5.6 Method of Substitution; Newton's Law of Cooling

OBJECTIVES

When you finish this section, you should be able to:

  1. Find an indefinite integral using substitution (p. 387)
  2. Find a definite integral using substitution (p. 391)
  3. Integrate even and odd functions (p. 393)
  4. Solve differential equations: Newton's Law of Cooling (p. 394)

1 Find an Indefinite Integral Using Substitution

Indefinite integrals that cannot be found using the formulas in Table 1 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.

NEED TO REVIEW?

Differentials are discussed in Section 3.4, pp. xx-xx.

For example, to find \(\int {(x^{2}+5)}^{3}2x\,dx,\) we use the substitution \(u\,=x^{2} + 5\). 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 \(du\), and integrate the simpler integral.

NOTE

When using the method of substitution, once the substitution \(u\) is chosen, we write the original integral in terms of \(u\) and \(du\).

\[ \begin{eqnarray*} \int (\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{x^{2}+5})^{3}}\underset{\color{#0066A7}{\hbox{\(du\)}}}{\underbrace{2x\,dx}}=\int u^{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. \[ {\dfrac{{d}}{{dx}}}{\left[ {{\dfrac{{({x^{2}+5})^{4}}}{{4}}}+C}\right] }={\dfrac{{1}}{{4}}}\left[ {4(x^{2}+5)^{3}(2x)}\right] =(x^{2}+5)^{3}2x \]

NEED TO REVIEW?

The Chain Rule is discussed in Section 3.1, pp. xx-xx.

The method of substitution is based on the Chain Rule, which states that if \(f\) and \(g\) are differentiable functions, then for the composite function \(f\circ g\), \[ \dfrac{d}{dx}\left( f\circ g\right) ={\dfrac{{d}}{{dx}}}f(g(x))=f^\prime (g(x))\, g^\prime (x) \]

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 \(du\) is called substitution. Substitution is a strategy for finding antiderivatives when the integrand is a composite function.

Finding Indefinite Integrals Using Substitution

Find:

  1. \(\int \sin (3x+2)\,dx \)
  2. \(\int x\sqrt{x^{2}+1}\,dx\)
  3. \(\int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx\)

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 \(\dfrac{dx}{\sqrt{x}}\) \(=2du\). \[ \int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx=\int e^{\sqrt{x}}\cdot \dfrac{dx}{\sqrt{x}}=\int e^{u}\cdot 2du=2e^{u}+C=2e^{\sqrt{{ x}}}+C \]

NOW WORK

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 the substitution \(u = x^{2} + 1\) work. On the other hand, if we try to use this same substitution 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 \(\int h(u)\,du\) that is simpler than the original integral \(\int {f(x)\,dx.}\) When a substitution does not simplify the integral, try other substitutions. If none of these work, other integration methods should be applied. Some of these methods are explored in Chapter 7.

Finding Indefinite Integrals Using Substitution

Find:

  1. \(\int \dfrac{5x^{2}dx}{4x^{3}-1}\)
  2. \( \int \dfrac{e^{x}}{e^{x}+4}dx\)

Solution (a) Notice that the numerator equals the derivative of the denominator, except for a constant factor. So, we try substitution. 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 the denominator. 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*} \]

NOW WORK

Problem 17.

Using Substitution To Establish an Integration Formula

Show that:

  1. \[ \boxed{\bbox[#FAF8ED,5pt]{ \int \tan x\,dx=-\ln \left\vert \cos x\right\vert +C=\ln \vert \sec x\vert +C }} \]
  2. \[ \boxed{\bbox[#FAF8ED,5pt]{ \int \sec x\,dx=\ln \vert \sec x+\tan x\vert +C }} \]

Solution (a) Since \(\tan x=\dfrac{\sin x}{\cos x},\) 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 \sec x\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 \(du=( \sec x\tan x+\sec ^{2}x) dx.\) \[ \int \sec x dx=\int \dfrac{du}{u}=\ln \left\vert u\right\vert +C=\ln \left\vert \sec x+\tan x\right\vert +C \]

NOW WORK

Problem 27.

Examples 2 and 3 illustrate a basic integration formula:

IN WORDS

If the numerator of the integrand equals the derivative of the denominator, then the integral equals a logarithmic function.

\[ \boxed{\bbox[#FAF8ED,5pt]{ \int \dfrac{g^\prime (x) }{g(x) }dx=\ln \left\vert g(x) \right\vert +C }}\tag {1} \]

Notice that in formula (1) the integral equals the natural logarithm of the absolute value of the function \(g\). The absolute value is necessary since the domain of the logarithm function is the set of positive real numbers. When \(g\) is known to be positive, as in Example 2(b), the absolute value is not required.

As we saw in Example 3(b), sometimes algebra is needed to transform an integral so that a basic integration formula can be used. Unlike differentiation, integration has no prescribed method; some ingenuity and a lot of practice are required. To illustrate, two different substitutions are used to solve Example 4.

Finding an Indefinite Integral Using Substitution

Find \(\int {x\sqrt{4+x}}\,dx\).

SolutionSubstitution 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*} \]

NOW WORK

Problem 35.

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Finding Indefinite Integrals Using Substitution

Find:

  1. \(\int \dfrac{dx}{\sqrt{4-x^{2}}}\)
  2. \(\int \dfrac{dx}{9+4x^{2}}\)

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 rewriting the 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 \(dx=\dfrac{3}{2} du\). \[ \begin{eqnarray*} \int \dfrac{dx}{9+4x^{2}}&=&\int \dfrac{dx}{9\left[ 1+\left( \dfrac{2x}{3} \right) ^{2}\right] }=\int \dfrac{\dfrac{3}{2}du}{9( 1+u^{2}) }=\dfrac{1}{6}\int \dfrac{du}{1+u^{2}} \\ &=& \dfrac{1}{6}\tan ^{-1}u+C=\dfrac{1}{6}\tan ^{-1}\left( \dfrac{2x}{3}\right) +C \end{eqnarray*} \]

NOW WORK

Problem 39.

2 Find a Definite Integral Using Substitution

Two approaches can be used to find a definite integral using substitution:

  • Method 1: Find the related indefinite integral using substitution, and then use the Fundamental Theorem of Calculus.
  • Method 2: Find the definite integral directly by making a substitution in the integrand and using the substitution to change the limits of integration.

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Finding a Definite Integral Using Substitution

Find \(\int_{0}^{2}x\sqrt{4-x^{2}}\,dx\).

Solution

Method 1: Use the related indefinite integral and then apply the Fundamental Theorem of Calculus. The related indefinite integral \(\int {x\sqrt{4-x^{2}}\,dx}\) can be found using the substitution \(u = 4 - x^{2}\). 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 a substitution 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.

  • The lower limit of integration is \(x=0\) so, in terms of \(u,\) it becomes \(u=4-0^{2}=4\).
  • The upper limit of integration is \(x=2\) so the upper limit becomes \(u=4-2^{2}=0\).

Then

CAUTION

When using substitution to find a definite integral directly, remember to change the limits of integration.

\[ \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*} \]

NOW WORK

Problem 45.

Finding a Definite Integral Using Substitution

Find \(\int_{0}^{\pi /2}\dfrac{{1-\cos}(2\theta)}{2}\,d\theta \).

Solution We use properties of integrals to simplify before integrating. \[ \begin{eqnarray*} \int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}\,d\theta &=&\dfrac{1}{2}\int_{0}^{\pi /2}\left[ 1-\cos (2\theta ) \right] \,d\theta \\ &=& \dfrac{1}{2}\left[ \int_{0}^{\pi /2}d\theta -\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \right]\\ & =& \dfrac{1}{2}\int_{0}^{\pi /2}d\theta -\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{1}{2} \Big[\theta\Big] _{0}^{\pi /2}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{\pi }{4}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \end{eqnarray*} \]

393

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:

  • when \( \theta = 0,\)   then \( u=2(0) =0 \)
  • when \( \theta =\dfrac{\pi }{2},\)   then \( u=2\left( \dfrac{\pi }{2}\right) =\pi \)

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, \[ \int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}d\theta =\dfrac{\pi}{4}-\dfrac{1}{2}\int_{0}^{\pi /2}{\cos }(2\theta ) {\,d\theta }=\dfrac{\pi }{4} \]

NOW WORK

Problem 53.

3 Integrate Even and Odd Functions

NEED TO REVIEW?

Even and odd functions are discussed in Section P.1, pp. xx-xx.

Integrals of even and odd functions can be simplified due to symmetry. Figure 32 illustrates the conclusions of the theorem that follows.

Figure 32

spanTHEOREMspan The Integrals of Even and Odd Functions

Let a function \(f\) be continuous on a closed interval \([-a,a]\), \(a>\,0 \).

  • If \(f\) is an even function, then \[ \boxed{\bbox[5pt]{ \int_{-a}^{a}f(x)\,dx=2\int_{{0}}^{a}f(x)\,dx }} \]
  • If \(f\) is an odd function, then \[ \boxed{\bbox[5pt]{ \int_{-a}^{a} f(x)\,dx=0 }} \]

The property for even functions is proved here; the proof for odd functions is left as an exercise. See Problem 123.

Proof

\(f\) 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:

Integrating an Even or Odd Function

Find:

  1. \(\int_{-3}^{3}{(x^{7}-4x^{3}+x)}\,dx\)
  2. \(\int_{-2}^{2}(x^{4}-x^{2}+3)\, dx\)

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, \[ \begin{eqnarray*} \int_{-2}^{2}( x^{4}-x^{2}+3) dx &=& 2\int_{0}^{2}(x^{4}-x^{2}+3)\, dx=2\left[ \dfrac{x^{5}}{5}-\dfrac{x^{3}}{3}+3x\right]_{0}^{2} \\ &=& 2\left[ \dfrac{32}{5}-\dfrac{8}{3}+6\right] =\dfrac{292}{15} \end{eqnarray*} \]

NOW WORK

Problems 63 and 67.

Using Properties of Integrals

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 \[ \int_{2}^{5}f(x)\, dx=\int_{2}^{0}f(x)\, dx+\int_{0}^{5}f(x)\, dx=6+8=14 \]

4 Solve Differential Equations: Newton's Law of Cooling

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 \[\boxed{\bbox[#FAF8ED,5pt]{ \dfrac{du}{dt}=k [ u(t) -T ] }}\tag {4} \]

395

where \(k\) is a constant that depends on the object. Since the ambient temperature \(T\) is lower than \(u(0) =u_{0},\) the object cools and 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 \[ \begin{eqnarray*} \ln \left\vert u-T\right\vert &=&k\,t+\ln \vert u_{0}-T\vert \\ \ln \left\vert u-T\right\vert -\ln \vert u_{0}-T\vert &=&k\,t \\ \ln \left\vert \dfrac{u-T}{u_{0}-T}\right\vert &=&k\,t \\ \dfrac{u-T}{u_{0}-T} &=&e^{kt} \\ u-T &=&( u_{0}-T) e^{kt} \\ \end{eqnarray*} \] \[ \boxed{\bbox[#FAF8ED,5pt]{ u= (u_{0}-T ) \,e^{kt}+T }}\tag{5} \]

Using Newton's Law of Cooling

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 \[ u (20) =70e ^{[0.1 \ln (4/7)] (20)} +20=70e^{2\ln (4/7)}+20\approx 42.86{}^{\circ}{\rm C} \]

NOW WORK

Problem 109.

396