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 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 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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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 $\int_{1}^{\infty}\frac{7}{x^{8}}dx$ .

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 $\int_{1}^{\infty}\frac{7}{x^{8}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{7}{x^{8}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{7}{x^{8}}dx$

$=7\int_{1}^{b}\frac{1}{x^{8}}dx$

$=7\int_{1}^{b}{x^{-8}}dx$

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:

$7\int_{1}^{b}{x^{-8}}dx$

$=7({\frac{x^{-8+1}}{-8+1}})_{1}^{b}$

$=7(\frac{x^{-7}}{-7}})_{1}^{b}$

$=7(-{\frac{1}{7x^{7}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }7(-{\frac{1}{7x^{7}}})_{1}^b$

$=\lim_{b\rightarrow \infty }7[-{\frac{1}{7b^{7}}-(-\frac{1}{7}})]$

$=\lim_{b\rightarrow \infty }7[-{\frac{1}{7b^{7}}+\frac{1}{7}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{7}}+1}]$ .

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Question

Evaluate:

$\int_{2}^{5}\frac{1}{2x-8} :dx$

Answer

To evaluate $\int_{2}^{5}\frac{1}{2x-8} :dx$ , notice that the denominator does not exist at . The integral must be rewritten so that it is in terms of a limit. This is an improper integral.

$\int_{2}^{5}\frac{1}{2x-8} :dx= \lim_{a \to 4^-}\int_{2}^{a}\frac{1}{2x-8} :dx+\lim_{b \to 4^+}\int_{b}^{5}\frac{1}{2x-8} :dx$

Evaluate the integrals.

$\frac{1}{2}du=dx$

$\frac{1}{2}\lim_{a \to 4^-}ln\left | \frac{1}{2x-8} \right |_{2}^{a} +\frac{1}{2}\lim_{b \to 4^+}ln\left | \frac{1}{2x-8} \right |_{b}^{5}$

By evaluating the terms and substituting the limits, we will notice that the integral diverges as a result since the terms $\frac{1}{2}ln\left | \frac{1}{0} \right |-\frac{1}{2}ln\left | \frac{1}{0} \right |$ cannot be cancelled as a result.

The correct answer is: Diverge

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Question

Evaluate the following integral:

$\int_{0}^{1}\frac{3}{x-1}dx$

Answer

The integral we were given in the problem statement is improper because the upper limit of integration creates 0 in the denominator of the fraction.

To fix this, we use the following technique:
$3\lim_{t\rightarrow 1^{-}}\int_{0}^{t}\frac{dx}{x-1}$

Note that we moved the three outside of the integral, and then outside of the limit altogether. This can be done with coefficients and makes things far easier to look at!

Next, keeping the coefficient and limit, integrate:

$3\lim_{t\rightarrow 1-}\ln \left | x-1 \right | |_{0}^{t}$

The integration was done using the following rule:

$\int \frac{dx}{x}=\ln\left | x \right |+C$

Next, solve the result:

$3\lim_{t\rightarrow 1-} \ln(t-1)-0=3\cdot -\infty =-\infty$

We chose 1 from the left in our limit because 1 is the upper limit of integration.

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