{"id":4104,"date":"2022-04-14T18:15:54","date_gmt":"2022-04-14T18:15:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-surface-integrals\/"},"modified":"2022-11-09T16:46:21","modified_gmt":"2022-11-09T16:46:21","slug":"skills-review-for-stokes-theorem-and-the-divergence-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-stokes-theorem-and-the-divergence-theorem\/","title":{"raw":"Skills Review for Stokes' Theorem and the Divergence Theorem","rendered":"Skills Review for Stokes&#8217; Theorem and the Divergence Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply basic derivative rules<\/li>\r\n \t<li>Use the product rule for finding the derivative of a product of functions<\/li>\r\n \t<li>Use the quotient rule for finding the derivative of a quotient of functions<\/li>\r\n \t<li>Apply the chain rule together with the power and product rule<\/li>\r\n \t<li>Evaluate indefinite integrals<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the section about Stokes' Theorem, we will learn about Stokes' Theorem, a higher-dimensional generalization of Green's Theorem. Then, we will explore the Divergence Theorem. Here we will review various differentiation techniques and rules along with basic integration techniques.\r\n<h2>Basic Derivative Rules<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-line-integrals\/\" target=\"_blank\" rel=\"noopener\">Module 6, Skills Review for Line Integrals and Conservative Vector Fields<\/a>)<\/em><\/strong>\r\n<h2>The Product Rule<\/h2>\r\n<strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong>\r\nAlthough it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the <strong>product rule<\/strong> does not follow this pattern. To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[\/latex], whose derivative is [latex]f^{\\prime}(x)=2x[\/latex] and not [latex]\\frac{d}{dx}(x)\\cdot \\frac{d}{dx}(x)=1\\cdot 1=1[\/latex].\r\n<div id=\"fs-id1169739298175\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Product Rule<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739298182\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\r\n\r\n<div id=\"fs-id1169739326645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(f(x)g(x))=\\frac{d}{dx}(f(x))\\cdot g(x)+\\frac{d}{dx}(g(x))\\cdot f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1169739187834\">That is,<\/p>\r\n\r\n<div id=\"fs-id1169739187837\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=f(x)g(x)[\/latex] then [latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1169739269717\">This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739273812\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Product Rule to Binomials<\/h3>\r\n<p id=\"fs-id1169739273822\">For [latex]j(x)=(x^2+2)(3x^3-5x)[\/latex], find [latex]j^{\\prime}(x)[\/latex] by applying the product rule.<\/p>\r\n[reveal-answer q=\"fs-id1169739301174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739301174\"]\r\n<p id=\"fs-id1169739301174\">If we set [latex]f(x)=x^2+2[\/latex] and [latex]g(x)=3x^3-5x[\/latex], then [latex]f^{\\prime}(x)=2x[\/latex] and [latex]g^{\\prime}(x)=9x^2-5[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169739343689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736612557\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736612560\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169736654821\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736654828\">Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[\/latex].<\/p>\r\n[reveal-answer q=\"034256\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"034256\"]\r\n<p id=\"fs-id1169736609959\">Set [latex]f(x)=2x^5[\/latex] and [latex]g(x)=4x^2+x[\/latex] and use the preceding example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169736654876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736654876\"]\r\n<p id=\"fs-id1169736654876\">[latex]j^{\\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Quotient Rule<\/h2>\r\n<strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong>\r\n<p id=\"fs-id1169739269461\">Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that<\/p>\r\n\r\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex], which is not the same as [latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169736662915\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Quotient Rule<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736662921\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\r\n\r\n<div id=\"fs-id1169739336009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\dfrac{f(x)}{g(x)}\\right)=\\dfrac{\\frac{d}{dx}(f(x))\\cdot g(x)-\\dfrac{d}{dx}(g(x))\\cdot f(x)}{(g(x))^2}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1169739190663\">That is,<\/p>\r\n\r\n<div id=\"fs-id1169739190666\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=\\dfrac{f(x)}{g(x)}[\/latex], then [latex]j^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739305225\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Quotient Rule<\/h3>\r\n<p id=\"fs-id1169739305235\">Use the quotient rule to find the derivative of [latex]k(x)=\\dfrac{5x^2}{4x+3}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739305276\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739305276\"]\r\n<p id=\"fs-id1169739305276\">Let [latex]f(x)=5x^2[\/latex] and [latex]g(x)=4x+3[\/latex]. Thus, [latex]f^{\\prime}(x)=10x[\/latex] and [latex]g^{\\prime}(x)=4[\/latex]. Substituting into the quotient rule, we have<\/p>\r\n\r\n<div id=\"fs-id1169739299200\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}=\\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739299784\">Simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739299787\" class=\"equation unnumbered\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739299850\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739299858\">Find the derivative of [latex]h(x)=\\dfrac{3x+1}{4x-3}[\/latex]<\/p>\r\n[reveal-answer q=\"336671\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"336671\"]\r\n<p id=\"fs-id1169739348450\">Apply the quotient rule with [latex]f(x)=3x+1[\/latex] and [latex]g(x)=4x-3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739348394\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739348394\"]\r\n<p id=\"fs-id1169739348394\">[latex]k^{\\prime}(x)=-\\dfrac{13}{(4x-3)^2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Chain Rule<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-line-integrals\/\" target=\"_blank\" rel=\"noopener\">Module 6, Skills Review for Line Integrals and Conservative Vector Fields<\/a>)<\/em><\/strong>\r\n<h2>Indefinite Integrals<\/h2>\r\n<strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong>\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-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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply basic derivative rules<\/li>\n<li>Use the product rule for finding the derivative of a product of functions<\/li>\n<li>Use the quotient rule for finding the derivative of a quotient of functions<\/li>\n<li>Apply the chain rule together with the power and product rule<\/li>\n<li>Evaluate indefinite integrals<\/li>\n<\/ul>\n<\/div>\n<p>In the section about Stokes&#8217; Theorem, we will learn about Stokes&#8217; Theorem, a higher-dimensional generalization of Green&#8217;s Theorem. Then, we will explore the Divergence Theorem. Here we will review various differentiation techniques and rules along with basic integration techniques.<\/p>\n<h2>Basic Derivative Rules<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-line-integrals\/\" target=\"_blank\" rel=\"noopener\">Module 6, Skills Review for Line Integrals and Conservative Vector Fields<\/a>)<\/em><\/strong><\/p>\n<h2>The Product Rule<\/h2>\n<p><strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong><br \/>\nAlthough it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the <strong>product rule<\/strong> does not follow this pattern. To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[\/latex], whose derivative is [latex]f^{\\prime}(x)=2x[\/latex] and not [latex]\\frac{d}{dx}(x)\\cdot \\frac{d}{dx}(x)=1\\cdot 1=1[\/latex].<\/p>\n<div id=\"fs-id1169739298175\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Product Rule<\/h3>\n<hr \/>\n<p id=\"fs-id1169739298182\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\n<div id=\"fs-id1169739326645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(f(x)g(x))=\\frac{d}{dx}(f(x))\\cdot g(x)+\\frac{d}{dx}(g(x))\\cdot f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1169739187834\">That is,<\/p>\n<div id=\"fs-id1169739187837\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=f(x)g(x)[\/latex] then [latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1169739269717\">This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.<\/p>\n<\/div>\n<div id=\"fs-id1169739273812\" class=\"textbook exercises\">\n<h3>Example: Applying the Product Rule to Binomials<\/h3>\n<p id=\"fs-id1169739273822\">For [latex]j(x)=(x^2+2)(3x^3-5x)[\/latex], find [latex]j^{\\prime}(x)[\/latex] by applying the product rule.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739301174\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739301174\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739301174\">If we set [latex]f(x)=x^2+2[\/latex] and [latex]g(x)=3x^3-5x[\/latex], then [latex]f^{\\prime}(x)=2x[\/latex] and [latex]g^{\\prime}(x)=9x^2-5[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739343689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736612557\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736612560\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736654821\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736654828\">Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q034256\">Hint<\/span><\/p>\n<div id=\"q034256\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736609959\">Set [latex]f(x)=2x^5[\/latex] and [latex]g(x)=4x^2+x[\/latex] and use the preceding example as a guide.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736654876\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736654876\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736654876\">[latex]j^{\\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Quotient Rule<\/h2>\n<p><strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong><\/p>\n<p id=\"fs-id1169739269461\">Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that<\/p>\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex], which is not the same as [latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169736662915\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Quotient Rule<\/h3>\n<hr \/>\n<p id=\"fs-id1169736662921\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\n<div id=\"fs-id1169739336009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\dfrac{f(x)}{g(x)}\\right)=\\dfrac{\\frac{d}{dx}(f(x))\\cdot g(x)-\\dfrac{d}{dx}(g(x))\\cdot f(x)}{(g(x))^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1169739190663\">That is,<\/p>\n<div id=\"fs-id1169739190666\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=\\dfrac{f(x)}{g(x)}[\/latex], then [latex]j^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169739305225\" class=\"textbook exercises\">\n<h3>Example: Applying the Quotient Rule<\/h3>\n<p id=\"fs-id1169739305235\">Use the quotient rule to find the derivative of [latex]k(x)=\\dfrac{5x^2}{4x+3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739305276\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739305276\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739305276\">Let [latex]f(x)=5x^2[\/latex] and [latex]g(x)=4x+3[\/latex]. Thus, [latex]f^{\\prime}(x)=10x[\/latex] and [latex]g^{\\prime}(x)=4[\/latex]. Substituting into the quotient rule, we have<\/p>\n<div id=\"fs-id1169739299200\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}=\\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739299784\">Simplifying, we obtain<\/p>\n<div id=\"fs-id1169739299787\" class=\"equation unnumbered\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739299850\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739299858\">Find the derivative of [latex]h(x)=\\dfrac{3x+1}{4x-3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336671\">Hint<\/span><\/p>\n<div id=\"q336671\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739348450\">Apply the quotient rule with [latex]f(x)=3x+1[\/latex] and [latex]g(x)=4x-3[\/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-id1169739348394\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739348394\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739348394\">[latex]k^{\\prime}(x)=-\\dfrac{13}{(4x-3)^2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Chain Rule<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-line-integrals\/\" target=\"_blank\" rel=\"noopener\">Module 6, Skills Review for Line Integrals and Conservative Vector Fields<\/a>)<\/em><\/strong><\/p>\n<h2>Indefinite Integrals<\/h2>\n<p><strong><em>(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)<\/em><\/strong><\/p>\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-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\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-4104\">\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 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