (b) As [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]M[\/latex]. The line [latex]y=M[\/latex] is a horizontal asymptote of [latex]f[\/latex].[\/caption]\r\n<p id=\"fs-id1165042647732\">A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as [latex]x[\/latex] approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function [latex]f(x)=\\frac{ \\cos x}{x}+1[\/latex] shown in Figure 3 intersects the horizontal asymptote [latex]y=1[\/latex] an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"529\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211031\/CNX_Calc_Figure_04_06_002.jpg\" alt=\"The function f(x) = (cos x)\/x + 1 is shown. It decreases from (0, \u221e) and then proceeds to oscillate around y = 1 with decreasing amplitude.\" width=\"529\" height=\"230\" \/> Figure 3. The graph of [latex]f(x)=\\cos x\/x+1[\/latex] crosses its horizontal asymptote [latex]y=1[\/latex] an infinite number of times.[\/caption]\r\n<div class=\"textbook exercises\">\r\n<h3>Example: Computing Limits at Infinity<\/h3>\r\n<p id=\"fs-id1165043262623\">For each of the following functions [latex]f[\/latex], evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042356111\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]f(x)=5-\\frac{2}{x^2}[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\dfrac{\\sin x}{x}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1165043183885\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043183885\"]\r\n<ol id=\"fs-id1165043183885\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Using the algebraic limit laws, we have [latex]\\underset{x\\to \\infty }{\\lim}(5-\\frac{2}{x^2})=\\underset{x\\to \\infty }{\\lim}5-2(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})=5-2 \\cdot 0=5[\/latex]. Similarly, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=5[\/latex].\r\n<div class=\"mceTemp\"><\/div><\/li>\r\n \t<li>nce [latex]-1\\le \\sin x\\le 1[\/latex] for all [latex]x[\/latex], we have\r\n<div id=\"fs-id1165043093355\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{-1}{x}\\le \\frac{\\sin x}{x}\\le \\frac{1}{x}[\/latex]<\/div>\r\nfor all [latex]x \\ne 0[\/latex]. Also, since\r\n<div id=\"fs-id1165043197153\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{-1}{x}=0=\\underset{x\\to \\infty }{\\lim}\\frac{1}{x}[\/latex],<\/div>\r\nwe can apply the squeeze theorem to conclude that\r\n<div id=\"fs-id1165043036581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\r\nSimilarly,\r\n<div id=\"fs-id1165043122536\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042320881\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043315935\">Evaluate [latex]\\underset{x\\to \u2212\\infty}{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"2473508\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2473508\"]\r\n<p id=\"fs-id1165042318511\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}1\/x=0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043390798\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043390798\"]\r\n<p id=\"fs-id1165043390798\">Both limits are 3.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042333169\" class=\"bc-section section\">\r\n<h2>Infinite Limits at Infinity<\/h2>\r\n<p id=\"fs-id1165042333174\">Sometimes the values of a function [latex]f[\/latex] become arbitrarily large as [latex]x\\to \\infty [\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex]. In this case, we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty [\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\\infty )[\/latex]. On the other hand, if the values of [latex]f[\/latex] are negative but become arbitrarily large in magnitude as [latex]x\\to \\infty [\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex], we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty [\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty )[\/latex].<\/p>\r\n<p id=\"fs-id1165042606820\">For example, consider the function [latex]f(x)=x^3[\/latex]. The [latex]\\underset{x\\to \\infty }{\\lim}x^3=\\infty[\/latex]. On the other hand, as [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f(x)=x^3[\/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\\underset{x\\to \u2212\\infty }{\\lim}x^3=\u2212\\infty[\/latex].<\/p>\r\n\r\n<table id=\"fs-id1165042406634\" class=\"column-header\" summary=\"The table has four rows and six columns. The first column is a header column and it reads x, x3, x, and x3. After the header, the first row reads 10, 20, 50, 100, and 1000. The second row reads 1000, 8000, 125000, 1,000,000, and 1,000,000,000. The third row reads \u221210, \u221220, \u221250, \u2212100, and \u22121000. The forth row reads \u22121000, \u22128000, \u2212125,000, \u22121,000,000, and \u22121,000,000,000.\"><caption>Values of a power function as [latex]x\\to \\pm \\infty [\/latex]<\/caption>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>10<\/td>\r\n<td>20<\/td>\r\n<td>50<\/td>\r\n<td>100<\/td>\r\n<td>1000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\r\n<td>1000<\/td>\r\n<td>8000<\/td>\r\n<td>125,000<\/td>\r\n<td>1,000,000<\/td>\r\n<td>1,000,000,000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>-10<\/td>\r\n<td>-20<\/td>\r\n<td>-50<\/td>\r\n<td>-100<\/td>\r\n<td>-1000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\r\n<td>-1000<\/td>\r\n<td>-8000<\/td>\r\n<td>-125,000<\/td>\r\n<td>-1,000,000<\/td>\r\n<td>-1,000,000,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"642\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211045\/CNX_Calc_Figure_04_06_022.jpg\" alt=\"The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.\" width=\"642\" height=\"272\" \/> For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1165043276353\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043276356\">(Informal) We say a function [latex]f[\/latex] has an infinite limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165043276364\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042709557\">if [latex]f(x)[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. We say a function has a negative infinite limit at infinity and write<\/p>\r\n\r\n<div id=\"fs-id1165042647077\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042327355\">if [latex]f(x)&lt;0[\/latex] and [latex]|f(x)|[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. Similarly, we can define infinite limits as [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042323710\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042320226\">Find [latex]\\underset{x\\to \\infty }{\\lim}3x^2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1165042708272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042708272\"]\r\n<p id=\"fs-id1165042383154\">[latex]\\infty[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16109[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the general antiderivative of a given function<\/li>\n<li>Evaluate trigonometric functions at specific angle measures<\/li>\n<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\n<li>Solve rational equations by clearing denominators<\/li>\n<li>Solve for a variable in a square root equation<\/li>\n<li>Convert between radical and exponent notations<\/li>\n<li>Solve for a variable in a complex radical or rational exponent equation<\/li>\n<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\n<\/ul>\n<\/div>\n<p>In the sections that discuss double and triple integrals, we will learn how to integrate functions with more than one independent variable over various types of regions in various coordinate systems. Here we will review basic integration techniques, evaluating trigonometric functions, evaluating definite integrals using the Second Fundamental Theorem of Calculus, manipulating equations, and how to find limits at infinity.<\/p>\n<h2>Indefinite Integrals<\/h2>\n<div class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165043393369\">Given a function [latex]f[\/latex], the <strong>indefinite integral<\/strong> of [latex]f[\/latex], denoted<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(x) dx[\/latex],<\/div>\n<div><\/div>\n<div><\/div>\n<p>is the most general antiderivative of [latex]f[\/latex]. If [latex]F[\/latex] is an antiderivative of [latex]f[\/latex], then<\/p>\n<div id=\"fs-id1165043119692\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int f(x) dx=F(x)+C[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<p id=\"fs-id1165043096049\">The expression [latex]f(x)[\/latex] is called the <em>integrand<\/em> and the variable [latex]x[\/latex] is the <em>variable of integration<\/em>.<\/p>\n<\/div>\n<div id=\"fs-id1165043041347\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Power Rule for Integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1165042514785\">For [latex]n \\ne \u22121[\/latex],<\/p>\n<div id=\"fs-id1165043250161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int x^n dx=\\dfrac{x^{n+1}}{n+1}+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1165043385541\">Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/appendix-b-table-of-derivatives\/\" target=\"_blank\" rel=\"noopener\">Appendix B: Table of Derivatives<\/a>.<\/p>\n<table summary=\"This is a table with two columns and fourteen rows, titled \u201cIntegration Formulas.\u201d The first row is a header row, and labels column one \u201cDifferentiation Formula\u201d and column two \u201cIndefinite Integral.\u201d The second row reads d\/dx (k) = 0, the integral of kdx = the integral of kx^0dx = kx + C. The third row reads d\/dx(x^n) = nx^(x-1), the integral of x^ndn = (x^n+1)\/(n+1) + C for n is not equal to negative 1. The fourth row reads d\/dx(ln(the absolute value of x))=1\/x, the integral of (1\/x)dx = ln(the absolute value of x) + C. The fifth row reads d\/dx(e^x) = e^x, the integral of e^xdx = e^x + C. The sixth row reads d\/dx(sinx) = cosx, the integral of cosxdx = sinx + C. The seventh row reads d\/dx(cosx) = negative sinx, the integral of sinxdx = negative cosx + C. The eighth row reads d\/dx(tanx) = sec squared x, the integral of sec squared xdx = tanx + C. The ninth row reads d\/dx(cscx) = negative cscxcotx, the integral of cscxcotxdx = negative cscx + C. The tenth row reads d\/dx(secx) = secxtanx, the integral of secxtanxdx = secx + C. The eleventh row reads d\/dx(cotx) = negative csc squared x, the integral of csc squared xdx = negative cot x + C. The twelfth row reads d\/dx(sin^-1(x)) = 1\/the square root of (1 \u2013 x^2), the integral of 1\/(the square root of (x^2 \u2013 1) = sin^-1(x) + C. The thirteenth row reads d\/dx (tan^-1(x)) = 1\/(1 + x^2), the integral of 1\/(1 + x^2)dx = tan^-1(x) + C. The fourteenth row reads d\/dx(sec^-1(the absolute value of x)) = 1\/x(the square root of x^2 \u2013 1), the integral of 1\/x(the square root of x^2 \u2013 1)dx = sec^-1(the absolute value of x) + C.\">\n<caption>Integration Formulas<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Differentiation Formula<\/th>\n<th>Indefinite Integral<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(k)=0[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int kdx=\\displaystyle\\int kx^0 dx=kx+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(x^n)=nx^{n-1}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int x^n dx=\\frac{x^{n+1}}{n+1}+C[\/latex] for [latex]n\\ne \u22121[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\ln |x|)=\\frac{1}{x}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x}dx=\\ln |x|+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(e^x)=e^x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int e^x dx=e^x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\cos x dx= \\sin x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cos x)=\u2212 \\sin x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan x)= \\sec^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec^2 x dx= \\tan x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\csc x)=\u2212\\csc x \\cot x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc x \\cot x dx=\u2212\\csc x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec x)= \\sec x \\tan x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec x \\tan x dx= \\sec x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cot x)=\u2212\\csc^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc^2 x dx=\u2212\\cot x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}( \\sin^{-1} x)=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-x^2}} dx= \\sin^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan^{-1} x)=\\frac{1}{1+x^2}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{1+x^2} dx= \\tan^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec^{-1} |x|)=\\frac{1}{x\\sqrt{x^2-1}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x\\sqrt{x^2-1}} dx= \\sec^{-1} |x|+C[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165043425485\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Properties of Indefinite Integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1165043395041\">Let [latex]F[\/latex] and [latex]G[\/latex] be antiderivatives of [latex]f[\/latex] and [latex]g[\/latex], respectively, and let [latex]k[\/latex] be any real number.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043393659\"><strong>Sums and Differences<\/strong><\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (f(x) \\pm g(x)) dx=F(x) \\pm G(x)+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042328714\"><strong>Constant Multiples<\/strong><\/p>\n<div id=\"fs-id1165042328717\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int kf(x) dx=kF(x)+C[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165043248811\" class=\"textbook exercises\">\n<h3>Example: Evaluating Indefinite Integrals<\/h3>\n<p>Evaluate each of the following indefinite integrals:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{x^2+4\\sqrt[3]{x}}{x} dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{4}{1+x^2} dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\tan x \\cos x dx[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1165042705917\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042552215\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042552215\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165042552215\" style=\"list-style-type: lower-alpha;\">\n<li>Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain\n<div id=\"fs-id1165042552227\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx=\\displaystyle\\int 5x^3 dx-\\displaystyle\\int 7x^2 dx+\\displaystyle\\int 3x dx+\\displaystyle\\int 4 dx[\/latex]<\/div>\n<p>From the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives<\/p>\n<div id=\"fs-id1165043312575\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 5x^3 dx-\\displaystyle\\int 7x^2 dx+\\displaystyle\\int 3x dx+\\displaystyle\\int 4 dx=5\\displaystyle\\int x^3 dx-7\\displaystyle\\int x^2 dx+3\\displaystyle\\int x dx+4\\displaystyle\\int 1 dx[\/latex]<\/div>\n<p>Using the power rule for integrals, we conclude that<\/p>\n<div id=\"fs-id1165042407363\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (5x^3-7x^2+3x+4) dx=\\frac{5}{4}x^4-\\frac{7}{3}x^3+\\frac{3}{2}x^2+4x+C[\/latex]<\/div>\n<\/li>\n<li>Rewrite the integrand as\n<div id=\"fs-id1165042371846\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{x^2+4\\sqrt[3]{x}}{x}=\\frac{x^2}{x}+\\frac{4\\sqrt[3]{x}}{x}[\/latex]<\/div>\n<p>Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have<\/p>\n<div id=\"fs-id1165043427498\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int (x+\\frac{4}{x^{2\/3}}) dx & =\\displaystyle\\int x dx+4\\displaystyle\\int x^{-2\/3} dx \\\\ & =\\frac{1}{2}x^2+4\\frac{1}{(\\frac{-2}{3})+1}x^{(-2\/3)+1}+C \\\\ & =\\frac{1}{2}x^2+12x^{1\/3}+C \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Using the properties of indefinite integrals, write the integral as\n<div id=\"fs-id1165043348665\" class=\"equation unnumbered\">[latex]4\\displaystyle\\int \\frac{1}{1+x^2} dx[\/latex].<\/div>\n<p>Then, use the fact that [latex]\\tan^{-1} (x)[\/latex] is an antiderivative of [latex]\\frac{1}{1+x^2}[\/latex] to conclude that<\/p>\n<div id=\"fs-id1165042374764\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{4}{1+x^2} dx=4 \\tan^{-1} (x)+C[\/latex]<\/div>\n<\/li>\n<li>Rewrite the integrand as\n<div class=\"equation unnumbered\">[latex]\\tan x \\cos x=\\frac{ \\sin x}{ \\cos x} \\cos x= \\sin x[\/latex].<\/div>\n<p>Therefore,<\/p>\n<div id=\"fs-id1165043317182\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\tan x \\cos x dx=\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\displaystyle\\int (4x^3-5x^2+x-7) dx[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4078823\">Hint<\/span><\/p>\n<div id=\"q4078823\" class=\"hidden-answer\" style=\"display: none\">\n<p>Integrate each term in the integrand separately, making use of the power rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043259694\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043259694\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x^4-\\frac{5}{3}x^3+\\frac{1}{2}x^2-7x+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm210143\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=210143&theme=oea&iframe_resize_id=ohm210143&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluate Trigonometric Functions<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Polar Coordinates)<\/em><\/strong><br \/>\nThe unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the [latex]\\left(x,y\\right)[\/latex] coordinates of all of the major angles in the first quadrant of the unit circle.<\/p>\n<p>Remember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function&#8217;s value at a given angle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Evaluating Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\n<p>If [latex]\\theta[\/latex] is an angle measure, then, using the unit circle,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan \\theta=\\frac{sin \\theta}{cos \\theta}\\\\ \\sec \\theta=\\frac{1}{cos \\theta}\\\\ \\csc t=\\frac{1}{sin \\theta}\\\\ \\cot \\theta=\\frac{cos \\theta}{sin \\theta}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Unit Circle to Find the Value of Trigonometric Functions<\/h3>\n<p>Find [latex]\\sin \\theta,\\cos \\theta,\\tan \\theta,\\sec \\theta,\\csc \\theta[\/latex], and [latex]\\cot \\theta[\/latex] when [latex]\\theta=\\frac{\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151548\">Show Solution<\/span><\/p>\n<div id=\"q151548\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &\\tan \\frac{\\pi }{3}=\\sqrt{3} \\text{ (From}\\frac{sin \\theta}{cos \\theta})\\\\ &\\sec \\frac{\\pi }{3}=2 \\text{ (From}\\frac{1}{cos \\theta})\\\\ &\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\text{ (From}\\frac{1}{sin \\theta} \\text{ - do not forget to rationalize})\\\\ &\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}(\\text{ (From}\\frac{cos \\theta}{sin \\theta} \\text{ - do not forget to rationalize})\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173354\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173354&theme=oea&iframe_resize_id=ohm173354\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Below are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.<\/p>\n<table id=\"Table_05_03_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Tangent<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Secant<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>2<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosecant<\/strong><\/td>\n<td>Undefined<\/td>\n<td>2<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Cotangent<\/strong><\/td>\n<td>Undefined<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<p><strong><em>(also in Module 3, Skills Review for Arc Length and Curvature)<\/em><\/strong><\/p>\n<p id=\"fs-id1170572622222\">The <strong>Fundamental Theorem of Calculus, Part 2<\/strong> (also known as the <strong>evaluation theorem<\/strong>) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\n<hr \/>\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\n<div id=\"fs-id1170571678913\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572173704\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572173704\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill & =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ & =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ & =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ & =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ & =\\frac{16}{3}-16\\hfill \\\\ & =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572130389\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572130389\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill & ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ & =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ & =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ & =18-6-\\frac{2}{3}+2\\hfill \\\\ & =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3887125\">Hint<\/span><\/p>\n<div id=\"q3887125\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571639103\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571639103\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{7}{24}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm211336\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=211336&theme=oea&iframe_resize_id=ohm211336&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Manipulate Equations<\/h2>\n<p>It will sometimes be necessary to change the bounds of an integral which will even require manipulating the given equation.<\/p>\n<h3>Manipulate Rational Equations<\/h3>\n<p>Equations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).<\/p>\n<p>One of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.<\/p>\n<p>Solve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.<\/p>\n<p>Multiply both sides of the equation by\u00a0[latex]4[\/latex], the common denominator of the fractional coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving a Rational Equation For a Specific Variable<\/h3>\n<p>Solve the equation [latex]\\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425621\">Show Solution<\/span><\/p>\n<div id=\"q425621\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the least common denominator of\u00a0[latex]8[\/latex] and\u00a0[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.<\/p>\n<p>Multiply both sides of the equation by the common denominator,\u00a0[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving a Rational Equation For a Specific Variable<\/h3>\n<p>Solve the equation [latex]x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331190\">Show Solution<\/span><\/p>\n<div id=\"q331190\" class=\"hidden-answer\" style=\"display: none\">\n<p>Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Manipulate Radical Equations<\/h3>\n<p><strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3y+18}=x & \\\\ \\sqrt{x+3}=y-3 & \\\\ \\sqrt{x+5}-\\sqrt{y - 3}=2\\end{array}[\/latex]<\/div>\n<p>Radical equations are manipulated by eliminating each radical, one at a time until you have solved for the indicated variable.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Radical Equations<\/h3>\n<p>An equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a radical equation, solve it<\/h3>\n<ol>\n<li>Isolate the radical expression containing your variable of interest on one side of the equal sign. Put all remaining terms on the other side.<\/li>\n<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\n<li>Solve the resulting equation for the variable of interest.<\/li>\n<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation with One Radical<\/h3>\n<p>Solve [latex]\\sqrt{15 - 2y}=x[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q503795\">Show Solution<\/span><\/p>\n<div id=\"q503795\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radical is already isolated on the left side of the equal sign, so proceed to square both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2y}=x & \\\\ {\\left(\\sqrt{15 - 2y}\\right)}^{2}={\\left(x\\right)}^{2} & \\\\ 15 - 2y={x}^{2}\\end{array}[\/latex]<\/div>\n<p>Now isolate [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll} 15 - 2y={x}^{2} \\\\ -2y={x}^{2}-15 \\\\ y=-\\frac{1}{2}{x}^{2}+\\frac{15}{2}\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the radical equation [latex]\\sqrt{y+3}=3x - 1[\/latex]\u00a0for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q719648\">Show Solution<\/span><\/p>\n<div id=\"q719648\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=9{x}^{2}-6x-2[\/latex] Note: When you square the binomial on the right side of the equation you get [latex]9{x}^{2}-6x+1[\/latex]. Then, take away [latex]3[\/latex] from both sides.<\/p><\/div>\n<\/div>\n<\/div>\n<p>Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex] and [latex]{8}^{\\frac{2}{3}}[\/latex] is another way of writing [latex]\\left(\\sqrt[3]{8}\\right)^2[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Rational Exponents<\/h3>\n<p>A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\n<\/div>\n<h2>Take Limits at Infinity<\/h2>\n<div id=\"fs-id1165043145047\" class=\"bc-section section\">\n<p id=\"fs-id1165043107285\">Recall that [latex]\\underset{x \\to a}{\\lim}f(x)=L[\/latex] means [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently close to [latex]a[\/latex]. We can extend this idea to limits at infinity. For example, consider the function [latex]f(x)=2+\\frac{1}{x}[\/latex]. As can be seen graphically below, as the values of [latex]x[\/latex] get larger, the values of [latex]f(x)[\/latex] approach 2. We say the limit as [latex]x[\/latex] approaches [latex]\\infty[\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=2[\/latex]. Similarly, for [latex]x<0[\/latex], as the values [latex]|x|[\/latex] get larger, the values of [latex]f(x)[\/latex] approaches 2. We say the limit as [latex]x[\/latex] approaches [latex]\u2212\\infty[\/latex] of [latex]f(x)[\/latex] is 2 and write [latex]\\underset{x\\to a}{\\lim}f(x)=2[\/latex].<\/p>\n<div style=\"width: 727px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211025\/CNX_Calc_Figure_04_06_019.jpg\" alt=\"The function f(x) 2 + 1\/x is graphed. The function starts negative near y = 2 but then decreases to \u2212\u221e near x = 0. The function then decreases from \u221e near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.\" width=\"717\" height=\"423\" \/><\/p>\n<p class=\"wp-caption-text\">The function approaches the asymptote [latex]y=2[\/latex] as [latex]x[\/latex] approaches [latex]\\pm \\infty[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165042936244\">More generally, for any function [latex]f[\/latex], we say the limit as [latex]x \\to \\infty[\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \\infty}{\\lim}f(x)=L[\/latex]. Similarly, we say the limit as [latex]x\\to \u2212\\infty[\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex] if [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] as long as [latex]x<0[\/latex] and [latex]|x|[\/latex] is sufficiently large. In that case, we write [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=L[\/latex]. We now look at the definition of a function having a limit at infinity.<\/p>\n<div id=\"fs-id1165042331960\" class=\"textbox shaded\">\n<div class=\"title\">\n<p style=\"text-align: center;\"><span style=\"font-size: 1.2em; font-weight: 600; text-align: center; text-transform: uppercase; background-color: #eeeeee;\">Definition<\/span><\/p>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165042970725\">(Informal) If the values of [latex]f(x)[\/latex] become arbitrarily close to [latex]L[\/latex] as [latex]x[\/latex] becomes sufficiently large, we say the function [latex]f[\/latex] has a limit at infinity and write<\/p>\n<div id=\"fs-id1165042986551\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042374662\">If the values of [latex]f(x)[\/latex] becomes arbitrarily close to [latex]L[\/latex] for [latex]x<0[\/latex] as [latex]|x|[\/latex] becomes sufficiently large, we say that the function [latex]f[\/latex] has a limit at negative infinity and write<\/p>\n<div id=\"fs-id1165043105208\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to -\\infty }{\\lim}f(x)=L[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1165043157752\">If the values [latex]f(x)[\/latex] are getting arbitrarily close to some finite value [latex]L[\/latex] as [latex]x\\to \\infty[\/latex] or [latex]x\\to \u2212\\infty[\/latex], the graph of [latex]f[\/latex] approaches the line [latex]y=L[\/latex]. In that case, the line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] (Figure 2). For example, for the function [latex]f(x)=\\frac{1}{x}[\/latex], since [latex]\\underset{x\\to \\infty }{\\lim}f(x)=0[\/latex], the line [latex]y=0[\/latex] is a horizontal asymptote of [latex]f(x)=\\frac{1}{x}[\/latex].<\/p>\n<div id=\"fs-id1165043262534\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165042973921\">If [latex]\\underset{x\\to \\infty }{\\lim}f(x)=L[\/latex] or [latex]\\underset{x \\to \u2212\\infty}{\\lim}f(x)=L[\/latex], we say the line [latex]y=L[\/latex] is a <strong>horizontal asymptote<\/strong> of [latex]f[\/latex].<\/p>\n<\/div>\n<div style=\"width: 776px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211028\/CNX_Calc_Figure_04_06_020.jpg\" alt=\"The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.\" width=\"766\" height=\"273\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. (a) As [latex]x\\to \\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]L[\/latex]. The line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex]. (b) As [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f[\/latex] are getting arbitrarily close to [latex]M[\/latex]. The line [latex]y=M[\/latex] is a horizontal asymptote of [latex]f[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165042647732\">A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as [latex]x[\/latex] approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function [latex]f(x)=\\frac{ \\cos x}{x}+1[\/latex] shown in Figure 3 intersects the horizontal asymptote [latex]y=1[\/latex] an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.<\/p>\n<div style=\"width: 539px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211031\/CNX_Calc_Figure_04_06_002.jpg\" alt=\"The function f(x) = (cos x)\/x + 1 is shown. It decreases from (0, \u221e) and then proceeds to oscillate around y = 1 with decreasing amplitude.\" width=\"529\" height=\"230\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The graph of [latex]f(x)=\\cos x\/x+1[\/latex] crosses its horizontal asymptote [latex]y=1[\/latex] an infinite number of times.<\/p>\n<\/div>\n<div class=\"textbook exercises\">\n<h3>Example: Computing Limits at Infinity<\/h3>\n<p id=\"fs-id1165043262623\">For each of the following functions [latex]f[\/latex], evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex].<\/p>\n<ol id=\"fs-id1165042356111\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x)=5-\\frac{2}{x^2}[\/latex]<\/li>\n<li>[latex]f(x)=\\dfrac{\\sin x}{x}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043183885\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043183885\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165043183885\" style=\"list-style-type: lower-alpha;\">\n<li>Using the algebraic limit laws, we have [latex]\\underset{x\\to \\infty }{\\lim}(5-\\frac{2}{x^2})=\\underset{x\\to \\infty }{\\lim}5-2(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})(\\underset{x\\to \\infty }{\\lim}\\frac{1}{x})=5-2 \\cdot 0=5[\/latex]. Similarly, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=5[\/latex].\n<div class=\"mceTemp\"><\/div>\n<\/li>\n<li>nce [latex]-1\\le \\sin x\\le 1[\/latex] for all [latex]x[\/latex], we have\n<div id=\"fs-id1165043093355\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{-1}{x}\\le \\frac{\\sin x}{x}\\le \\frac{1}{x}[\/latex]<\/div>\n<p>for all [latex]x \\ne 0[\/latex]. Also, since<\/p>\n<div id=\"fs-id1165043197153\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{-1}{x}=0=\\underset{x\\to \\infty }{\\lim}\\frac{1}{x}[\/latex],<\/div>\n<p>we can apply the squeeze theorem to conclude that<\/p>\n<div id=\"fs-id1165043036581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\n<p>Similarly,<\/p>\n<div id=\"fs-id1165043122536\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u2212\\infty}{\\lim}\\frac{\\sin x}{x}=0[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042320881\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043315935\">Evaluate [latex]\\underset{x\\to \u2212\\infty}{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}\\left(3+\\frac{4}{x}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2473508\">Hint<\/span><\/p>\n<div id=\"q2473508\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042318511\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}1\/x=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043390798\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043390798\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043390798\">Both limits are 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042333169\" class=\"bc-section section\">\n<h2>Infinite Limits at Infinity<\/h2>\n<p id=\"fs-id1165042333174\">Sometimes the values of a function [latex]f[\/latex] become arbitrarily large as [latex]x\\to \\infty[\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex]. In this case, we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\\infty )[\/latex]. On the other hand, if the values of [latex]f[\/latex] are negative but become arbitrarily large in magnitude as [latex]x\\to \\infty[\/latex] (or as [latex]x\\to \u2212\\infty )[\/latex], we write [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex] (or [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=\u2212\\infty )[\/latex].<\/p>\n<p id=\"fs-id1165042606820\">For example, consider the function [latex]f(x)=x^3[\/latex]. The [latex]\\underset{x\\to \\infty }{\\lim}x^3=\\infty[\/latex]. On the other hand, as [latex]x\\to \u2212\\infty[\/latex], the values of [latex]f(x)=x^3[\/latex] are negative but become arbitrarily large in magnitude. Consequently, [latex]\\underset{x\\to \u2212\\infty }{\\lim}x^3=\u2212\\infty[\/latex].<\/p>\n<table id=\"fs-id1165042406634\" class=\"column-header\" summary=\"The table has four rows and six columns. The first column is a header column and it reads x, x3, x, and x3. After the header, the first row reads 10, 20, 50, 100, and 1000. The second row reads 1000, 8000, 125000, 1,000,000, and 1,000,000,000. The third row reads \u221210, \u221220, \u221250, \u2212100, and \u22121000. The forth row reads \u22121000, \u22128000, \u2212125,000, \u22121,000,000, and \u22121,000,000,000.\">\n<caption>Values of a power function as [latex]x\\to \\pm \\infty[\/latex]<\/caption>\n<tbody>\n<tr valign=\"top\">\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>10<\/td>\n<td>20<\/td>\n<td>50<\/td>\n<td>100<\/td>\n<td>1000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\n<td>1000<\/td>\n<td>8000<\/td>\n<td>125,000<\/td>\n<td>1,000,000<\/td>\n<td>1,000,000,000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>-10<\/td>\n<td>-20<\/td>\n<td>-50<\/td>\n<td>-100<\/td>\n<td>-1000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td><strong>[latex]x^3[\/latex]<\/strong><\/td>\n<td>-1000<\/td>\n<td>-8000<\/td>\n<td>-125,000<\/td>\n<td>-1,000,000<\/td>\n<td>-1,000,000,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>\n<div style=\"width: 652px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211045\/CNX_Calc_Figure_04_06_022.jpg\" alt=\"The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.\" width=\"642\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">For this function, the functional values approach infinity as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043276353\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1165043276356\">(Informal) We say a function [latex]f[\/latex] has an infinite limit at infinity and write<\/p>\n<div id=\"fs-id1165043276364\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042709557\">if [latex]f(x)[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. We say a function has a negative infinite limit at infinity and write<\/p>\n<div id=\"fs-id1165042647077\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty }{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042327355\">if [latex]f(x)<0[\/latex] and [latex]|f(x)|[\/latex] becomes arbitrarily large for [latex]x[\/latex] sufficiently large. Similarly, we can define infinite limits as [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165042323710\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042320226\">Find [latex]\\underset{x\\to \\infty }{\\lim}3x^2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042708272\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042708272\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042383154\">[latex]\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16109\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16109&theme=oea&iframe_resize_id=ohm16109&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4098\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4098","chapter","type-chapter","status-publish","hentry"],"part":4135,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4098","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4098\/revisions"}],"predecessor-version":[{"id":6487,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4098\/revisions\/6487"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4135"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4098\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4098"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4098"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4098"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4098"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}