Improper Integrals - AP Calculus BC

Card 0 of 40

Evaluate:

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

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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

, 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,

.

$\int_${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}$$

Evaluate:

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

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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:

and , so by L'Hospital's rule,

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}$$

Evaluate:

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Rewrite the integral as

.

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 , so

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

Since , the above is equal to

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

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

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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 $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

.

So,

.

Evaluate:

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

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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

, 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,

.

$\int_${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}$$

Evaluate:

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

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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:

and , so by L'Hospital's rule,

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}$$

Evaluate:

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Rewrite the integral as

.

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 , so

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

Since , the above is equal to

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

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

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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 $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

.

So,

.

Evaluate:

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

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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

, 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,

.

$\int_${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}$$

Evaluate:

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

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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:

and , so by L'Hospital's rule,

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}$$

Evaluate:

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Rewrite the integral as

.

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 , so

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

Since , the above is equal to

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

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

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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 $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

.

So,

.

Evaluate:

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

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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

, 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,

.

$\int_${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}$$

Evaluate:

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

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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:

and , so by L'Hospital's rule,

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}$$

Evaluate:

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Rewrite the integral as

.

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 , so

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

Since , the above is equal to

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

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

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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 $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

.

So,

.

Evaluate:

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

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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

, 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,

.

$\int_${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}$$

Evaluate:

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

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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:

and , so by L'Hospital's rule,

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}$$

Evaluate:

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Rewrite the integral as

.

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 , so

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

Since , the above is equal to

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

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

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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 $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

.

So,

.