AP Calculus BC - Improper Integrals

Evaluate the following integral:

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

$\infty$

$-\infty$

Explanation

This integral is improper because the lower bound creates a zero in the denominator. To integrate, we must use a limit:

$\lim_{t\rightarrow 1^-}\int_{t}^{2}\frac{dx}{x-1}=\lim_{t\rightarrow 1^-} \ln\left | x-1\right |\left.\begin{matrix} 2\ t \end{matrix}\right|$

The following rule was used:

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

Now, the definite part of the integral:

$\lim_{t\rightarrow 1^-} \ln\left | 1\right |-\ln\left | t-1\right |=\infty$

The natural logarithm of 1 equals zero, and as the natural logarithm approaches zero, the function goes to negative infinity. The negative sign in front of the function makes it go to infinity.

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

$\infty$

Explanation

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

Evaluate the following integral:

$\int_{0}^{2}\left(\frac{x}{x^2-4}\right)dx$

$-\infty$

$\infty$

$\ln 4$

$-\ln4$

Explanation

The integral is improper because of the upper limit of integration (creates zero in the denominator of the function being integrated). So, we have to do the following:

$\lim_{t\rightarrow 2^{-}}\int_{0}^{t}\left(\frac{x}{x^2-4}\right)dx$

We evaluate the limit from the left side because the upper limit of integration was the one that caused problems.

Now, to integrate we must do a substitution, but this also means changing the limit:

The derivative was perfomed using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, rearrange, and rewrite the limit in terms of our new t value, which originally was :

$\frac{1}{2}\lim_{t\rightarrow 4^{-}}\int_{0}^{t}\frac{du}{u-4}$

Next, perform the definite integration, keeping the limit:

$\frac{1}{2}\lim_{t\rightarrow 4^{-}}[\ln\left |(t-4) \right |-\ln\left |4 \right |]$

The integration was performed using the following rule:

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

Now, when we evaluate the limit, we notice that the natural log function, as it approaches zero, approaches negative infinity. The fact that there is a coefficient on the limit or the additional term being subtracted are insignificant compared to the negative infinity term. So, our final answer is $-\infty$ .

he Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function into its Laplace Transform is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where , where and are constants and is the imaginary number.

Give the Laplace Transform of $\frac{\mathrm{d} }{\mathrm{d} t}[f(t)]$ .

Explanation

The Laplace Transform of the derivative is given by:

$\int_{0}^{\infty}e^{-st}*f'(t) dt$

Using integration by parts,

$\int_{0}^{\infty} UdV= [UV]_{0}^{\infty}-\int_{0}^{\infty} VdU$

Let and $U=e^{-st}$

$[e^{-st}*f(t) ]_{0}^{\infty}-\int_{0}^{\infty}(-s)e^{-st}*f(t) dt$

The first term becomes

and the second term becomes

The Laplace Transform therefore becomes:

Evaluate the improper integral:

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

$\frac{1}{4}$

The integral does not converge.

$\frac{1}{2}$

$\frac{e}{4}$

$\frac{e}{2}$

Explanation

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.

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

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

Evaluate:

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

$\frac{1}{2}$

$\frac{1}{e}$

$\frac{1}{2e}$

$\frac{\sqrt{2}}{2e}$

Explanation

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,

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

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

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

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

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

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

Evaluate:

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

$-\frac{1}{4}$

$\frac{1}{4}$

The integral does not converge

$\ln 4$

$-\ln 4$

Explanation

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,

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

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

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

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

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