Improper Integrals - AP Calculus BC

Card 0 of 132

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 $n\neq-1$ 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 $n\neq-1$ 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

Determine if the following improper integral converges or diverges. If it converges, find the solution.

$\int_{0}^{\infty }\frac{\arctan x}{1 + x^2} dx$

Answer

Because this is an improper integral, you must rewrite it as so:

$\lim_{t \rightarrow \infty } \int_{0}^{t} \frac{\arctan x}{1+x^2} dx$ .
You can now solve the integral by substuting $u = \arctan x$ and $du = \frac{1}{1+x^2} dx$ to recreate the integral as $\lim_{t \rightarrow \infty } \int_{0}^{t} u du$ which then becomes after integration $\frac{u^2}{2}$ evaluated at t and 0.

Substituting arctanx back in for u and then evaluating at the bounds gives you the following:

$\frac{\arctan t^2}{2} - \frac{\arctan 0^2}{2}$ .

The arc tangent of 0 is just 0. But remember that t is approaching infinity. Below is the graph of arctan.

As you can see, as x (or in this case t) approaches infinity, the graph slowly approaches its horizontal asymptote at pi/2. And so, $\arctan \infty$ can be seen to be $\frac{\pi }{2}$ .

So the answer to the problem is

$\frac{(\frac{\pi }{2})^{2}}{2}$ which simplifies to $\frac{\pi^2}{8}$

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Question

Calculate the following integral:

$y=\int_{2}^{\infty }\frac{1}{x^{2}}dx$

Answer

When evaluating an improper integral, we substitue infinity for a constant, and then evaluate the integral as the constant goes towards infinity, which looks like this:

$y=\int_{2}^{\infty }\frac{1}{x^{2}}dx$ , let "t" = infinity, which gives a new integral: $y=\int_{2}^{t}\frac{1}{x^{2}}dx$

evaluating the integral, we get: $y=\lim_{t\rightarrow \infty }\frac{-1}{t}-\lim_{t\rightarrow \infty }\frac{-1}{2}$

Since $\lim_{t\rightarrow\infty }\frac{-1}{\infty }$ equals 0, and $\frac{-1}{2}$ is a constant, unaffected by what happens to "t", our integral is $0-(\frac{-1}{2})$ ,which equals $\frac{1}{2}$

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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:

$\int_{1}^{\infty } \frac{x; \mathrm{d}x}{e^{2x}}$

Answer

First, we will find the indefinite integral

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

using integration by parts.

We will let and $\mathrm{d}v= e^{-2x}; \mathrm{d}x$ .

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

Also,

$v= \int e^{-2x}; \mathrm{d}x$ .

To find , we can substitute for . Then $\frac{\mathrm{d} t}{\mathrm{d} x} = -2$ or $\mathrm{d} x = -\frac{1}{2} \mathrm{d} t$ , so the antiderivative is

$v= - \frac{1}{2} \int e^{t}; \mathrm{d}t = - \frac{1}{2} e^{t} = - \frac{1}{2} e^{-2x} = - \frac{1}{2e^{2x}}$ .

Now we can integrate by parts:

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

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

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

$= x \left ( - \frac{1}{2e^{2x}} \right ) - \int\left ( - \frac{1}{2e^{2x}} \right ); \mathrm{d} x$

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

$= - \frac{x}{2e^{2x}} + \frac{1}{2}\left ( - \frac{1}{2e^{2x}} \right )$

$= - \frac{x}{2e^{2x}} - \frac{1}{4e^{2x}}$

$= - \frac{2x}{4e^{2x}} - \frac{1}{4e^{2x}}$

$= - \frac{2x+1}{4e^{2x}}$

By L'Hospital's rule, since both the numerator and the denominator approach $\infty$ as $x\rightarrow \infty$ ,

$\lim_{x\rightarrow \infty } - \frac{2x+1}{4e^{2x}} = \lim_{x\rightarrow \infty } - \frac{2}{8e^{2x}} = \lim_{x\rightarrow \infty } - \frac{1}{4e^{2x}} = 0$ .

So:

$\int_{1}^{\infty } \frac{x; \mathrm{d}x}{e^{2x}}$

$= \lim_{b\rightarrow \infty } \int_{1}^{b } \frac{x; \mathrm{d}x}{e^{2x}}$

$= \lim_{b\rightarrow \infty } \left.\begin{matrix} - \frac{2x+1}{4e^{2x}} \ \textrm{ } \end{matrix}\right| \begin{matrix} b \ 1 \end{matrix}$

$= 0 -\left ( - \frac{2 \cdot 1 +1}{4e^{2\cdot 1 }} \ \textrm{ } \right ) = \frac{3}{4e^{2 }}$

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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:

$\int_{1}^{\infty } \frac{4; \mathrm{d}x}{x^{2}}$

Answer

$\int_{1}^{\infty } \frac{4; \mathrm{d}x}{x^{2}} = 4 \lim_{b\rightarrow \infty } \int_{1}^{b}\frac{; \mathrm{d}x}{x^{2}}$

$\int \frac{ ; \mathrm{d}x}{x^{2}} = \int x^{-2} \mathrm{d}x} = \frac{ x^{-1}}{-1}= \left ( - \frac{1}{x} \right )= - \frac{1}{x}$

$\lim_{x\rightarrow \infty } \frac{1}{x} = 0$ , so

$4 \lim_{b\rightarrow \infty } \int_{1}^{b}\frac{; \mathrm{d}x}{x^{2}}$

$= 4 \lim_{b\rightarrow \infty } \left.\begin{matrix} - \frac{1}{x}\ \textrm{ } \ \textrm{ } \end{matrix}\right|\begin{matrix} b\ \1 \end{matrix}$

$=4 \left [ -0 -\left ( - \frac{1}{1}\right ) \right ]= 4 \cdot 1 = 4$

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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:

$\int_{0}^{\infty} \frac{ x ; \mathrm{d} x}{1 +x^{ 2 }}$

Answer

Substitute $u = 1 + x^{2}$ .

$\frac{\mathrm{d} u}{\mathrm{d} x} = 2 x$

$\frac{1}{2}\mathrm{d} u = x ; \mathrm{d} x$

The lower bound of integration becomes $u = 1 + 0^{2} = 1$ ;

the upper bound becomes $\lim_{x\rightarrow \infty }\left ( x^{2}+ 1 \right )= \infty$ .

The integral therefore becomes

$\int_{0}^{\infty} \frac{ x ; \mathrm{d} x}{1 +x^{ 2 }}$

$= \int_{1}^{\infty} \frac{ \mathrm{d} u}{2u} = \frac{1}{2} \int_{1}^{\infty} \frac{ \mathrm{d} u}{u}$

$= \frac{1}{2} \lim_{b\rightarrow \infty} \int_{1}^{b} \frac{ \mathrm{d} u}{u}$

$= \lim_{b\rightarrow \infty} \left.\begin{matrix} \frac{1}{2} \ln u \ \ \end{matrix}\right| \begin{matrix} b \ \ 1 \end{matrix}$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b - \frac{1}{2} \ln 1$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b - \frac{1}{2} \cdot0$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b = \infty$

The integral, therefore, does not converge.

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Question

Evaluate the improper integral:

$\int_{0}^{\infty} xe^{-2x} \mathrm{d}x$

Answer

First, we will perform an integration by parts on the indefinite integral

$\int xe^{-2x} \mathrm{d}x$ .

Let and $\mathrm {d}v = e^{-2x} ; \mathrm {d}x$ .

Then,

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

Also,

$\mathrm {d}v = e^{-2x} ; \mathrm {d}x$ .

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

$v = \frac{e^{-2x}}{-2} =- \frac{1}{2}e^{-2x}$

Therefore,

$\int xe^{-2x} \mathrm{d}x$

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

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

$= x \cdot - \frac{1}{2}e^{-2x}- \int \left (- \frac{1}{2}e^{-2x} \right ) \cdot ; \mathrm{d}x$

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

$= - \frac{x}{2}e^{-2x} + \frac{1}{2}\left ( - \frac{1}{2}e^{-2x} \right )$

$= - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x}$

The antiderivative of $f(x) = xe^{-2x}$ is

$F(x) = - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x}$

and

$\int_{0}^{ \infty} xe^{-2x} \mathrm{d}x = \lim_{b\rightarrow \infty }\int_{0}^{b} xe^{-2x} \mathrm{d}x= \lim_{b\rightarrow \infty } F(b) - F (0)$ .

$\lim_{b\rightarrow \infty } F(b) = -\lim_{b\rightarrow \infty } \frac{x}{2}e^{-2x} -\lim_{b\rightarrow \infty } - \frac{1}{4} e^{-2x}$

$\lim_{b\rightarrow \infty } F(b) = -\lim_{b\rightarrow \infty } \frac{x}{2e^{2x}} -\lim_{b\rightarrow \infty } \frac{1}{4e^{2x}} = 0$ , as can be proved by L'Hospital's rule.

$F(0) = - \frac{0}{2}\cdot e^{-2 \cdot 0} - \frac{1}{4}\cdot e^{-2 \cdot 0} = 0 - \frac{1}{4} = - \frac{1}{4}$

$\int_{0}^{ \infty} xe^{-2x} \mathrm{d}x = 0 - \left ( - \frac{1}{4}\right ) = \frac{1}{4}$

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Question

Evaluate the improper integral:

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

Answer

First, we will perform an integration by parts on the indefinite integral:

$\int x^2 e^{-x} \mathrm{d}x$

Let and $\mathrm {d}v = e^{-x} ; \mathrm {d}x$ .

Then,

$\frac{\mathrm{d} u}{\mathrm{d} x} = 2x$ , $\mathrm{d} u =2 x ; \mathrm{d} x$ ,

and

$\mathrm {d}v = e^{-x} ; \mathrm {d}x$ .

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

$v = -e^{-x}$

$\int x^2 e^{-x} \mathrm{d}x$

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

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

$= x^2 (-e^{-x})- \int (-e^{-x}) \cdot 2x ; \mathrm{d}x$

$= -x^2 e^{-x}+ 2\int xe^{-x} \mathrm{d}x$

We do another integration by parts, setting

$\widetilde{u }= x$ and $\mathrm {d}\widetilde{v} = e^{-x} ; \mathrm {d}x$ .

Then,

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

Also,

$\mathrm {d}\widetilde{v} = e^{-x} ; \mathrm {d}x$ and, again, $\widetilde{v }= -e^{-x}$ .

$=- x^2 e^{-x}+ 2\int xe^{-x} \mathrm{d}x$

$=- x^2 e^{-x}+ 2 \int \widetilde{u} ; \mathrm{d}\widetilde{v}$

$=- x^2 e^{-x}+ 2 \left ( \widetilde{u} \widetilde{v} - \int \widetilde{v} \mathrm{d}\widetilde{u} \right )$

$= -x^2 e^{-x}+ 2 \left [ x \left ( -e^{-x} \right ) - \int \left ( -e^{-x} \right ) \mathrm{d}x\right]$

$=- x^2 e^{-x}- 2 x e^{-x} +2 \int e^{-x} \mathrm{d}x\right]$

$=- x^2 e^{-x}- 2 x e^{-x} + 2\left ( - e^{-x} \right )$

$=- x^2 e^{-x}- 2 x e^{-x} - 2e^{-x}$

$=\left (- x^2 - 2 x - 2 \right )e^{-x}$

$=\frac{ -x^2 - 2 x -2 }{e^x}$

Therefore, the antiderivative of $f(x ) = x^2 e^{-x}$ is $F(x)=\frac{ x^2 - 2 x -2 }{e^x}$ .

$\int_{0}^{\infty} x^2 e^{-x} \mathrm{d}x = \lim_{b\rightarrow \infty } \int_{0}^{b} x^2 e^{-x} \mathrm{d}x =\lim_{b\rightarrow \infty } F(b) - F(0)$

$\lim_{b\rightarrow \infty } F(b) = \lim_{x\rightarrow \infty } \frac{ x^2 - 2 x -2 }{e^x} = 0$ , which two applications of L'Hospital's rule can easily reveal.

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

Therefore,

$\int_{0}^{\infty} x^2 e^{-x} \mathrm{d}x = 0 - (-2) = 2$ .

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Question

Evaluate:

$\int_{1}^{\infty} \frac{2}{x^3}: dx$

Answer

Write the formula rule for the case:

$\int_{a}^{\infty} F(x):dx = \lim_{b \to \infty} \int_{a}^{b}f(x):dx$

Apply this rule for the following question.

$\int_{1}^{\infty} \frac{2}{x^3}: dx= 2\int_{1}^{\infty} \frac{1}{x^3}: dx$

$2\int_{1}^{\infty} \frac{1}{x^3}: dx = 2\lim_{b \to \infty} \int_{1}^{b}{x}^-^3}:dx= 2\left[-\frac{x^-^2}{2}\right]_{1}^{b}$

$\lim_{b \to \infty} 2\left[-\frac{x^-^2}{2}\right]_{1}^{b}= \lim_{b \to \infty}2\left[-\frac{1}{2b^2}+\frac{1}{2}\right]$

$=\lim_{b \to \infty}{-\frac{1}{b^2}}+1 = 0+1 = 1$

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Question

Evaluate $\int_{1}^{\infty}\frac{3}{x^{2}}dx$ .

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 $\int_{1}^{\infty}\frac{3}{x^{2}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{3}{x^{2}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{3}{x^{2}}dx$

$=3\int_{1}^{b}\frac{1}{x^{2}}dx$

$=3\int_{1}^{b}{x^{-2}}dx$

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:

$3\int_{1}^{b}{x^{-2}}dx$

$=3({\frac{x^{-2+1}}{-2+1}})_{1}^{b}$

$=3(\frac{x^{-1}}{-1}})_{1}^{b}$

$=3(-{\frac{1}{x}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }3[-{\frac{1}{x}}]_{1}^b$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}-(-\frac{1}{1}})]$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]$

As $b\rightarrow \infty$ , $\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]=3(0+1)=3$

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