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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #147 : Definite Integrals

$\int_{0}^{4}\frac{4-x}{x}dx$

Possible Answers:

1.5474

1.5430

1.9992

1.5452

1.3826

Correct answer:

1.5452

Explanation:

First, chop up the fraction into two separate terms:
$\int \frac{4}{x}-1dx$

Now, integrate:

$4ln\left | x \right |-x$

Now, evaluate at 4 and then 0. Subtract the results:

(4ln(4)-4)-(4ln(0)-0)=1.5452

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Example Question #148 : Definite Integrals

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

Possible Answers:

Correct answer:

Explanation:

First, integrate. Remember to add one to the exponent and then also put that result on the denominator:

$\frac{2x^4}{4}-\frac{3x^3}{3}+\frac{x^2}{2}-x=\frac{x^4}{2}-x^3+\frac{x^2}{2}-x$

Now, evaluate at 3 and then 1. Subtract the results:

$(\frac{81}{2}-27+\frac{9}{2}-3)-(\frac{1}{2}-1+\frac{1}{2}-1)$

Simplify:

$(\frac{90}{2}-30)-(-1)=16$

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Example Question #149 : Definite Integrals

$\int_{1}^{2}x^5-x^3+2x-1dx$

Possible Answers:

$\frac{35}{4}$

$\frac{35}{3}$

$\frac{37}{4}$

$\frac{1}{4}$

$\frac{39}{4}$

Correct answer:

$\frac{35}{4}$

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^6}{6}-\frac{x^4}{4}+\frac{2x^2}{2}-x=\frac{x^6}{6}-\frac{x^4}{4}+x^2-x$

Now, evaluate at 2 and then 1:

$(\frac{64}{6}-\frac{16}{4}+4-2)-(\frac{1}{6}-\frac{1}{4}+1-1)$

Simplify to get your answer:

$(\frac{64}{6}-2)-(-\frac{1}{12})=\frac{35}{4}$

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Example Question #151 : Definite Integrals

$\int_{0}^{5}\frac{x^2+x-1}{x}dx$

Possible Answers:

15.8906

15.8726

18.5906

15.8986

15.1906

Correct answer:

15.8906

Explanation:

First, chop up the fraction into three separate terms:

$\int x+1-\frac{1}{x}dx$

Next, integrate:

$\frac{x^2}{2}+x-ln\left | x \right |$

Evaluate at 5 and then 0. Subtract the results:

$(\frac{25}{2}+5-ln5)-(0)=15.8906$

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Example Question #152 : Definite Integrals

What is the integral of $\int_{4}^{8} \frac{2}{x^2}dx$ ?

Possible Answers:

$\frac{1}{2}$

$\frac{3}{4}$

Correct answer:

$\frac{3}{4}$

Explanation:

Rewriting the integral as an easy-to-integrate indefinite integral gives us $\int 2(x^{-2})$ , for which we can just use the power rule to get: $-2x^{-1}=-\frac{2}{x}$ .

Plugging in the values gives us $-\frac{2}{8}+\frac{2}{4}=-\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ .

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Example Question #153 : Definite Integrals

Which of the following is $\int_{0}^{1} x^{\tfrac{1}{3}}dx$ ?

Possible Answers:

$\frac{3}{4}$

$\frac{1}{3}$

$\frac{4}{3}$

Correct answer:

$\frac{3}{4}$

Explanation:

If we solve for the indefinite integral $\int x^\frac{1}{3}$ using the power rule, we get $\frac{3}{4}x^\frac{4}{3}$ . If we then plug in the values, we get $\frac{3}{4}(1)^\frac{4}{3}-\frac{3}{4}(0)^\frac{4}{3}=\frac{3}{4}(1)-\frac{3}{4}(0)=\frac{3}{4}-0=\frac{3}{4}$ .

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Example Question #154 : Definite Integrals

Evaluate the definite integral

$\int_0^{\pi/4}sec^2x\,dx$

Possible Answers:

$\frac{1}{2}$

$\sqrt{2}$

Correct answer:

Explanation:

The antiderivative of sec^2x is $tan\,x$ .

Using the Fundamental Theorem of Calculus,

$\int_0^{\pi/4}sec^2x\,dx=tan\,x|_0^{\pi/4}$

$=tan\,(\frac{\pi}{4})-tan(0)$

=1-0

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Example Question #155 : Definite Integrals

Evaluate the definite integral

$\int_0^{1}\frac{1}{x^2+1}\,dx$

Possible Answers:

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{3}{2}$

$\pi$

Correct answer:

$\frac{\pi}{4}$

Explanation:

For this problem we use the fact that

$\int_a^b\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}(x)|_a^b$

As such,

$\int_0^1\frac{1}{x^2+1}dx=\frac{1}{1}tan^{-1}(x)|_0^1$

$=tan^{-1}(x)|_0^1$

$=tan^{-1}(1)-tan^{-1}(0)$

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

$=\frac{\pi}{4}$

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Example Question #156 : Definite Integrals

Evaluate

$\small \int_0^1 e^{5x+1}dx$

Possible Answers:

$\small \frac{1}{5}(e^5-e)$

$\small \frac{1}{5}(e^6-e)$

$\small \frac{1}{5}(e^6-e^2)$

$\small \frac{1}{5}(e^6-e^5)$

Correct answer:

$\small \frac{1}{5}(e^6-e)$

Explanation:

To evaluate

$\small \int_0^1 e^{5x+1}dx$

we take its antiderivative $\small F(x)$ and calculate $\small F(1)-F(0)$ . With $\small F(x)=\frac{1}{5}e^{5x+1}$ , we get

$\small \small \int_0^1 e^{5x+1}dx=\frac{1}{5}(e^{5+1}-e^{0+1})=\frac{1}{5}(e^6-e)$

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Example Question #157 : Definite Integrals

Evaluate

$\small \small \int_0^\pi 6x^2dx$

Possible Answers:

$6\pi^3$

$\small 3\pi^2$

$2\pi^3$

$\small 6\pi^2$

Correct answer:

$2\pi^3$

Explanation:

To evaluate

$\small \small \int_0^\pi 6x^2dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small F(\pi)-F(0)$ . With $\small \small F(x)=2x^3$ , we get

$\small \small \small \int_0^\pi 6x^2dx=2(\pi^3-0^3)=2\pi^3$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #147 : Definite Integrals

Example Question #148 : Definite Integrals

Example Question #149 : Definite Integrals

Example Question #151 : Definite Integrals

Example Question #152 : Definite Integrals

Example Question #153 : Definite Integrals

Example Question #154 : Definite Integrals

Example Question #155 : Definite Integrals

Example Question #156 : Definite Integrals

Example Question #157 : Definite Integrals

All Calculus 2 Resources

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