{"id":408,"date":"2025-02-13T19:44:40","date_gmt":"2025-02-13T19:44:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-involving-exponential-and-logarithmic-functions-learn-it-1\/"},"modified":"2025-02-13T19:44:40","modified_gmt":"2025-02-13T19:44:40","slug":"integrals-involving-exponential-and-logarithmic-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-involving-exponential-and-logarithmic-functions-learn-it-1\/","title":{"raw":"Integrals Involving Exponential and Logarithmic Functions: Learn It 1","rendered":"Integrals Involving Exponential and Logarithmic Functions: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Perform integrations on functions that include exponential terms<\/li>\n\t<li>Solve integrals that feature logarithmic functions<\/li>\n<\/ul>\n<\/section>\n<h2>Integrals of Exponential Functions<\/h2>\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<section class=\"textbox keyTakeaway\">\n<h3>integrals of exponential functions<\/h3>\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} &amp; {=} &amp; {{e}^{x}+C} \\\\ {\\displaystyle\\int{a}^{x}dx} &amp; {=} &amp; {\\dfrac{{a}^{x}}{\\text{ln}a}+C}\\end{array}[\/latex]<\/div>\n<\/section>\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].&nbsp;<\/p>\n<section class=\"textbox proTip\">\n<p>If only one [latex]e[\/latex] exists, choose the exponent of [latex]e[\/latex] as [latex]u[\/latex]. If more than one [latex]e[\/latex] exists, choose the more complicated function involving [latex]e[\/latex] as [latex]u[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the exponential function [latex]e^{-x}[\/latex].<\/p>\n<p>[reveal-answer q=\"26617708\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"26617708\"]<\/p>\n<p>Use substitution, setting [latex]u=\\text{\u2212}x,[\/latex] and then [latex]du=-1dx.[\/latex]<\/p>\n<p>Multiply the <em>du<\/em> equation by [latex]\u22121[\/latex], so you now have [latex]\\text{\u2212}du=dx.[\/latex]<\/p>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[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]<br>\n[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox proTip\">\n<p>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<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the exponential function [latex]{e}^{x}\\sqrt{1+{e}^{x}}.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170572551693\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572551693\"]\n\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]<\/p>\n<p>Then, [latex]du={e}^{x}dx.[\/latex] We have,<\/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\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\/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]\n\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to evaluate the indefinite integral [latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572216534\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572216534\"]<\/p>\n<p id=\"fs-id1170572216534\">Here we choose to let [latex]u[\/latex] equal the expression in the exponent on [latex]e[\/latex].<\/p>\n<p>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^2[\/latex], not 6[latex]x^2[\/latex].<\/p>\n<p>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].<\/p>\n<p>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][\/hidden-answer]<\/div>\n<\/section>\n<p id=\"fs-id1170571733816\">Exponential functions are used in many real-life applications. The number <em data-effect=\"italics\">e<\/em>&nbsp;is 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<section class=\"textbox connectIt\">\n<p>A&nbsp;<span id=\"term238\" class=\"no-emphasis\" data-type=\"term\">price\u2013demand function<\/span> tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases.<\/p>\n<p>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.<\/p>\n<p>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<\/section>\n<section class=\"textbox example\">\n<p>Find the price\u2013demand equation for a particular brand of toothpaste at a supermarket chain when the demand is [latex]50[\/latex] tubes per week at [latex]$2.35[\/latex] 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 id=\"fs-id1170571595414\">If the supermarket chain sells [latex]100[\/latex] tubes per week, what price should it set?<\/p>\n<p>[reveal-answer q=\"fs-id1170571543163\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571543163\"]<\/p>\n<p id=\"fs-id1170571543163\" style=\"text-align: left;\">To find the price\u2013demand equation, integrate the marginal price\u2013demand function. First find the antiderivative, then look at the particulars.<\/p>\n<p style=\"text-align: left;\">Thus,<\/p>\n<div id=\"fs-id1170572216339\" class=\"equation unnumbered\" style=\"text-align: center;\">[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>\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 [latex]\u22120.01[\/latex].<\/p>\n<p>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 &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>\n<p id=\"fs-id1170572111467\">The next step is to solve for [latex]C[\/latex]. We know that when the price is [latex]$2.35[\/latex] per tube, the demand is [latex]50[\/latex] tubes per week. This means<\/p>\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>\n<p id=\"fs-id1170571569187\">Now, just solve for [latex]C[\/latex]:<\/p>\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>\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 id=\"fs-id1170571547617\">If the supermarket sells [latex]100[\/latex] 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 id=\"fs-id1170572393399\">The supermarket should charge [latex]$1.99[\/latex] per tube if it is selling [latex]100[\/latex] tubes per week.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\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\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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 \"5.6.1\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral [latex]{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx.[\/latex]<br>\n[reveal-answer q=\"fs-id1170572247799\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572247799\"]<\/p>\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\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1-(1)=0\\hfill \\\\ u=1-(2)=-1.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571645662\">The integral then becomes<\/p>\n<div id=\"fs-id1170571645665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx\\hfill &amp; =\\text{\u2212}{\\displaystyle\\int }_{0}^{-1}{e}^{u}du\\hfill&nbsp; \\\\ &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>\n\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]\n[\/hidden-answer]<\/section>\n<section class=\"textbox example\">\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 [latex]10,000[\/latex] 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 [latex]2[\/latex] hours?<\/p>\n<p>[reveal-answer q=\"fs-id1170571699005\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571699005\"]<\/p>\n<p>We have,<\/p>\n<p style=\"text-align: center;\">[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<\/p>\n<p style=\"text-align: center;\">[latex]Q(0)=10=\\frac{1}{\\text{ln}3}+C,[\/latex]&nbsp;<\/p>\n<p>so [latex]C\\approx 9.090[\/latex] and we get,<\/p>\n<p style=\"text-align: center;\">[latex]Q(t)=\\frac{{3}^{t}}{\\text{ln}3}+9.090.[\/latex]<\/p>\n<p><em>Note: We are using [latex]10[\/latex] in place of [latex]10,000[\/latex] since [latex]10 ,000[\/latex] bacteria are [latex]10[\/latex] thousands of bacteria. We will multiple our final answer by a power of [latex]1000[\/latex] at the end of our calculation to account for this.&nbsp;<br>\n<\/em><br>\nAt time [latex]t=2,[\/latex] we have,<\/p>\n<div id=\"fs-id1170571698223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]Q(2)=\\frac{{3}^{2}}{\\text{ln}3}+9.090=17.282[\/latex]<\/div>\n<p id=\"fs-id1170571637508\">After [latex]2[\/latex] hours, there are [latex]17,282[\/latex] bacteria in the dish.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral using substitution:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{2}\\dfrac{{e}^{1\\text{\/}x}}{{x}^{2}}dx.[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572587719\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572587719\"]<\/p>\n<p id=\"fs-id1170572587719\">This problem requires some rewriting to simplify applying the properties.<\/p>\n<p>First, rewrite the exponent on [latex]e[\/latex] as a power of [latex]x[\/latex], then bring the [latex]x^2[\/latex] in the denominator up to the numerator using a negative exponent.<\/p>\n<p>We have,<\/p>\n<div id=\"fs-id1170572587738\" class=\"equation unnumbered\" style=\"text-align: center;\">[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 id=\"fs-id1170571636291\">Let [latex]u={x}^{-1},[\/latex] the exponent on [latex]e[\/latex].<\/p>\n<p>Then,<\/p>\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>\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 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 [latex]\u22121 [\/latex]and interchange the limits.<\/p>\n<p>Thus,<\/p>\n<div id=\"fs-id1170572293466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\\\ \\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>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288436[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Perform integrations on functions that include exponential terms<\/li>\n<li>Solve integrals that feature logarithmic functions<\/li>\n<\/ul>\n<\/section>\n<h2>Integrals of Exponential Functions<\/h2>\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<section class=\"textbox keyTakeaway\">\n<h3>integrals of exponential functions<\/h3>\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<\/section>\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].&nbsp;<\/p>\n<section class=\"textbox proTip\">\n<p>If only one [latex]e[\/latex] exists, choose the exponent of [latex]e[\/latex] as [latex]u[\/latex]. If more than one [latex]e[\/latex] exists, choose the more complicated function involving [latex]e[\/latex] as [latex]u[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the exponential function [latex]e^{-x}[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q26617708\">Show Solution<\/button><\/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]<\/p>\n<p>Multiply the <em>du<\/em> equation by [latex]\u22121[\/latex], so you now have [latex]\\text{\u2212}du=dx.[\/latex]<\/p>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[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<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\">\n<p>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<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the exponential function [latex]{e}^{x}\\sqrt{1+{e}^{x}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572551693\">Show Solution<\/button><\/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]<\/p>\n<p>Then, [latex]du={e}^{x}dx.[\/latex] We have,<\/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<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 1. The graph shows an exponential function times the square root of an exponential function.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Use substitution to evaluate the indefinite integral [latex]\\displaystyle\\int 3{x}^{2}{e}^{2{x}^{3}}dx.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572216534\">Show Solution<\/button><\/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].<\/p>\n<p>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^2[\/latex], not 6[latex]x^2[\/latex].<\/p>\n<p>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].<\/p>\n<p>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<\/section>\n<p id=\"fs-id1170571733816\">Exponential functions are used in many real-life applications. The number <em data-effect=\"italics\">e<\/em>&nbsp;is 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<section class=\"textbox connectIt\">\n<p>A&nbsp;<span id=\"term238\" class=\"no-emphasis\" data-type=\"term\">price\u2013demand function<\/span> tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases.<\/p>\n<p>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.<\/p>\n<p>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<\/section>\n<section class=\"textbox example\">\n<p>Find the price\u2013demand equation for a particular brand of toothpaste at a supermarket chain when the demand is [latex]50[\/latex] tubes per week at [latex]$2.35[\/latex] 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 id=\"fs-id1170571595414\">If the supermarket chain sells [latex]100[\/latex] tubes per week, what price should it set?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571543163\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571543163\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571543163\" style=\"text-align: left;\">To find the price\u2013demand equation, integrate the marginal price\u2013demand function. First find the antiderivative, then look at the particulars.<\/p>\n<p style=\"text-align: left;\">Thus,<\/p>\n<div id=\"fs-id1170572216339\" class=\"equation unnumbered\" style=\"text-align: center;\">[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 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 [latex]\u22120.01[\/latex].<\/p>\n<p>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 id=\"fs-id1170572111467\">The next step is to solve for [latex]C[\/latex]. We know that when the price is [latex]$2.35[\/latex] per tube, the demand is [latex]50[\/latex] 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 id=\"fs-id1170571569187\">Now, just solve for [latex]C[\/latex]:<\/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 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 id=\"fs-id1170571547617\">If the supermarket sells [latex]100[\/latex] 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 id=\"fs-id1170572393399\">The supermarket should charge [latex]$1.99[\/latex] per tube if it is selling [latex]100[\/latex] tubes per week.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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<\/section>\n<section class=\"textbox example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572247799\">Show Solution<\/button><\/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\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1-(1)=0\\hfill \\\\ u=1-(2)=-1.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571645662\">The integral then becomes<\/p>\n<div id=\"fs-id1170571645665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{2}{e}^{1-x}dx\\hfill & =\\text{\u2212}{\\displaystyle\\int }_{0}^{-1}{e}^{u}du\\hfill&nbsp; \\\\ & ={\\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<figure style=\"width: 325px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 2. The indicated area can be calculated by evaluating a definite integral using substitution.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\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 [latex]10,000[\/latex] 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 [latex]2[\/latex] hours?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571699005\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571699005\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have,<\/p>\n<p style=\"text-align: center;\">[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<\/p>\n<p style=\"text-align: center;\">[latex]Q(0)=10=\\frac{1}{\\text{ln}3}+C,[\/latex]&nbsp;<\/p>\n<p>so [latex]C\\approx 9.090[\/latex] and we get,<\/p>\n<p style=\"text-align: center;\">[latex]Q(t)=\\frac{{3}^{t}}{\\text{ln}3}+9.090.[\/latex]<\/p>\n<p><em>Note: We are using [latex]10[\/latex] in place of [latex]10,000[\/latex] since [latex]10 ,000[\/latex] bacteria are [latex]10[\/latex] thousands of bacteria. We will multiple our final answer by a power of [latex]1000[\/latex] at the end of our calculation to account for this.&nbsp;<br \/>\n<\/em><br \/>\nAt time [latex]t=2,[\/latex] we have,<\/p>\n<div id=\"fs-id1170571698223\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]Q(2)=\\frac{{3}^{2}}{\\text{ln}3}+9.090=17.282[\/latex]<\/div>\n<p id=\"fs-id1170571637508\">After [latex]2[\/latex] hours, there are [latex]17,282[\/latex] bacteria in the dish.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate the definite integral using substitution:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{2}\\dfrac{{e}^{1\\text{\/}x}}{{x}^{2}}dx.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572587719\">Show Solution<\/button><\/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.<\/p>\n<p>First, rewrite the exponent on [latex]e[\/latex] as a power of [latex]x[\/latex], then bring the [latex]x^2[\/latex] in the denominator up to the numerator using a negative exponent.<\/p>\n<p>We have,<\/p>\n<div id=\"fs-id1170572587738\" class=\"equation unnumbered\" style=\"text-align: center;\">[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 id=\"fs-id1170571636291\">Let [latex]u={x}^{-1},[\/latex] the exponent on [latex]e[\/latex].<\/p>\n<p>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 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 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 [latex]\u22121[\/latex]and interchange the limits.<\/p>\n<p>Thus,<\/p>\n<div id=\"fs-id1170572293466\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\\\ \\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<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288436\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288436&theme=lumen&iframe_resize_id=ohm288436&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.6.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":399,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"5.6.1","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/408"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/408\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/408\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=408"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=408"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=408"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=408"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}