Improper Integrals - AP Calculus BC

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Question

Evaluate:

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

Answer

First, we will find the indefinite integral $\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$ using integration by parts.

We will let $u = \cos x$ and $\mathrm{d}v= e^{-x}; \mathrm{d}x$ .

Then $\frac{\mathrm{d} u}{\mathrm{d} x} = - \sin x$ and $\mathrm{d} u = - \sin x ; \mathrm{d} x$ .

$v= \int e^{-x}; \mathrm{d}x = - e^{-x} = - \frac{1}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \cos x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \left (- \sin x \right ); \mathrm{d} x$

$= - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

To find $\int e^{-x} \sin x ; \mathrm{d} x$ , we use another integration by parts:

$u = \sin x$ , which means that $\mathrm{d} u = \cos x ; \mathrm{d} x$ , and

$\mathrm{d}v= e^{-x}; \mathrm{d}x$ , which means that, again, $v= - \frac{1}{e^{x} }$ .

$\int e^{-x} \sin x ; \mathrm{d} x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \sin x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \cos x ; \mathrm{d} x$

$= - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \left ( - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} } \right )$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} } - \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{2e^{x} }$

Since

$\frac{-2}{2e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{2}{2e^{x} }$ , or,

$- \frac{1}{e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{1}{e^{x} }$

for all real , and

$- \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = 0$ ,

by the Squeeze Theorem,

$\lim_{x\rightarrow \infty } \frac{\sin x -\cos x}{2e^{x} } = 0$ .

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{b\rightarrow \infty }\int_{0}^{b } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{ b\rightarrow \infty }$ $\left.\begin{matrix} \frac{\sin x -\cos x}{2e^{x} } \ \textrm{ } \end{matrix}\right|$ $\begin{matrix} b\ \ 0 \end{matrix}$

$= 0 - \frac{\sin 0 -\cos 0}{2e^{0} }$

$= - \frac{ 0 -1}{2 \cdot 1 } = \frac{1}{2}$

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Question

Evaluate:

$\int_{0}^{1} x \ln x ; \mathrm{d}x$

Answer

First, we will find the indefinite integral, $\int x \ln x ; \mathrm{d}x$ .

We will let $u = \ln x$ and $\mathrm{d}v= x ; \mathrm{d}x$ .

Then,

$\frac{\mathrm{d} u }{\mathrm{d} x}= \frac{1}{x}$ and $\mathrm{d} u = \frac{\mathrm{d} x}{x}$ .

$v= \int x ; \mathrm{d}x =\frac{ x ^{2}}{2}$

and

$\int x \ln x ; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \ln{x} \cdot \frac{ x ^{2}}{2} - \int \left (\frac{ x ^{2}}{2} \right ) \frac{\mathrm{d} x}{x}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2}\int x; \mathrm{d} x$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2} \cdot \frac{x^{2}}{2}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4}$

Now, this expression evaluated at is equal to

$\frac{ 1 ^{2}\ln{1}}{2} - \frac{1^{2}}{4} = 0 - \frac{1}{4} = - \frac{1}{4}$ .

At it is undefined, because $\ln 0$ does not exist.

We can use L'Hospital's rule to find its limit as $x\rightarrow 0$ , as follows:

$\lim_{x\rightarrow 0}\frac{2 x ^{2}\ln{x} - x^{2}}{4} =\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}$

$\lim_{x\rightarrow 0} \left (2 \ln{x} - 1 \right ) = - \infty$ and $\lim_{x\rightarrow 0} \frac{4}{x ^{2}} = \infty$ , so by L'Hospital's rule,

$\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}=\lim_{x\rightarrow 0} \frac{\frac {2}{x} }{\frac{-8}{x ^{3}}} = \lim_{x\rightarrow 0}\left ( -\frac{x^{2}}{4} \right ) = 0$

Therefore,

$\int_{0}^{1} x \ln x ; \mathrm{d}x$

$= \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ 0 \end{matrix}$

$=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]$

$= -\frac{1}{4} - 0$

$= -\frac{1}{4}$

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Question

Evaluate:

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}}$

Answer

Rewrite the integral as

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }$ .

Substitute $u = - x^{2}$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = - 2x$ and $-\frac{1}{2} \mathrm{d} u = x ; \mathrm{d} x$ . The lower bound of integration stays $u = - 0^{2} = 0$ , and the upper bound becomes $\lim_{x\rightarrow \infty } - x^{2} = - \infty$ , so

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }=-\frac{1}{2} \int_{0 }^{-\infty}e^{u} \mathrm{d}u}= \frac{1}{2} \int_{-\infty }^{0}e^{u} \mathrm{d}u}$

$=\frac{1}{2} \lim_{b\rightarrow -\infty } \left.\begin{matrix} e ^{u}\ \textrm{ } \ \textrm{ } \end{matrix}\right| \begin{matrix} 0\ \ b \end{matrix}$

Since $\lim_{b\rightarrow -\infty } e ^{u} = 0$ , the above is equal to

$=\frac{1}{2} (e^{0}-0) =\frac{1}{2} (1) = \frac{1}{2}$ .

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Question

Evaluate $\int_{1}^{\infty}\frac{4}{x^{5}}dx$ .

Answer

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{4}{x^{5}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{4}{x^{5}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{4}{x^{5}}dx$

$=4\int_{1}^{b}\frac{1}{x^{5}}dx$

$=4\int_{1}^{b}{x^{-5}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all with an arbitrary constant of integration . Therefore:

$4\int_{1}^{b}{x^{-5}}dx$

$=4({\frac{x^{-5+1}}{-5+1}})_{1}^{b}$

$=4(\frac{x^{-4}}{-4}})_{1}^{b}$

$=4(-{\frac{1}{4x^{4}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }4(-{\frac{1}{4x^{4}}})_{1}^b$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{4}}+1}]$ .

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Question

Evaluate:

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

Answer

First, we will find the indefinite integral $\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$ using integration by parts.

We will let $u = \cos x$ and $\mathrm{d}v= e^{-x}; \mathrm{d}x$ .

Then $\frac{\mathrm{d} u}{\mathrm{d} x} = - \sin x$ and $\mathrm{d} u = - \sin x ; \mathrm{d} x$ .

$v= \int e^{-x}; \mathrm{d}x = - e^{-x} = - \frac{1}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \cos x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \left (- \sin x \right ); \mathrm{d} x$

$= - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

To find $\int e^{-x} \sin x ; \mathrm{d} x$ , we use another integration by parts:

$u = \sin x$ , which means that $\mathrm{d} u = \cos x ; \mathrm{d} x$ , and

$\mathrm{d}v= e^{-x}; \mathrm{d}x$ , which means that, again, $v= - \frac{1}{e^{x} }$ .

$\int e^{-x} \sin x ; \mathrm{d} x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \sin x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) ; \cdot \cos x ; \mathrm{d} x$

$= - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x ; \mathrm{d} x$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \left ( - \frac{ \sin x }{e^{x} } + \int \frac{\cos x ; \mathrm{d} x}{e^{x} } \right )$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} } - \int \frac{\cos x ; \mathrm{d} x}{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} }$

$2 \int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{e^{x} }$

$\int \frac{\cos x; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{2e^{x} }$

Since

$\frac{-2}{2e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{2}{2e^{x} }$ , or,

$- \frac{1}{e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{1}{e^{x} }$

for all real , and

$- \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = 0$ ,

by the Squeeze Theorem,

$\lim_{x\rightarrow \infty } \frac{\sin x -\cos x}{2e^{x} } = 0$ .

$\int_{0}^{\infty } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{b\rightarrow \infty }\int_{0}^{b } \frac{\cos x; \mathrm{d}x}{e^{x}}$

$= \lim_{ b\rightarrow \infty }$ $\left.\begin{matrix} \frac{\sin x -\cos x}{2e^{x} } \ \textrm{ } \end{matrix}\right|$ $\begin{matrix} b\ \ 0 \end{matrix}$

$= 0 - \frac{\sin 0 -\cos 0}{2e^{0} }$

$= - \frac{ 0 -1}{2 \cdot 1 } = \frac{1}{2}$

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Question

Evaluate:

$\int_{0}^{1} x \ln x ; \mathrm{d}x$

Answer

First, we will find the indefinite integral, $\int x \ln x ; \mathrm{d}x$ .

We will let $u = \ln x$ and $\mathrm{d}v= x ; \mathrm{d}x$ .

Then,

$\frac{\mathrm{d} u }{\mathrm{d} x}= \frac{1}{x}$ and $\mathrm{d} u = \frac{\mathrm{d} x}{x}$ .

$v= \int x ; \mathrm{d}x =\frac{ x ^{2}}{2}$

and

$\int x \ln x ; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \ln{x} \cdot \frac{ x ^{2}}{2} - \int \left (\frac{ x ^{2}}{2} \right ) \frac{\mathrm{d} x}{x}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2}\int x; \mathrm{d} x$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2} \cdot \frac{x^{2}}{2}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4}$

Now, this expression evaluated at is equal to

$\frac{ 1 ^{2}\ln{1}}{2} - \frac{1^{2}}{4} = 0 - \frac{1}{4} = - \frac{1}{4}$ .

At it is undefined, because $\ln 0$ does not exist.

We can use L'Hospital's rule to find its limit as $x\rightarrow 0$ , as follows:

$\lim_{x\rightarrow 0}\frac{2 x ^{2}\ln{x} - x^{2}}{4} =\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}$

$\lim_{x\rightarrow 0} \left (2 \ln{x} - 1 \right ) = - \infty$ and $\lim_{x\rightarrow 0} \frac{4}{x ^{2}} = \infty$ , so by L'Hospital's rule,

$\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}=\lim_{x\rightarrow 0} \frac{\frac {2}{x} }{\frac{-8}{x ^{3}}} = \lim_{x\rightarrow 0}\left ( -\frac{x^{2}}{4} \right ) = 0$

Therefore,

$\int_{0}^{1} x \ln x ; \mathrm{d}x$

$= \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ 0 \end{matrix}$

$=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right ]$

$= -\frac{1}{4} - 0$

$= -\frac{1}{4}$

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Question

Evaluate:

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}}$

Answer

Rewrite the integral as

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }$ .

Substitute $u = - x^{2}$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = - 2x$ and $-\frac{1}{2} \mathrm{d} u = x ; \mathrm{d} x$ . The lower bound of integration stays $u = - 0^{2} = 0$ , and the upper bound becomes $\lim_{x\rightarrow \infty } - x^{2} = - \infty$ , so

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }=-\frac{1}{2} \int_{0 }^{-\infty}e^{u} \mathrm{d}u}= \frac{1}{2} \int_{-\infty }^{0}e^{u} \mathrm{d}u}$

$=\frac{1}{2} \lim_{b\rightarrow -\infty } \left.\begin{matrix} e ^{u}\ \textrm{ } \ \textrm{ } \end{matrix}\right| \begin{matrix} 0\ \ b \end{matrix}$

Since $\lim_{b\rightarrow -\infty } e ^{u} = 0$ , the above is equal to

$=\frac{1}{2} (e^{0}-0) =\frac{1}{2} (1) = \frac{1}{2}$ .

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Question

Evaluate $\int_{1}^{\infty}\frac{4}{x^{5}}dx$ .

Answer

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{4}{x^{5}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{4}{x^{5}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{4}{x^{5}}dx$

$=4\int_{1}^{b}\frac{1}{x^{5}}dx$

$=4\int_{1}^{b}{x^{-5}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all with an arbitrary constant of integration . Therefore:

$4\int_{1}^{b}{x^{-5}}dx$

$=4({\frac{x^{-5+1}}{-5+1}})_{1}^{b}$

$=4(\frac{x^{-4}}{-4}})_{1}^{b}$

$=4(-{\frac{1}{4x^{4}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }4(-{\frac{1}{4x^{4}}})_{1}^b$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{4}}+1}]$ .

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