When you finish this section, you should be able to:
Indefinite integrals that cannot be found using the formulas in Table # on page 380 sometimes can be found using the method of substitution. In the method of substitution, we use a change of variables to transform the integrand so one of the formulas in the table applies.
Differentials are discussed in Section 3.4, pp. xx-xx.
When using the method of substitution, once the substitution \(u\)is chosen, we write the original integral in terms of \(u\) and\(du\).
For example, to find \(\int {(x^{2}+5)}^{3}2x\,dx,\) we use thesubstitution \(u\,=x^{2} + 5\). The differential of \(u=x^{2} + 5\) is \(du =2x\,dx.\) Now we write \(( x^{2}+5) ^{3}2x\,dx\) in terms of \(u\) and \(du\), andintegrate the simpler integral.\[\begin{eqnarray*}\int(\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{x^{2}+5})^{3}}\underset{\color{#0066A7}{\hbox{\(du\)}}}{\underbrace{2x\,dx}}=\intu^{3}du=\dfrac{u^{4}}{4}+C \underset{\underset{{\color{#0066A7}{\hbox{\(u=x^{2}{+5 }\)}}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{(x^{2}+5)^{4}}{4}+C \\\end{eqnarray*}\]
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We can verify the answer by differentiating using the Power Rule for Functions. \[{\dfrac{{d}}{{dx}}}{\left[ {{\dfrac{{({x^{2}+5})^{4}}}{{4}}}+C}\right] }={\dfrac{{1}}{{4}}}\left[ {4(x^{2}+5)^{3}(2x)}\right] =(x^{2}+5)^{3}2x\]
The Chain Rule is discussed in Section 3.1, pp. xx-xx.
The method of substitution is based on the Chain Rule, which statesthat if \(f\) and \(g\) are differentiable functions, then for the compositefunction \(f\circ g\), \[\dfrac{d}{dx}\left( f\circ g\right) ={\dfrac{{d}}{{dx}}}f(g(x))=f^\prime (g(x))\, g^\prime (x)\]
The Chain Rule provides a template for finding integrals of the form \[\int f^\prime (g(x))g^\prime (x)\,dx\]
If, in the integral, we let \(u=g(x)\), then the differential \(du=g^\prime(x)\,dx\), and we have \[\int f^\prime(\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{g(x)}})\underset{\color{#0066A7}{{\hbox{\(du\)}}}}{\underbrace{g^{\prime }(x)\,dx}} =\int f^\prime (u)\,du=f(u)+C=f(g(x))+C\]
Replacing \(g(x)\) by \(u\) and \(g^\prime (x) dx\) by \(du\) iscalled substitution. Substitution is a strategy forfinding antiderivatives when the integrand is a compositefunction.
Find:
Solution (a) Since we know \(\int\sin x\,dx,\) we let \(u = 3x + 2\). Then \(du = 3\,dx\) so \(dx = \dfrac{du}{3}\).\[\begin{eqnarray*}\int \sin (\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{(3x+2)}}\enspace\underset{\color{#0066A7}{\tfrac{du}{2}}}{\underbrace{dx}}&=& \int {\sin }\,u\dfrac{du}{3}=\dfrac{1}{3}\int {\sin }\,u\,du\\ &=&\dfrac{1}{3}({-}\cos u)+C \underset{\underset{{\color{#0066A7}{\hbox{\(u=3 x+2\)}}}}{\color{#0066A7}{\uparrow }}}{=} -\dfrac{1}{3}{\cos }(3x+2)+C \end{eqnarray*}\]
(b) We let \(u = x^{2} + 1\). Then \(du = 2x\,dx\), so \(x\,dx = \dfrac{du}{2}\). \[\begin{eqnarray*}\int x\sqrt{x^{2}+1}\,dx &=& \int\underset{\color{#0066A7}{\hbox{\(u\)}}}{{\underbrace{\sqrt{x^{2}+1}}}\,}\underset{\color{#0066A7}{\hbox{\(\tfrac{du}{2}\)}}}{\underbrace{x\,dx}}=\int\sqrt{u}\,\dfrac{du}{2}=\dfrac{1}{2}\int u^{1/2}du=\dfrac{1}{2}\left(\dfrac{u^{3/2}}{\dfrac{3}{2}}\right) +C \\&= & \dfrac{(x^{2}+1) ^{3/2}}{3} +C\end{eqnarray*}\]
(c) We let \(u=\sqrt{x}=x^{1/2}\). Then \(du=\dfrac{1}{2}x^{-1/2}dx=\dfrac{dx}{2\sqrt{x}}\), so \(\dfrac{dx}{\sqrt{x}}\) \(=2du\). \[\int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx=\int e^{\sqrt{x}}\cdot \dfrac{dx}{\sqrt{x}}=\int e^{u}\cdot 2du=2e^{u}+C=2e^{\sqrt{{ x}}}+C \]
Problems 5 and 11.
When an integrand equals the product of an expression involving a function and its derivative (or a multiple of its derivative), then substitution isoften a good strategy. For example, for \(\int \dfrac{e^{\sqrt{x}}}{\sqrt{x}}dx,\) we used the substitution \(u=\sqrt{x}\), since \(\dfrac{du}{dx}=\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}\) is a multiple of \(\dfrac{1}{\sqrt{x}}.\)
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Similarly, in (b) the factor \(x\) in the integrand makes thesubstitution \(u = x^{2} + 1\) work. On the other hand, if we try to use this samesubstitution to integrate \(\int \sqrt{x^{2}\,{+}\,1}\,dx\), then \[\begin{eqnarray*}\int \sqrt{x^{2}+1}\,dx=\int \sqrt{u} {\dfrac{{du}}{{2x}}} \underset{\underset{{\color{#0066A7}{\hbox{\(x=\sqrt{{u-1}}\)}}}}{\color{#0066A7}{\uparrow }}}{=}\int {{\dfrac{\sqrt{u}}{{2\sqrt{u-1}}}}\,du}\end{eqnarray*}\] and the resulting integral is more complicated than the original integral.
The idea behind substitution is to obtain an integral \(\int h(u)\,du\) thatis simpler than the original integral \(\int {f(x)\,dx.}\) When a substitutiondoes not simplify the integral, try other substitutions. If none of thesework, other integration methods should be applied. Some of these methods areexplored in Chapter 7.
Find:
Solution (a) Notice that the numerator equals the derivative ofthe denominator, except for a constant factor. So, we trysubstitution. Let \(\ u=4x^{3}-1.\) Then \(du=12x^{2}dx\) so\(5x^{2}dx=\dfrac{5}{12}\,du\). \[\int \dfrac{5x^{2}dx}{4x^{3}-1}=\int \dfrac{\dfrac{5}{12}\,du}{u}=\dfrac{5}{12}\int \dfrac{du}{u}=\dfrac{5}{12}\ln \left\vert u\right\vert +C=\dfrac{5}{12}\ln \left\vert 4x^{3}-1\right\vert +C\]
(b) Here, the numerator equals the derivative of thedenominator. So, we use the substitution \(u=e^{x}+4\). Then\(du=e^{x}dx\).\[\begin{eqnarray*}\int \dfrac{e^{x}}{e^{x}+4}dx=\int \dfrac{1}{e^{x}+4}\cdot e^{x}dx=\int \dfrac{1}{u}\,du=\ln \vert u\vert +C \underset{\underset{{\color{#0066A7}{\hbox{\(u=e^{x}+4>0\)}}}}{\color{#0066A7}{\uparrow}}}{=} \ln (e^{x}+4) +C \\\end{eqnarray*}\]
Problem 17.
Show that:
Solution (a) Since \(\tanx=\dfrac{\sin x}{\cos x},\) we let \(u=\cos x.\) Then\(du=-\sin x\, dx\) and\[\begin{eqnarray*}\int \tan x dx = \int \dfrac{\sin x}{\cos x}dx=\int -\dfrac{du}{u}=-\ln\left\vert u\right\vert +C=-\ln \left\vert \cos x\right\vert +C \\\underset{\underset{\color{#0066A7}{\hbox{\(r\ln x=\ln x^{r}\)}}}{\color{#0066A7}{\uparrow}}} {=} \ln \left\vert \cos x\right\vert^{-1}+C =\ln \left\vert \dfrac{1}{\cos x}\right\vert + C =\ln \left\vert \secx\right\vert +C\end{eqnarray*}\]
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(b) To find \(\int \sec x dx\), we multiply the integrand by \(\dfrac{\sec x+\tan x}{\sec x+\tan x}.\)\[\int \sec x dx=\int \sec x\cdot \dfrac{\sec x+\tan x}{\sec x+\tan x}dx=\int \dfrac{\sec ^{2}x+\sec x\tan x}{\sec x+\tan x} dx \]
Now the numerator equals the derivative of the denominator. So if\(u=\sec x+\tan x,\) then \(du=( \sec x\tan x+\sec ^{2}x) dx.\)\[\int \sec x dx=\int \dfrac{du}{u}=\ln \left\vert u\right\vert +C=\ln \left\vert \sec x+\tan x\right\vert +C \]
Problem 27.
Examples 2 and 3 illustrate a basic integration formula: \[\boxed{\bbox[#FAF8ED,5pt]{\int \dfrac{g^\prime (x) }{g(x) }dx=\ln \left\vert g(x) \right\vert +C }}\tag {1}\]
If the numerator of the integrand equals the derivative of thedenominator, then the integral equals a logarithmic function.
Notice that in formula (1) the integral equals the natural logarithm of the absolute value of the function \(g\). The absolute value is necessarysince the domain of the logarithm function is the set of positive realnumbers. When \(g\) is known to be positive, as in Example 2(b), the absolutevalue is not required.
As we saw in Example 3(b), sometimes algebra is needed to transform anintegral so that a basic integration formula can be used. Unlikedifferentiation, integration has no prescribed method; some ingenuity and alot of practice are required. To illustrate, two different substitutions areused to solve Example 4.
Find \(\int {x\sqrt{4+x}}\,dx\).
Solution Substitution I Let \(u =4 + x\). Then \(du = dx.\) Since \(u=4+x,\) \(x=u-4.\) Substituting gives \[\begin{eqnarray*}\int {x\sqrt{4+x}}\,dx \underset{\underset{{\color{#0066A7}{\hbox{\(u=x+4\)}}}}{\color{#0066A7}{\uparrow }}} =&& \int \underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{( u-4) }} \underset{\underset{{\color{#0066A7}{\hbox{\(4+x\)}}}}{\color{#0066A7}{\uparrow }}}{{\sqrt{u}}}\,\,\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{du}}\, =\int {({u^{3/2}-4u^{1/2}})}\,du \\&=& {\dfrac{{u^{5/2}}}{{{\dfrac{{5}}{{2}}}}}}-4\cdot {\dfrac{{u^{3/2}}}{{{\dfrac{{3}}{{2}}}}}}+C \\&=&{\dfrac{{2({4+x})^{5/2}}}{{5}}}-{\dfrac{{8({4+x})^{3/2}}}{{3}}}+C\end{eqnarray*}\]
Substitution II Let \(u = \sqrt{4\,{+}\,x}\), so \(u^{2} = 4 + x\) and \(x = u^{2} - 4\). Then \(dx=2u\,du\) and \[\begin{eqnarray*}\int {x\sqrt{4+x}}\,dx & = &\int{(}\underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{{u^{2}-4}}}{)(u)(}\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{{2u\,du}}}{)= 2 \int {(u^{4}-4u^{2})\,du}}=2\left[ {\dfrac{{u^{5}}}{{5}}}-{\dfrac{{4u^{3}}}{{3}}}\right] +C \\&=&\dfrac{2}{5} \big(\sqrt{4+x}\,\big) ^{5} -\dfrac{8}{3} \big(\sqrt{4+x}\,\big) ^{3}+C= \dfrac{2(4+x)^{5/2}}{5} -\dfrac{8(4+x)^{3/2}}{3}+C \end{eqnarray*}\]
Problem 35.
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Find:
Solution (a) \(\int \dfrac{dx}{\sqrt{4-x^{2}}}\) resembles \(\int \dfrac{1}{\sqrt{1-x^{2}}}\,dx=\) \(\sin ^{-1}x+C.\) We begin by rewritingthe integrand as\[\dfrac{1}{\sqrt{4-x^{2}}}=\dfrac{1}{\sqrt{4\left( 1-\dfrac{x^{2}}{4}\right) }}=\dfrac{1}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}}\]
Now we let \(u=\dfrac{x}{2}\). Then \(du=\dfrac{dx}{2}\), so \(dx=2\,du.\) \[\begin{eqnarray*}\int \dfrac{dx}{\sqrt{4-x^{2}}}&=&\int \dfrac{dx}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}} \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(dx=2du\)}}}}{\color{#0066A7}{\hbox{\(u=\tfrac{x}{2}\)}}}}}{\color{#0066A7}{\uparrow}}} = \int \dfrac{2 du}{2\sqrt{1-u^{2}}}=\int \dfrac{du}{\sqrt{1-u^{2}}}=\sin ^{-1}u+C \\&=&\sin ^{-1}\left( \dfrac{x}{2}\right) +C \end{eqnarray*}\]
(b) \(\int \dfrac{dx}{9+4x^{2}}\) resembles \(\int \dfrac{1}{1+x^{2}}dx=\tan ^{-1}x+C\). We rewrite the integrand as\[\dfrac{1}{9+4x^{2}}=\dfrac{1}{9\left( 1+\dfrac{4x^{2}}{9}\right) }=\dfrac{1}{9\left[ 1+\left( \dfrac{2x}{3}\right) ^{2}\right] }\]
Now let \(u=\dfrac{2x}{3}.\) Then \(du=\dfrac{2}{3}dx\), so \(dx=\dfrac{3}{2} du\).\[\begin{eqnarray*}\int \dfrac{dx}{9+4x^{2}}&=&\int \dfrac{dx}{9\left[ 1+\left( \dfrac{2x}{3}\right) ^{2}\right] }=\int \dfrac{\dfrac{3}{2}du}{9( 1+u^{2}) }=\dfrac{1}{6}\int \dfrac{du}{1+u^{2}} \\&=& \dfrac{1}{6}\tan ^{-1}u+C=\dfrac{1}{6}\tan ^{-1}\left( \dfrac{2x}{3}\right) +C \end{eqnarray*}\]
Problem 39.
Two approaches can be used to find a definite integral using substitution:
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Find \(\int_{0}^{2}x\sqrt{4-x^{2}}\,dx\).
Solution
Method 1: Use the related indefinite integral and thenapply the Fundamental Theorem of Calculus. The relatedindefinite integral \(\int {x\sqrt{4-x^{2}}\,dx}\) can be found using thesubstitution \(u = 4 - x^{2}\). Then \(du\,=-2x\,dx\), so \(x\,dx=-\dfrac{du}{2}.\) \[\begin{eqnarray*}\int {x\sqrt{4-x^{2}}\,dx}&=&\int {\sqrt{u}\left( {-}{\dfrac{{du}}{{2}}}\right) }={-}{\dfrac{{1}}{{2}}}\int u^{1/2}du=-\dfrac{1}{2}\cdot {{\dfrac{{u^{3/2}}}{{{\dfrac{{3}}{{2}}}}}}}+C \\&=&-\dfrac{1}{3}{{(4-x^{2})^{3/2}}}+C\end{eqnarray*}\]
Then by the Fundamental Theorem of Calculus, \[\int_{0}^{2} x\sqrt{4-x^{2}}\,dx = -\dfrac{1}{3} \Big[(4-x^{2})^{3/2}\Big]_{0}^{2} = -\dfrac{1}{3}\big[0-4^{3/2}\big] =\dfrac{8}{3}\]
Method 2: Find the definite integral directly by making asubstitution in the integrand and changing the limits of integration. We let \(u=4-x^{2}\); then \(du=-2x\,dx.\) Now use the function\(u=4-x^{2}\) to change the limits of integration.
Then \[\begin{eqnarray*}{\int_{0}^{2}{x\sqrt{4-x^{2}}\,dx}} && \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(x dx=-\tfrac{1}{2}du\)}}}}{{\color{#0066A7}{\hbox{\(u=4-x^{2} \)}}}}}}{\color{#0066A7}{\uparrow }}}{=} \int_{4}^{0}\sqrt{u}\left( -\dfrac{du}{2}\right) =-\dfrac{1}{2}\int_{4}^{0}\sqrt{u} du=-\dfrac{1}{2}\cdot \left[ \dfrac{u^{3/2}}{\dfrac{3}{2}}\right]_{4}^{0} \\{\;\;} &&= -{\dfrac{1}{3}}(0-8)={\dfrac{{8}}{{3}}}\end{eqnarray*}\]
When using substitution to find a definite integral directly, remember to change the limits of integration.
Problem 45.
Find \(\int_{0}^{\pi /2}\dfrac{{1-\cos}(2\theta)}{2}\,d\theta \).
Solution We use properties of integrals to simplify before integrating. \[\begin{eqnarray*}\int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}\,d\theta &=&\dfrac{1}{2}\int_{0}^{\pi /2}\left[ 1-\cos (2\theta ) \right] \,d\theta \\&=& \dfrac{1}{2}\left[ \int_{0}^{\pi /2}d\theta -\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \right]\\& =& \dfrac{1}{2}\int_{0}^{\pi /2}d\theta -\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\&=& \dfrac{1}{2} \Big[\theta\Big] _{0}^{\pi /2}-\dfrac{1}{2}\int_{0}^{\pi/2}\cos (2\theta ) \,d\theta \\&=& \dfrac{\pi }{4}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta )\,d\theta\end{eqnarray*}\]
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In the integral on the right, we use the substitution \(u = 2\theta \). Then \(du = 2\,d\theta \) so \(d\theta =\dfrac{du}{2}\). Now we change the limits of integration:
Now \[\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta =\int_{0}^{\pi }{\cos u\,\dfrac{{du}}{{2}}}= \dfrac{1}{2}\Big[\sin u\Big] _{0}^{\pi }=\dfrac{1}{2}\left( \sin \pi -\sin 0\right) =0\]
Then,\[\int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}d\theta =\dfrac{\pi}{4}-\dfrac{1}{2}\int_{0}^{\pi /2}{\cos }(2\theta ) {\,d\theta }=\dfrac{\pi }{4} \]
Problem 53.
Integrals of even and odd functions can be simplified due to symmetry. Figure # illustrates the conclusions of the theorem that follows.
Even and odd functions are discussed in Section P.1, pp. xx-xx.
Let a function \(f\) be continuous on a closed interval\([-a,a]\), \(a>\,0 \).
The property for even functions is proved here; the proof for odd functions is left as an exercise. See Problem 123.
\(f\) is an even function: Since \(f\)is continuous on the closed interval \([ -a,\,a] ,\) \(a>0,\) and\(0\) is in the interval \(\left[ -a,a\right] \), we have \[\begin{equation*}\int_{-a}^{a} f(x)\,dx=\int_{-a}^{0} f(x)\,dx+\int_{0}^{a}f(x)\,dx=-\int_{0}^{-a}f(x)\,dx+\int_{0}^{a}f(x)\,dx \tag {2}\end{equation*}\]
In \(- {\int_{0}^{-a}{f(x)\,dx}}\), we use the substitution \(u = -x\). Then \(du = -dx\). Also, if \(x=0\), then \(u=0\), and if \(x=-a\), then \(u = a\). Therefore,\[\begin{eqnarray*}-{\int_{0}^{-a}{f(x)\,dx={\int_{0}^{a}{f({-}u)\,du}}}} \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(f(-u) =f(u) \)}}}}{\color{#0066A7}{\hbox{\(f\) is even}}}}}{\color{#0066A7}{\uparrow }}}= \int_{0}^{a}f(u) du=\int_{0}^{a}f(x)\, dx \tag{3}\end{eqnarray*}\]
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Combining (2) and (3), we obtain \[{\int_{{-}a}^{a}{f(x)\,dx={\int_{0}^{a}{f(x)\,dx}}}}+{\int_{0}^{a}{f(x)\,dx}}=2{\int_{0}^{a}{f(x)\,dx}} \]
To use the theorem involving even and odd functions, three conditions must be met:
Find:
Solution (a) If\(f(x)={{x^{7}-4x^{3}+x,}}\) then\(f({-}x)=({-}x)^{7}-4({-}x)^{3}+({-}x)={-}(x^{7}-4x^{3}+x)={-}f(x).\)Since \(f\) is an odd function, \[\int_{-3}^{3}{(x^{7}-4x^{3}+x)}\,dx=0\]
(b) If \(g(x) =x^{4}-x^{2}+3,\) then \(g(-x) =(-x) ^{4}-(-x) ^{2}+3=x^{4}-x^{2}+3=g(x).\) Since \(g\) is an even function, \[\begin{eqnarray*}\int_{-2}^{2}( x^{4}-x^{2}+3) dx &=& 2\int_{0}^{2}(x^{4}-x^{2}+3)\, dx=2\left[ \dfrac{x^{5}}{5}-\dfrac{x^{3}}{3}+3x\right]_{0}^{2} \\&=& 2\left[ \dfrac{32}{5}-\dfrac{8}{3}+6\right] =\dfrac{292}{15} \end{eqnarray*}\]
Problems 63 and 67.
If \(f\) is an even function and \(\int_{0}^{2}f(x)\, dx=-6\) and \(\int_{-5}^{0}f(x)\, dx=8,\) find \(\int_{2}^{5}f(x)\, dx.\)
Solution \(\int_{2}^{5}f(x)\, dx=\int_{2}^{0}f(x)\, dx+\int_{0}^{5}f(x)\, dx\)
Now \(\int_{2}^{0}f(x)\, dx=-\int_{0}^{2}f(x)\, dx=6.\)
Since \(f\) is even, \(\int_{0}^{5}f(x)\, dx=\) \(\int_{-5}^{0}f(x)\, dx \) \(\,=8.\) Then \[\int_{2}^{5}f(x)\, dx=\int_{2}^{0}f(x)\, dx+\int_{0}^{5}f(x)\, dx=6+8=14 \]
Suppose an object is heated to a temperature \(u_{0}.\) Then at time \(t=0\), the object is put into a medium with a constant lower temperature causing the object to cool. Newton's Law of Cooling states that the rate of change of the temperature of the object with respect to time is continuous and proportional to the difference between the temperature of the object and the ambient temperature (the temperature of the surrounding medium). That is, if \(u=u(t) \) is the temperature of the object at time \(t\) and if \(T\) is the (constant) ambient temperature, then Newton's Law of Cooling is modeled by the differential equation \[\boxed{\bbox[#FAF8ED,5pt]{\dfrac{du}{dt}=k [ u(t) -T ] }}\tag {4}\]
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where \(k\) is a constant that depends on the object. Since the ambienttemperature \(T\) is lower than \(u(0) =u_{0},\) the object coolsand its temperature decreases so that \(\dfrac{du}{dt}\lt0\). Then,since \(u(t) >T\), \(k\) is a negative constant.
We find \(u\) as a function of \(t\) by solving the differential equation \(\dfrac{du}{dt}=k\left( u-T\right) \). We rewrite the differential equation as \(\dfrac{du}{u-T}=k\, dt\) and integrate both sides.\[\begin{eqnarray*}\int \dfrac{du}{u-T} &=& \int k\,dt \\\ln \vert u-T \vert &=& k\,t+C\end{eqnarray*}\]
To find \(C,\) we use the boundary condition that at time\(t=0\), the initial temperature of the object is \(u(0) =u_{0}\), Then\[\begin{eqnarray*}\ln \left\vert u_{0}-T\right\vert &=&k\cdot 0+C \\C &=&\ln \left\vert u_{0}-T\right\vert\end{eqnarray*}\]
Using this expression for \(C,\) we obtain \[\begin{eqnarray*}\ln \left\vert u-T\right\vert &=&k\,t+\ln \vert u_{0}-T\vert \\\ln \left\vert u-T\right\vert -\ln \vert u_{0}-T\vert &=&k\,t \\\ln \left\vert \dfrac{u-T}{u_{0}-T}\right\vert &=&k\,t \\\dfrac{u-T}{u_{0}-T} &=&e^{kt} \\u-T &=&( u_{0}-T) e^{kt} \\\end{eqnarray*}\]\[\boxed{\bbox[#FAF8ED,5pt]{u= (u_{0}-T ) \,e^{kt}+T }}\tag{5}\]
An object is heated to \(90{}^{\circ}{\rm C}\) and allowed to cool in a room with a constant ambient temperature of \(20{}^{\circ}{\rm C}\). If after \(10\) min the temperature of the object is \(60{}^{\circ}{\rm C}\), what will its temperature be after \(20\) min?
Solution When \(t=0\), \(u(0) =u_{0}=90{}^{\circ}{\rm C}\), and when \(t=10\) min, \(u\left( 10\right) =60{}^{\circ}{\rm C}\). Given that the ambient temperature \(T\) is \(20{}^{\circ}{\rm C}\), we substitute these values into equation (5).\[\begin{eqnarray*}\begin{array}{rcl@{\qquad}l@{\hspace*{-6pc}}}u(t) &=&( u_{0}-T) e^{kt}+T \\60 &=&( 90-20) e^{10k}+20 & {\color{#0066A7}{\hbox{\(u=60 \hbox{ when}t=10; \ T=20;\ u_{0} =90\)}}} \\\dfrac{40}{70} &=&e^{10k} \\k &=&\dfrac{1}{10}\ln \dfrac{4}{7} = 0.1 0 \ln \dfrac{4}{7}\end{array}\end{eqnarray*}\]
The temperature \(u\) is \[u(t) =70 e^{[0.1 \ln (4/7)]t} +20\]
Then when \(t=20\), the temperature \(u\) of the object is \[u (20) =70e ^{[0.1 \ln (4/7)] (20)} +20=70e^{2\ln (4/7)}+20\approx 42.86{}^{\circ}{\rm C} \]
Problem 109.
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