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Calculus 2 : Definite Integrals

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Example Questions

Example Question #21 : Definite Integrals

Evaluate the indefinite integral $\int x^3+\pi^2dx$ .

Possible Answers:

None of the other answers

$\frac{x^4}{4}+\frac{\pi^3}{3} +C$

$x^3+\pi^2+C$

$4x^3+3\pi^2+C$

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{x^4}{4}+\pi^2x+C$ .

The integral itself is not too difficult to take, simply use the Power Rule on the x^3 and $\pi^2$ terms. The trick is to be careful when integrating $\pi^2$ . $\pi^2$ is a constant value (about 9.869 ) not a variable, so it must be integrated accordingly.

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Example Question #22 : Definite Integrals

Evaluate $\int^{1}_{0} 5\tan^{-1}(x)dx$ .

Possible Answers:

$\frac{\pi+\ln2}{2}$

None of the other answers

$\frac{\pi^2-2}{\ln(5)}$

$\frac{5\pi-10\ln2}{4}$

$\frac{3\pi}{2}$

Correct answer:

$\frac{5\pi-10\ln2}{4}$

Explanation:

This integral requires integration by parts followed by u-substitution. Here are the details

$\int^{1}_{0} 5\tan^{-1}(x)dx$ . Start

$=5\int^{1}_{0} \tan^{-1}(x)dx$ . Factor out the

Set up integration by parts with $u = \tan^{-1}(x)$ , dv = dx . We then have $du = \frac{1}{1+x^2}dx$ and v=x . Afterward, we use the integration by parts formula $uv - \int vdu$ .

$= 5\{[x\tan^{-1}(x)]^{1}_{0}-\int^{1}_{0} \frac{x}{1+x^2}dx\}$ .

Now at this point we use u-substitution to evaluate the 2nd integral. Let u =1+x^2 , then du = 2xdx, and therefore $\frac{1}{2}du = xdx$ . Substituting into the integral we have

$= 5\{[x\tan^{-1}(x)]^{1}_{0}-\frac{1}{2}\int^{2}_{1} \frac{1}{u}du\}$ . (Don't forget to change the bounds of integration by plugging them into for our equation for .)

$= 5\{[x\tan^{-1}(x)]^{1}_{0}-[\frac{1}{2}\ln(u)]^{2}_{1}\}$

$=5((\frac{\pi}{4}) - (\frac{1}{2}\ln2))$

$= \frac{5\pi-10\ln2}{4}$

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Example Question #23 : Definite Integrals

Evaluate $\int^{1}_{-1}xe^{x^4}dx$ .

Possible Answers:

$\ln2-\frac{1}{2}$

$\frac{1}{2}$

Not possible without a calculator

Correct answer:

Explanation:

This integral isn't possible to integrate directly using antiderivatives, but we can still find its value by noticing that $xe^{x^4}$ is an odd function (f(-x)=-f(x)) , and that our limits of integration are negatives of each other.

Hence

$\int^{1}_{-1}xe^{x^4}dx = \int^{1}_{0}xe^{x^4}dx +\int^{0}_{-1}xe^{x^4}dx$

$=\int^{1}_{0}xe^{x^4}dx - \int^{-1}_{0}xe^{x^4}dx$

$=\int^{1}_{0}xe^{x^4}dx - \int^{1}_{0}xe^{x^4}dx$ . (Since $xe^{x^4}$ is an odd function)

= 0

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Example Question #21 : Definite Integrals

Evaluate $\int_{0}^{\pi/2} \sin^{99}(x)\cos(x)dx$

Possible Answers:

$\frac{\pi^{99}}{99}$

$\frac{1}{100}$

None of the other answers.

$\frac{\pi^2}{2}$

Correct answer:

$\frac{1}{100}$

Explanation:

We can use u-substitution for this integral

$\int_{0}^{\pi/2} \sin^{99}(x)\cos(x)dx$ . Start

Let $u= \sin(x)$ , then $du = \cos(x)dx$ , and our integral becomes

$\int_{0}^{1}u^{99}du$ . (Don't forget to change the bounds of integration by plugging them into our equation for )

$= [\frac{u^{100}}{100}]_{0}^{1}$

$=\frac{1}{100}$

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Example Question #25 : Definite Integrals

Evaluate $\int_{1}^{e} \frac{\ln(x)}{x^2}dx$ .

Possible Answers:

e^2-8

$-\infty$

None of the other answers

$\frac{e-2}{e}$

Correct answer:

$\frac{e-2}{e}$

Explanation:

We proceed by using integration by parts.

$\int_{1}^{e} \frac{\ln(x)}{x^2}dx$ . Start

Let $u= \ln(x), dv=\frac{1}{x^2}dx$ , then we get $du = \frac{1}{x}dx, v=-\frac{1}{x}$
. Then using the integration by parts formula $uv - \int vdu$ , we get

$=[-\frac{\ln(x)}{x}]_{1}^{e} - \int_{1}^{e}-\frac{1}{x^2}dx$

$=(-\frac{\ln(e)}{e}+\frac{\ln(1)}{1}) + \int_{1}^{e}\frac{1}{x^2}dx$

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

$=-\frac{1}{e} +[-x^{-1}]^{e}_{1}$

$= -\frac{1}{e}+(-\frac{1}{e}+1)$

$=1-\frac{2}{e}$

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

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Example Question #26 : Definite Integrals

Evaluate the following integral:

$\int_{0}^{2\pi}7x\sin(x)dx$

Possible Answers:

$7\pi$

$-14\pi$

$-15\pi$

$14\pi$

Correct answer:

$-14\pi$

Explanation:

To evaluate this integral, we must integrate by parts, according to the following formula:

$\int u dv=uv-\int vdu$

So, we must assign our u and dv, and differentiate and integrate to find du and v, respectively:

u=7x, du=7dx

$dv=\sin(x)dx, v=-\cos(x)$

The derivative and integral were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\int \sin(x)dx=-\cos(x)+C$

Note that we ignore the constant of integration.

Now, use the above formula:

$[-7(2\pi)\cos(2\pi)-7(0)\cos(0)]+7\int_{0}^{\2\pi} \cos(x)dx=-14\pi+[7\sin(2\pi)-7\sin(0)]$

Note that both the product of u and v and the integral are being evaluated from zero to $2\pi$ .

The integral was performed using the following rule:

$\int \cos(x)dx=\sin(x)+C$

Simplifying the above results, we get $[-14\pi-0]+[0-0]=-14\pi$ .

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Example Question #21 : Definite Integrals

Find the area between y=sin(x) and y=sec^2(x) between $x:(0, \frac{\pi}{4})$

Possible Answers:

$\frac{\sqrt2}{2}+2$

$\frac{\sqrt2}{2}$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

We can write this problem as:

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx$

Integrating:

$\int(sec^2(x)-sin(x))dx=tan(x)+cos(x)+C=F(x)$

By the fundamental theorem of calculus:

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx=F(\frac{\pi}{4})-F(0)$

$=[tan(\frac{\pi}{4})+cos(\frac{\pi}{4})]-[tan(0)+cos(0)]=[1+\frac{\sqrt2}{2}]-[1]$

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx=\frac{\sqrt2}{2}$

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Example Question #391 : Integrals

$Evaluate:\int_{0}^{5}(x^2+x)dx$

Possible Answers:

$\frac{325}{6}$

$\frac{175}{6}$

$\frac{200}{6}$

$\frac{325}{2}$

$\frac{250}{3}$

Correct answer:

$\frac{325}{6}$

Explanation:

Compute the Indefinite Integral

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

Evaluate the integral

$(\frac{5^3}{3}+\frac{5^2}{2})-(\frac{0^3}{3}+\frac{0^2}{2})$

$(\frac{125}{3}+\frac{25}{2})-(0+0)$

$\frac{250}{6}+\frac{75}{6}=\frac{325}{6}$

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Example Question #29 : Definite Integrals

Suppose u=ax , where is a constant

Find such that $\int_{0}^{2} 5u dx =2$

Possible Answers:

u=x

u=5x

$u=\frac{1}{5}x$

$u=\frac{5}{2}x$

Correct answer:

$u=\frac{1}{5}x$

Explanation:

$\int 5ax dx= \frac{5}{2}ax^2 +C= F(x)$

By the fundamental theorem of calculus:

$\int_{0}^{2} 5u dx =F(2)-F(0)=2$

$\frac{5}{2}a*2^2-0=2$

10a=2

$a=\frac{1}{5}$

$u=\frac{1}{5}x$

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Example Question #21 : Definite Integrals

Evaluate this integral.

$\int_{2}^{0} 3+4x dx$

Possible Answers:

Answer not listed

Correct answer:

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

In this case, f(x) = 3 + 4x .

The antiderivative is F(x) = 3x + 2x^2 .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{2}^{0} 3 + 4x dx = \left (3x + 2x^2 \right ) _{2}^{0}$

$= 3(0) + 2(0)^2 - \left (3(2) + 2(2)^2 \right )$

= - 14

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Definite Integrals

Example Question #22 : Definite Integrals

Example Question #23 : Definite Integrals

Example Question #21 : Definite Integrals

Example Question #25 : Definite Integrals

Example Question #26 : Definite Integrals

Example Question #21 : Definite Integrals

Example Question #391 : Integrals

Example Question #29 : Definite Integrals

Example Question #21 : Definite Integrals

All Calculus 2 Resources

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