AP Calculus BC - Fundamental Theorem of Calculus and Techniques of Antidifferentiation

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:

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

$\infty$

$-\infty$

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 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:

$\int\frac{x-1}{(x-3)(x+4)} dx=$

$\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

$\frac{(x-3)^2}{}7+\frac{2(x+4)^2}{7}+C$

$\frac{1}{7}\ln|x-2|+\frac{1}{7}\ln|x-4|+C$

$\frac{5}{7(x+4)}\ln |x|+\frac{2}{7(x-3)}\ln |x|+C$

$\frac{5}{7(x+4)}+\frac{2}{7(x+2)}+C$

Explanation

In order to evaluate this integral, we will need to use partial fraction decomposition.

$\frac{x-1}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}$

Multiply both sides of the equation by the common denominator, which is

This means that must equal 1, and

$\frac{x-1}{(x-3)(x+4)}=\frac{2/7}{x-3}+\frac{5/7}{x+4}$

$\int(\frac{2}{7}\cdot \frac{1}{x-3}+\frac{5}{7}\cdot \frac{1}{x+4})dx$

$=\int\frac{2}{7}\cdot\frac{1}{x-3} dx +\int\frac{5}{7}\cdot\frac{1}{x+4}dx$

$=\frac{2}{7}\int\frac{dx}{x-3}+\frac{5}{7}\int\frac{dx}{x+4}$

$=\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

The answer is $\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$ .

Evaluate $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

$\int\frac{x-1}{(x-3)(x+4)} dx=$

$\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

$\frac{(x-3)^2}{}7+\frac{2(x+4)^2}{7}+C$

$\frac{1}{7}\ln|x-2|+\frac{1}{7}\ln|x-4|+C$

$\frac{5}{7(x+4)}\ln |x|+\frac{2}{7(x-3)}\ln |x|+C$

$\frac{5}{7(x+4)}+\frac{2}{7(x+2)}+C$

Explanation

In order to evaluate this integral, we will need to use partial fraction decomposition.

$\frac{x-1}{(x-3)(x+4)}=\frac{A}{x-3}+\frac{B}{x+4}$

Multiply both sides of the equation by the common denominator, which is

This means that must equal 1, and

$\frac{x-1}{(x-3)(x+4)}=\frac{2/7}{x-3}+\frac{5/7}{x+4}$

$\int(\frac{2}{7}\cdot \frac{1}{x-3}+\frac{5}{7}\cdot \frac{1}{x+4})dx$

$=\int\frac{2}{7}\cdot\frac{1}{x-3} dx +\int\frac{5}{7}\cdot\frac{1}{x+4}dx$

$=\frac{2}{7}\int\frac{dx}{x-3}+\frac{5}{7}\int\frac{dx}{x+4}$

$=\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$

The answer is $\frac{2}{7}\ln|x-3|+\frac{5}{7}\ln| x+4|+C$ .

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 $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

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,

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

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

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

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

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

Fundamental Theorem of Calculus and Techniques of Antidifferentiation

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