{"id":1139,"date":"2021-06-30T17:01:58","date_gmt":"2021-06-30T17:01:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-of-exponential-functions\/"},"modified":"2022-03-19T03:22:13","modified_gmt":"2022-03-19T03:22:13","slug":"integrals-of-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/integrals-of-exponential-functions\/","title":{"raw":"Integrals of Exponential Functions","rendered":"Integrals of Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Integrate functions involving exponential functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170571602573\">The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, [latex]y={e}^{x},[\/latex] is its own derivative and its own integral.<\/p>\r\n\r\n<div id=\"fs-id1170572370390\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Integrals of Exponential Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572375243\">Exponential functions can be integrated using the following formulas.<\/p>\r\n\r\n<div id=\"fs-id1170572374424\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc} {\\displaystyle\\int{e}^{x}dx} &amp; {=} &amp; {{e}^{x}+C} \\\\ {\\displaystyle\\int{a}^{x}dx} &amp; {=} &amp; {\\dfrac{{a}^{x}}{\\text{ln}a}+C}\\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\nThe nature of the antiderivative of [latex]{e}^{x}[\/latex] makes it fairly easy to identify what to choose as [latex]u[\/latex]. If only one\u00a0[latex]e[\/latex] exists, choose the exponent of [latex]e[\/latex] as\u00a0[latex]u[\/latex]. If more than one [latex]e[\/latex] exists, choose the more complicated function involving [latex]e[\/latex] as\u00a0[latex]u[\/latex].\r\n<div id=\"fs-id11705722115180\" class=\"textbook exercises\">\r\n<h3>Example: Finding an Antiderivative of an Exponential Function<\/h3>\r\nFind the antiderivative of the exponential function [latex]e^{-x}[\/latex].\r\n\r\n[reveal-answer q=\"26617708\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"26617708\"]\r\n\r\nUse substitution, setting [latex]u=\\text{\u2212}x,[\/latex] and then [latex]du=-1dx.[\/latex] Multiply the <em>du<\/em> equation by \u22121, so you now have [latex]\\text{\u2212}du=dx.[\/latex] Then,\r\n\r\n[latex]\\begin{array}{cc}{\\displaystyle\\int {e}^{\\text{\u2212}x}dx}\\hfill &amp; =\\text{\u2212}{\\displaystyle\\int {e}^{u}du}\\hfill \\\\ \\\\ &amp; =\\text{\u2212}{e}^{u}+C\\hfill \\\\ &amp; =\\text{\u2212}{e}^{\\text{\u2212}x}+C.\\hfill \\end{array}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572248011\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of the function using substitution: [latex]{x}^{2}{e}^{-2{x}^{3}}.[\/latex]\r\n\r\n[reveal-answer q=\"20881654\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"20881654\"]\r\n<p id=\"fs-id1170571622632\">Let [latex]u[\/latex] equal the exponent on [latex]e[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572269592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572269592\"]\r\n<p id=\"fs-id1170572269592\">[latex]\\displaystyle\\int {x}^{2}{e}^{-2{x}^{3}}dx=-\\frac{1}{6}{e}^{-2{x}^{3}}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=88&amp;end=170&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_88to170_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.1\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1170572607911\">A common mistake when dealing with exponential expressions is treating the exponent on [latex]e[\/latex] the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on [latex]e[\/latex]. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we\u2019re using the right rules for the functions we\u2019re integrating.<\/p>\r\n\r\n<div id=\"fs-id1170572209244\" class=\"textbook exercises\">\r\n<h3>Example: Square Root of an Exponential Function<\/h3>\r\nFind the antiderivative of the exponential function [latex]{e}^{x}\\sqrt{1+{e}^{x}}.[\/latex]\r\n<div id=\"fs-id1170571602248\" class=\"exercise\">[reveal-answer q=\"fs-id1170572551693\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572551693\"]\r\n<p id=\"fs-id1170572551693\">First rewrite the problem using a rational exponent:<\/p>\r\n\r\n<div id=\"fs-id1170572549246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}\\sqrt{1+{e}^{x}}dx=\\displaystyle\\int {e}^{x}{(1+{e}^{x})}^{1\\text{\/}2}dx.[\/latex]<\/div>\r\nUsing substitution, choose [latex]u=1+{e}^{x}.u=1+{e}^{x}.[\/latex] Then, [latex]du={e}^{x}dx.[\/latex] We have (Figure 1)\r\n<div id=\"fs-id1170572373474\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}{(1+{e}^{x})}^{1\\text{\/}2}dx=\\displaystyle\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170572307548\">Then<\/p>\r\n\r\n<div id=\"fs-id1170572560605\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {u}^{1\\text{\/}2}du=\\frac{{u}^{3\\text{\/}2}}{3\\text{\/}2}+C=\\frac{2}{3}{u}^{3\\text{\/}2}+C=\\frac{2}{3}{(1+{e}^{x})}^{3\\text{\/}2}+C.[\/latex]<\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204252\/CNX_Calc_Figure_05_06_001.jpg\" alt=\"A graph of the function f(x) = e^x * sqrt(1 + e^x), which is an increasing concave up curve, over [-3, 1]. It begins close to the x-axis in quadrant two, crosses the y-axis at (0, sqrt(2)), and continues to increase rapidly.\" width=\"325\" height=\"208\" \/> Figure 1. The graph shows an exponential function times the square root of an exponential function.[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572207056\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the antiderivative of [latex]{e}^{x}{(3{e}^{x}-2)}^{2}.[\/latex]\r\n\r\n[reveal-answer q=\"366490\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"366490\"]\r\n\r\nLet [latex]u=3{e}^{x}-2u=3{e}^{x}-2.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"6711088\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"6711088\"]\r\n<div id=\"fs-id1170572451630\" class=\"exercise\">\r\n<div class=\"solution\">\r\n<p id=\"fs-id1170572209127\">[latex]\\displaystyle\\int {e}^{x}{(3{e}^{x}-2)}^{2}dx=\\frac{1}{9}{(3{e}^{x}-2)}^{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=257&amp;end=336&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_257to336_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.1\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbook exercises\">\r\n<div id=\"fs-id1170572240614\" class=\"exercise\">\r\n<h3>Example: Using Substitution with an Exponential Function<\/h3>\r\nUse substitution to evaluate the indefinite integral [latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572216534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572216534\"]\r\n<p id=\"fs-id1170572216534\">Here we choose to let [latex]u[\/latex] equal the expression in the exponent on [latex]e[\/latex]. Let [latex]u=2{x}^{3}[\/latex] and [latex]du=6{x}^{2}dx..[\/latex] Again, <em>du<\/em> is off by a constant multiplier; the original function contains a factor of 3[latex]x[\/latex]<sup>2<\/sup>, not 6[latex]x[\/latex]<sup>2<\/sup>. Multiply both sides of the equation by [latex]\\frac{1}{2}[\/latex] so that the integrand in [latex]u[\/latex] equals the integrand in [latex]x[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572470447\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx=\\frac{1}{2}\\displaystyle\\int {e}^{u}du.[\/latex]<\/div>\r\n<p id=\"fs-id1170572101855\">Integrate the expression in [latex]u[\/latex] and then substitute the original expression in [latex]x[\/latex] back into the [latex]u[\/latex] integral:<\/p>\r\n\r\n<div id=\"fs-id1170572448333\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}\\displaystyle\\int {e}^{u}du=\\frac{1}{2}{e}^{u}+C=\\frac{1}{2}{e}^{2{x}^{3}}+C.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572551647\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate the indefinite integral [latex]\\displaystyle\\int 2{x}^{3}{e}^{{x}^{4}}dx.[\/latex]\r\n\r\n[reveal-answer q=\"212787664\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"212787664\"]\r\n<p id=\"fs-id1170572232621\">Let [latex]u={x}^{4}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572242344\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572242344\"]\r\n<p id=\"fs-id1170572242344\">[latex]\\displaystyle\\int 2{x}^{3}{e}^{{x}^{4}}dx=\\frac{1}{2}{e}^{{x}^{4}}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170572294922\" class=\"exercise\">\r\n<div id=\"fs-id1170573414530\" class=\"commentary\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571733816\">As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number\u00a0<em data-effect=\"italics\">e<\/em>\u00a0is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let\u2019s look at an example in which integration of an exponential function solves a common business application.<\/p>\r\n<p id=\"fs-id1170571571226\">A\u00a0<span id=\"term238\" class=\"no-emphasis\" data-type=\"term\">price\u2013demand function<\/span>\u00a0tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal price\u2013demand function is the derivative of the price\u2013demand function and it tells us how fast the price changes at a given level of production. These functions are used in business to determine the price\u2013elasticity of demand, and to help companies determine whether changing production levels would be profitable.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Price\u2013Demand Equation<\/h3>\r\nFind the price\u2013demand equation for a particular brand of toothpaste at a supermarket chain when the demand is 50 tubes per week at $2.35 per tube, given that the marginal price\u2014demand function, [latex]{p}^{\\prime }(x),[\/latex] for [latex]x[\/latex] number of tubes per week, is given as\r\n<div id=\"fs-id1170572141865\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p\\text{'}(x)=-0.015{e}^{-0.01x}.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571595414\">If the supermarket chain sells 100 tubes per week, what price should it set?<\/p>\r\n[reveal-answer q=\"fs-id1170571543163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571543163\"]\r\n<p id=\"fs-id1170571543163\" style=\"text-align: center;\">To find the price\u2013demand equation, integrate the marginal price\u2013demand function. First find the antiderivative, then look at the particulars. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572216339\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ p(x)\\hfill &amp; =\\int -0.015{e}^{-0.01x}dx\\hfill \\\\ &amp; =-0.015\\int {e}^{-0.01x}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572507579\">Using substitution, let [latex]u=-0.01x[\/latex] and [latex]du=-0.01dx.[\/latex] Then, divide both sides of the <em>du<\/em> equation by \u22120.01. This gives<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{-0.015}{-0.01}\\int {e}^{u}du\\hfill &amp; =1.5\\int {e}^{u}du\\hfill \\\\ \\\\ &amp; =1.5{e}^{u}+C\\hfill \\\\ &amp; =1.5{e}^{-0.01x}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572111467\">The next step is to solve for <em>C<\/em>. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ p(50)\\hfill &amp; =1.5{e}^{-0.01(50)}+C\\hfill \\\\ &amp; =2.35.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571569187\">Now, just solve for <em>C<\/em>:<\/p>\r\n\r\n<div id=\"fs-id1170572415392\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ C\\hfill &amp; =2.35-1.5{e}^{-0.5}\\hfill \\\\ &amp; =2.35-0.91\\hfill \\\\ &amp; =1.44.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572140846\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572375703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p(x)=1.5{e}^{-0.01x}+1.44.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571547617\">If the supermarket sells 100 tubes of toothpaste per week, the price would be<\/p>\r\n\r\n<div id=\"fs-id1170571653359\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p(100)=1.5{e}^{-0.01(100)}+1.44=1.5{e}^{-1}+1.44\\approx 1.99.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572393399\">The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding a Price\u2013Demand Equation.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=465&amp;end=815&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_465to815_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.1\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571637964\" class=\"textbook exercises\">\r\n<div id=\"fs-id1170572204890\" class=\"exercise\">\r\n<h3>example: Evaluating a Definite Integral Involving an Exponential Function<\/h3>\r\nEvaluate the definite integral [latex]{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx.[\/latex]\r\n[reveal-answer q=\"fs-id1170572247799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572247799\"]\r\n<p id=\"fs-id1170572247799\">Again, substitution is the method to use. Let [latex]u=1-x,[\/latex] so [latex]du=-1dx[\/latex] or [latex]\\text{\u2212}du=dx.[\/latex] Then [latex]\\displaystyle\\int {e}^{1-x}dx=\\text{\u2212}\\displaystyle\\int {e}^{u}du.[\/latex] Next, change the limits of integration. Using the equation [latex]u=1-x,[\/latex] we have<\/p>\r\n\r\n<div id=\"fs-id1170571652053\" class=\"equation unnumbered\">[latex]\\begin{array}{c}u=1-(1)=0\\hfill \\\\ u=1-(2)=-1.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571645662\">The integral then becomes<\/p>\r\n\r\n<div id=\"fs-id1170571645665\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx\\hfill &amp; =\\text{\u2212}{\\displaystyle\\int }_{0}^{-1}{e}^{u}du\\hfill \\\\ \\\\ \\\\ &amp; ={\\displaystyle\\int }_{-1}^{0}{e}^{u}du\\hfill \\\\ &amp; ={{e}^{u}|}_{-1}^{0}\\hfill \\\\ &amp; ={e}^{0}-({e}^{-1})\\hfill \\\\ &amp; =\\text{\u2212}{e}^{-1}+1.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572495875\">See Figure 2.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204254\/CNX_Calc_Figure_05_06_002.jpg\" alt=\"A graph of the function f(x) = e^(1-x) over [0, 3]. It crosses the y-axis at (0, e) as a decreasing concave up curve and symptotically approaches 0 as x goes to infinity.\" width=\"325\" height=\"208\" \/> Figure 2. The indicated area can be calculated by evaluating a definite integral using substitution.[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571599287\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]{\\displaystyle\\int }_{0}^{2}{e}^{2x}dx.[\/latex]\r\n\r\n[reveal-answer q=\"3355576\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3355576\"]\r\n\r\nLet [latex]u=2x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572274760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572274760\"]\r\n\r\n[latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{4}{e}^{u}du=\\frac{1}{2}({e}^{4}-1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Growth of Bacteria in a Culture<\/h3>\r\nSuppose the rate of <span class=\"no-emphasis\">growth of bacteria<\/span> in a Petri dish is given by [latex]q(t)={3}^{t},[\/latex] where [latex]t[\/latex] is given in hours and [latex]q(t)[\/latex] is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function [latex]Q(t)[\/latex] that gives the number of bacteria in the Petri dish at any time [latex]t[\/latex]. How many bacteria are in the dish after 2 hours?\r\n\r\n[reveal-answer q=\"fs-id1170571699005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571699005\"]\r\n\r\nWe have\r\n\r\n[latex]Q(t)=\\int {3}^{t}dt=\\frac{{3}^{t}}{\\text{ln}3}+C.[\/latex]\r\n\r\nThen, at [latex]t=0[\/latex] we have [latex]Q(0)=10=\\frac{1}{\\text{ln}3}+C,[\/latex] so [latex]C\\approx 9.090[\/latex] and we get\r\n\r\n[latex]Q(t)=\\frac{{3}^{t}}{\\text{ln}3}+9.090.[\/latex]\r\n\r\nAt time [latex]t=2,[\/latex] we have\r\n<div id=\"fs-id1170571698223\" class=\"equation unnumbered\">[latex]Q(2)=\\frac{{3}^{2}}{\\text{ln}3}+9.090[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1170571637497\" class=\"equation unnumbered\">[latex]=17.282.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571637508\">After 2 hours, there are 17,282 bacteria in the dish.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571637514\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572437454\">From <a class=\"autogenerated-content\" href=\"#fs-id1170572624841\">(Figure)<\/a>, suppose the bacteria grow at a rate of [latex]q(t)={2}^{t}.[\/latex] Assume the culture still starts with 10,000 bacteria. Find [latex]Q(t).[\/latex] How many bacteria are in the dish after 3 hours?<\/p>\r\n[reveal-answer q=\"680248\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"680248\"]\r\n<p id=\"fs-id1170572444227\">Use the procedure from <a class=\"autogenerated-content\" href=\"#fs-id1170572624841\">(Figure)<\/a> to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572444236\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572444236\"]\r\n<p id=\"fs-id1170572444236\">[latex]Q(t)=\\frac{{2}^{t}}{\\text{ln}2}+8.557.[\/latex] There are 20,099 bacteria in the dish after 3 hours.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571807209\" class=\"textbox exercises\">\r\n<h3>Fruit Fly Population Growth<\/h3>\r\n<div id=\"fs-id1170571807211\" class=\"exercise\">\r\n<p id=\"fs-id1170571807219\">Suppose a population of <span class=\"no-emphasis\">fruit flies<\/span> increases at a rate of [latex]g(t)=2{e}^{0.02t},[\/latex] in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days?<\/p>\r\n[reveal-answer q=\"fs-id1170572183855\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572183855\"]\r\n<p id=\"fs-id1170572183855\">Let [latex]G(t)[\/latex] represent the number of flies in the population at time [latex]t[\/latex]. Applying the net change theorem, we have<\/p>\r\n\r\n<div id=\"fs-id1170571712220\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ G(10)\\hfill &amp; =G(0)+{\\int }_{0}^{10}2{e}^{0.02t}dt\\hfill \\\\ &amp; =100+{\\left[\\frac{2}{0.02}{e}^{0.02t}\\right]|}_{0}^{10}\\hfill \\\\ &amp; =100+{\\left[100{e}^{0.02t}\\right]|}_{0}^{10}\\hfill \\\\ &amp; =100+100{e}^{0.2}-100\\hfill \\\\ &amp; \\approx 122.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572338467\">There are 122 flies in the population after 10 days.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose the rate of growth of the fly population is given by [latex]g(t)={e}^{0.01t},[\/latex] and the initial fly population is 100 flies. How many flies are in the population after 15 days?\r\n\r\n[reveal-answer q=\"572927\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"572927\"]Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170571807209\">(Figure)<\/a> to solve the problem.[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571653080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653080\"]There are 116 flies.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbook exercises\">\r\n<div id=\"fs-id1170571653091\" class=\"exercise\">\r\n<h3>example: Evaluating a Definite Integral Using Substitution<\/h3>\r\nEvaluate the definite integral using substitution: [latex]{\\displaystyle\\int }_{1}^{2}\\dfrac{{e}^{1\\text{\/}x}}{{x}^{2}}dx.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572587719\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587719\"]\r\n<p id=\"fs-id1170572587719\">This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on [latex]e[\/latex] as a power of [latex]x[\/latex], then bring the [latex]x[\/latex]<sup>2<\/sup> in the denominator up to the numerator using a negative exponent. We have<\/p>\r\n\r\n<div id=\"fs-id1170572587738\" class=\"equation unnumbered\">[latex]{\\displaystyle\\int }_{1}^{2}\\frac{{e}^{1\\text{\/}x}}{{x}^{2}}dx={\\displaystyle\\int }_{1}^{2}{e}^{{x}^{-1}}{x}^{-2}dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571636291\">Let [latex]u={x}^{-1},[\/latex] the exponent on [latex]e[\/latex]. Then<\/p>\r\n\r\n<div id=\"fs-id1170571636314\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill du&amp; =\\text{\u2212}{x}^{-2}dx\\hfill \\\\ \\hfill -du&amp; ={x}^{-2}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571599633\">Bringing the negative sign outside the integral sign, the problem now reads<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{\u2212}\\displaystyle\\int {e}^{u}du.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571599663\">Next, change the limits of integration:<\/p>\r\n\r\n<div id=\"fs-id1170571599666\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ u={(1)}^{-1}=1\\hfill \\\\ u={(2)}^{-1}=\\frac{1}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\nNotice that now the limits begin with the larger number, meaning we must multiply by \u22121 and interchange the limits. Thus,\r\n<div id=\"fs-id1170572293466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\\\ \\text{\u2212}{\\displaystyle\\int }_{1}^{1\\text{\/}2}{e}^{u}du\\hfill &amp; ={\\displaystyle\\int }_{1\\text{\/}2}^{1}{e}^{u}du\\hfill \\\\ &amp; ={e}^{u}{|}_{1\\text{\/}2}^{1}\\hfill \\\\ &amp; =e-{e}^{1\\text{\/}2}\\hfill \\\\ &amp; =e-\\sqrt{e}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572554374\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate the definite integral using substitution: [latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx.[\/latex]\r\n<div id=\"fs-id1170572554378\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"43778229\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"43778229\"]\r\n<p id=\"fs-id1170572554429\">Let [latex]u=4{x}^{-2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572628419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572628419\"]\r\n<p id=\"fs-id1170572628419\">[latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx=\\frac{1}{8}\\left[{e}^{4}-e\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571419835\" class=\"commentary\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20047[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Integrate functions involving exponential functions<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170571602573\">The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, [latex]y={e}^{x},[\/latex] is its own derivative and its own integral.<\/p>\n<div id=\"fs-id1170572370390\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Integrals of Exponential Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1170572375243\">Exponential functions can be integrated using the following formulas.<\/p>\n<div id=\"fs-id1170572374424\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc} {\\displaystyle\\int{e}^{x}dx} & {=} & {{e}^{x}+C} \\\\ {\\displaystyle\\int{a}^{x}dx} & {=} & {\\dfrac{{a}^{x}}{\\text{ln}a}+C}\\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<p>The nature of the antiderivative of [latex]{e}^{x}[\/latex] makes it fairly easy to identify what to choose as [latex]u[\/latex]. If only one\u00a0[latex]e[\/latex] exists, choose the exponent of [latex]e[\/latex] as\u00a0[latex]u[\/latex]. If more than one [latex]e[\/latex] exists, choose the more complicated function involving [latex]e[\/latex] as\u00a0[latex]u[\/latex].<\/p>\n<div id=\"fs-id11705722115180\" class=\"textbook exercises\">\n<h3>Example: Finding an Antiderivative of an Exponential Function<\/h3>\n<p>Find the antiderivative of the exponential function [latex]e^{-x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q26617708\">Show Solution<\/span><\/p>\n<div id=\"q26617708\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use substitution, setting [latex]u=\\text{\u2212}x,[\/latex] and then [latex]du=-1dx.[\/latex] Multiply the <em>du<\/em> equation by \u22121, so you now have [latex]\\text{\u2212}du=dx.[\/latex] Then,<\/p>\n<p>[latex]\\begin{array}{cc}{\\displaystyle\\int {e}^{\\text{\u2212}x}dx}\\hfill & =\\text{\u2212}{\\displaystyle\\int {e}^{u}du}\\hfill \\\\ \\\\ & =\\text{\u2212}{e}^{u}+C\\hfill \\\\ & =\\text{\u2212}{e}^{\\text{\u2212}x}+C.\\hfill \\end{array}[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572248011\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of the function using substitution: [latex]{x}^{2}{e}^{-2{x}^{3}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q20881654\">Hint<\/span><\/p>\n<div id=\"q20881654\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571622632\">Let [latex]u[\/latex] equal the exponent on [latex]e[\/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-id1170572269592\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572269592\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572269592\">[latex]\\displaystyle\\int {x}^{2}{e}^{-2{x}^{3}}dx=-\\frac{1}{6}{e}^{-2{x}^{3}}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=88&amp;end=170&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_88to170_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.1&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1170572607911\">A common mistake when dealing with exponential expressions is treating the exponent on [latex]e[\/latex] the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on [latex]e[\/latex]. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we\u2019re using the right rules for the functions we\u2019re integrating.<\/p>\n<div id=\"fs-id1170572209244\" class=\"textbook exercises\">\n<h3>Example: Square Root of an Exponential Function<\/h3>\n<p>Find the antiderivative of the exponential function [latex]{e}^{x}\\sqrt{1+{e}^{x}}.[\/latex]<\/p>\n<div id=\"fs-id1170571602248\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572551693\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572551693\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572551693\">First rewrite the problem using a rational exponent:<\/p>\n<div id=\"fs-id1170572549246\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}\\sqrt{1+{e}^{x}}dx=\\displaystyle\\int {e}^{x}{(1+{e}^{x})}^{1\\text{\/}2}dx.[\/latex]<\/div>\n<p>Using substitution, choose [latex]u=1+{e}^{x}.u=1+{e}^{x}.[\/latex] Then, [latex]du={e}^{x}dx.[\/latex] We have (Figure 1)<\/p>\n<div id=\"fs-id1170572373474\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {e}^{x}{(1+{e}^{x})}^{1\\text{\/}2}dx=\\displaystyle\\int {u}^{1\\text{\/}2}du.[\/latex]<\/div>\n<p id=\"fs-id1170572307548\">Then<\/p>\n<div id=\"fs-id1170572560605\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {u}^{1\\text{\/}2}du=\\frac{{u}^{3\\text{\/}2}}{3\\text{\/}2}+C=\\frac{2}{3}{u}^{3\\text{\/}2}+C=\\frac{2}{3}{(1+{e}^{x})}^{3\\text{\/}2}+C.[\/latex]<\/div>\n<div>\n<div style=\"width: 335px\" 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\/11204252\/CNX_Calc_Figure_05_06_001.jpg\" alt=\"A graph of the function f(x) = e^x * sqrt(1 + e^x), which is an increasing concave up curve, over &#091;-3, 1&#093;. It begins close to the x-axis in quadrant two, crosses the y-axis at (0, sqrt(2)), and continues to increase rapidly.\" width=\"325\" height=\"208\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The graph shows an exponential function times the square root of an exponential function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572207056\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the antiderivative of [latex]{e}^{x}{(3{e}^{x}-2)}^{2}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q366490\">Hint<\/span><\/p>\n<div id=\"q366490\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]u=3{e}^{x}-2u=3{e}^{x}-2.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6711088\">Show Solution<\/span><\/p>\n<div id=\"q6711088\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572451630\" class=\"exercise\">\n<div class=\"solution\">\n<p id=\"fs-id1170572209127\">[latex]\\displaystyle\\int {e}^{x}{(3{e}^{x}-2)}^{2}dx=\\frac{1}{9}{(3{e}^{x}-2)}^{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=257&amp;end=336&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_257to336_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.1&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbook exercises\">\n<div id=\"fs-id1170572240614\" class=\"exercise\">\n<h3>Example: Using Substitution with an Exponential Function<\/h3>\n<p>Use substitution to evaluate the indefinite integral [latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572216534\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572216534\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572216534\">Here we choose to let [latex]u[\/latex] equal the expression in the exponent on [latex]e[\/latex]. Let [latex]u=2{x}^{3}[\/latex] and [latex]du=6{x}^{2}dx..[\/latex] Again, <em>du<\/em> is off by a constant multiplier; the original function contains a factor of 3[latex]x[\/latex]<sup>2<\/sup>, not 6[latex]x[\/latex]<sup>2<\/sup>. Multiply both sides of the equation by [latex]\\frac{1}{2}[\/latex] so that the integrand in [latex]u[\/latex] equals the integrand in [latex]x[\/latex]. Thus,<\/p>\n<div id=\"fs-id1170572470447\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx=\\frac{1}{2}\\displaystyle\\int {e}^{u}du.[\/latex]<\/div>\n<p id=\"fs-id1170572101855\">Integrate the expression in [latex]u[\/latex] and then substitute the original expression in [latex]x[\/latex] back into the [latex]u[\/latex] integral:<\/p>\n<div id=\"fs-id1170572448333\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}\\displaystyle\\int {e}^{u}du=\\frac{1}{2}{e}^{u}+C=\\frac{1}{2}{e}^{2{x}^{3}}+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572551647\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate the indefinite integral [latex]\\displaystyle\\int 2{x}^{3}{e}^{{x}^{4}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q212787664\">Hint<\/span><\/p>\n<div id=\"q212787664\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572232621\">Let [latex]u={x}^{4}.[\/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-id1170572242344\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572242344\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572242344\">[latex]\\displaystyle\\int 2{x}^{3}{e}^{{x}^{4}}dx=\\frac{1}{2}{e}^{{x}^{4}}[\/latex]<\/p>\n<div id=\"fs-id1170572294922\" class=\"exercise\">\n<div id=\"fs-id1170573414530\" class=\"commentary\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571733816\">As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number\u00a0<em data-effect=\"italics\">e<\/em>\u00a0is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let\u2019s look at an example in which integration of an exponential function solves a common business application.<\/p>\n<p id=\"fs-id1170571571226\">A\u00a0<span id=\"term238\" class=\"no-emphasis\" data-type=\"term\">price\u2013demand function<\/span>\u00a0tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal price\u2013demand function is the derivative of the price\u2013demand function and it tells us how fast the price changes at a given level of production. These functions are used in business to determine the price\u2013elasticity of demand, and to help companies determine whether changing production levels would be profitable.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Price\u2013Demand Equation<\/h3>\n<p>Find the price\u2013demand equation for a particular brand of toothpaste at a supermarket chain when the demand is 50 tubes per week at $2.35 per tube, given that the marginal price\u2014demand function, [latex]{p}^{\\prime }(x),[\/latex] for [latex]x[\/latex] number of tubes per week, is given as<\/p>\n<div id=\"fs-id1170572141865\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p\\text{'}(x)=-0.015{e}^{-0.01x}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571595414\">If the supermarket chain sells 100 tubes per week, what price should it set?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571543163\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571543163\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571543163\" style=\"text-align: center;\">To find the price\u2013demand equation, integrate the marginal price\u2013demand function. First find the antiderivative, then look at the particulars. Thus,<\/p>\n<div id=\"fs-id1170572216339\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ \\\\ p(x)\\hfill & =\\int -0.015{e}^{-0.01x}dx\\hfill \\\\ & =-0.015\\int {e}^{-0.01x}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572507579\">Using substitution, let [latex]u=-0.01x[\/latex] and [latex]du=-0.01dx.[\/latex] Then, divide both sides of the <em>du<\/em> equation by \u22120.01. This gives<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\frac{-0.015}{-0.01}\\int {e}^{u}du\\hfill & =1.5\\int {e}^{u}du\\hfill \\\\ \\\\ & =1.5{e}^{u}+C\\hfill \\\\ & =1.5{e}^{-0.01x}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572111467\">The next step is to solve for <em>C<\/em>. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ p(50)\\hfill & =1.5{e}^{-0.01(50)}+C\\hfill \\\\ & =2.35.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571569187\">Now, just solve for <em>C<\/em>:<\/p>\n<div id=\"fs-id1170572415392\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ C\\hfill & =2.35-1.5{e}^{-0.5}\\hfill \\\\ & =2.35-0.91\\hfill \\\\ & =1.44.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572140846\">Thus,<\/p>\n<div id=\"fs-id1170572375703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p(x)=1.5{e}^{-0.01x}+1.44.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571547617\">If the supermarket sells 100 tubes of toothpaste per week, the price would be<\/p>\n<div id=\"fs-id1170571653359\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]p(100)=1.5{e}^{-0.01(100)}+1.44=1.5{e}^{-1}+1.44\\approx 1.99.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572393399\">The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding a Price\u2013Demand Equation.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V9NlSl17duk?controls=0&amp;start=465&amp;end=815&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.1_465to815_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.1&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571637964\" class=\"textbook exercises\">\n<div id=\"fs-id1170572204890\" class=\"exercise\">\n<h3>example: Evaluating a Definite Integral Involving an Exponential Function<\/h3>\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572247799\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572247799\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572247799\">Again, substitution is the method to use. Let [latex]u=1-x,[\/latex] so [latex]du=-1dx[\/latex] or [latex]\\text{\u2212}du=dx.[\/latex] Then [latex]\\displaystyle\\int {e}^{1-x}dx=\\text{\u2212}\\displaystyle\\int {e}^{u}du.[\/latex] Next, change the limits of integration. Using the equation [latex]u=1-x,[\/latex] we have<\/p>\n<div id=\"fs-id1170571652053\" class=\"equation unnumbered\">[latex]\\begin{array}{c}u=1-(1)=0\\hfill \\\\ u=1-(2)=-1.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571645662\">The integral then becomes<\/p>\n<div id=\"fs-id1170571645665\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx\\hfill & =\\text{\u2212}{\\displaystyle\\int }_{0}^{-1}{e}^{u}du\\hfill \\\\ \\\\ \\\\ & ={\\displaystyle\\int }_{-1}^{0}{e}^{u}du\\hfill \\\\ & ={{e}^{u}|}_{-1}^{0}\\hfill \\\\ & ={e}^{0}-({e}^{-1})\\hfill \\\\ & =\\text{\u2212}{e}^{-1}+1.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572495875\">See Figure 2.<\/p>\n<div style=\"width: 335px\" 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\/11204254\/CNX_Calc_Figure_05_06_002.jpg\" alt=\"A graph of the function f(x) = e^(1-x) over &#091;0, 3&#093;. It crosses the y-axis at (0, e) as a decreasing concave up curve and symptotically approaches 0 as x goes to infinity.\" width=\"325\" height=\"208\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The indicated area can be calculated by evaluating a definite integral using substitution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571599287\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]{\\displaystyle\\int }_{0}^{2}{e}^{2x}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3355576\">Hint<\/span><\/p>\n<div id=\"q3355576\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]u=2x[\/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-id1170572274760\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572274760\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{4}{e}^{u}du=\\frac{1}{2}({e}^{4}-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Growth of Bacteria in a Culture<\/h3>\n<p>Suppose the rate of <span class=\"no-emphasis\">growth of bacteria<\/span> in a Petri dish is given by [latex]q(t)={3}^{t},[\/latex] where [latex]t[\/latex] is given in hours and [latex]q(t)[\/latex] is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function [latex]Q(t)[\/latex] that gives the number of bacteria in the Petri dish at any time [latex]t[\/latex]. How many bacteria are in the dish after 2 hours?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571699005\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571699005\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have<\/p>\n<p>[latex]Q(t)=\\int {3}^{t}dt=\\frac{{3}^{t}}{\\text{ln}3}+C.[\/latex]<\/p>\n<p>Then, at [latex]t=0[\/latex] we have [latex]Q(0)=10=\\frac{1}{\\text{ln}3}+C,[\/latex] so [latex]C\\approx 9.090[\/latex] and we get<\/p>\n<p>[latex]Q(t)=\\frac{{3}^{t}}{\\text{ln}3}+9.090.[\/latex]<\/p>\n<p>At time [latex]t=2,[\/latex] we have<\/p>\n<div id=\"fs-id1170571698223\" class=\"equation unnumbered\">[latex]Q(2)=\\frac{{3}^{2}}{\\text{ln}3}+9.090[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1170571637497\" class=\"equation unnumbered\">[latex]=17.282.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571637508\">After 2 hours, there are 17,282 bacteria in the dish.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571637514\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572437454\">From <a class=\"autogenerated-content\" href=\"#fs-id1170572624841\">(Figure)<\/a>, suppose the bacteria grow at a rate of [latex]q(t)={2}^{t}.[\/latex] Assume the culture still starts with 10,000 bacteria. Find [latex]Q(t).[\/latex] How many bacteria are in the dish after 3 hours?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q680248\">Hint<\/span><\/p>\n<div id=\"q680248\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572444227\">Use the procedure from <a class=\"autogenerated-content\" href=\"#fs-id1170572624841\">(Figure)<\/a> to solve the problem.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572444236\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572444236\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572444236\">[latex]Q(t)=\\frac{{2}^{t}}{\\text{ln}2}+8.557.[\/latex] There are 20,099 bacteria in the dish after 3 hours.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571807209\" class=\"textbox exercises\">\n<h3>Fruit Fly Population Growth<\/h3>\n<div id=\"fs-id1170571807211\" class=\"exercise\">\n<p id=\"fs-id1170571807219\">Suppose a population of <span class=\"no-emphasis\">fruit flies<\/span> increases at a rate of [latex]g(t)=2{e}^{0.02t},[\/latex] in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572183855\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572183855\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572183855\">Let [latex]G(t)[\/latex] represent the number of flies in the population at time [latex]t[\/latex]. Applying the net change theorem, we have<\/p>\n<div id=\"fs-id1170571712220\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ G(10)\\hfill & =G(0)+{\\int }_{0}^{10}2{e}^{0.02t}dt\\hfill \\\\ & =100+{\\left[\\frac{2}{0.02}{e}^{0.02t}\\right]|}_{0}^{10}\\hfill \\\\ & =100+{\\left[100{e}^{0.02t}\\right]|}_{0}^{10}\\hfill \\\\ & =100+100{e}^{0.2}-100\\hfill \\\\ & \\approx 122.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572338467\">There are 122 flies in the population after 10 days.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose the rate of growth of the fly population is given by [latex]g(t)={e}^{0.01t},[\/latex] and the initial fly population is 100 flies. How many flies are in the population after 15 days?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q572927\">Hint<\/span><\/p>\n<div id=\"q572927\" class=\"hidden-answer\" style=\"display: none\">Use the process from <a class=\"autogenerated-content\" href=\"#fs-id1170571807209\">(Figure)<\/a> to solve the problem.<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571653080\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571653080\" class=\"hidden-answer\" style=\"display: none\">There are 116 flies.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook exercises\">\n<div id=\"fs-id1170571653091\" class=\"exercise\">\n<h3>example: Evaluating a Definite Integral Using Substitution<\/h3>\n<p>Evaluate the definite integral using substitution: [latex]{\\displaystyle\\int }_{1}^{2}\\dfrac{{e}^{1\\text{\/}x}}{{x}^{2}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572587719\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572587719\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572587719\">This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on [latex]e[\/latex] as a power of [latex]x[\/latex], then bring the [latex]x[\/latex]<sup>2<\/sup> in the denominator up to the numerator using a negative exponent. We have<\/p>\n<div id=\"fs-id1170572587738\" class=\"equation unnumbered\">[latex]{\\displaystyle\\int }_{1}^{2}\\frac{{e}^{1\\text{\/}x}}{{x}^{2}}dx={\\displaystyle\\int }_{1}^{2}{e}^{{x}^{-1}}{x}^{-2}dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571636291\">Let [latex]u={x}^{-1},[\/latex] the exponent on [latex]e[\/latex]. Then<\/p>\n<div id=\"fs-id1170571636314\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill du& =\\text{\u2212}{x}^{-2}dx\\hfill \\\\ \\hfill -du& ={x}^{-2}dx.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571599633\">Bringing the negative sign outside the integral sign, the problem now reads<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{\u2212}\\displaystyle\\int {e}^{u}du.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571599663\">Next, change the limits of integration:<\/p>\n<div id=\"fs-id1170571599666\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ u={(1)}^{-1}=1\\hfill \\\\ u={(2)}^{-1}=\\frac{1}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Notice that now the limits begin with the larger number, meaning we must multiply by \u22121 and interchange the limits. Thus,<\/p>\n<div id=\"fs-id1170572293466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\\\ \\text{\u2212}{\\displaystyle\\int }_{1}^{1\\text{\/}2}{e}^{u}du\\hfill & ={\\displaystyle\\int }_{1\\text{\/}2}^{1}{e}^{u}du\\hfill \\\\ & ={e}^{u}{|}_{1\\text{\/}2}^{1}\\hfill \\\\ & =e-{e}^{1\\text{\/}2}\\hfill \\\\ & =e-\\sqrt{e}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572554374\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate the definite integral using substitution: [latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx.[\/latex]<\/p>\n<div id=\"fs-id1170572554378\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q43778229\">Hint<\/span><\/p>\n<div id=\"q43778229\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572554429\">Let [latex]u=4{x}^{-2}[\/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-id1170572628419\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572628419\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572628419\">[latex]{\\displaystyle\\int }_{1}^{2}\\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx=\\frac{1}{8}\\left[{e}^{4}-e\\right][\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170571419835\" class=\"commentary\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20047\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20047&theme=oea&iframe_resize_id=ohm20047&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-1139\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.6.1. <strong>Authored by<\/strong>: Ryan Melton. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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":17533,"menu_order":26,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.6.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"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-1139","chapter","type-chapter","status-publish","hentry"],"part":1113,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1139","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1139\/revisions"}],"predecessor-version":[{"id":2647,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1139\/revisions\/2647"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1113"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1139\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1139"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1139"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1139"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}