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Calculus 2 : Definite Integrals

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Example Questions

Example Question #141 : Definite Integrals

Evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ .

Possible Answers:

The integral cannot be evaluated.

$\pi$

$\frac{9\pi}{2}$

$\frac{7\pi}{2}$

Correct answer:

$\frac{9\pi}{2}$

Explanation:

This integral cannot be evaluated using the Fundamental Theorem of Calculus, since an antiderivative does not exist for $\sqrt{9-x^2}$ . Instead, we need to find the area under the curve $y=\sqrt{9-x^2}$ bounded by the axis.

$y=\sqrt{9-x^2}$ from -3<x<3 , is the graph of the upper half of the circle y^2+x^2=3^2 . The area of this upper half is

$\frac{\pi r^2}{2} = \frac{\pi(3)^2}{2} = \frac{9\pi}{2}$ .

Hence $\int_{-3}^3 \sqrt{9-x^2}dx=\frac{9\pi}{2}$ .

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Example Question #142 : Definite Integrals

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

Possible Answers:

1.5863

3.3898

3.3863

3.3274

2.3863

Correct answer:

3.3863

Explanation:

First, chop up the fraction into two separate fractions:

$\int \frac{2}{x}+1dx$

Now, integrate:

$2ln\left | x \right |+x$

Next, evaluate at 2 and then at 0. Subtract the results:

2ln(2)+2-(2ln(0)+0)=3.3863 .

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Example Question #143 : Definite Integrals

$\int_{1}^{3}x^2-4xdx$

Possible Answers:

$\frac{94}{3}$

$\frac{14}{3}$

$\frac{26}{3}$

$\frac{74}{3}$

$\frac{22}{3}$

Correct answer:

$\frac{74}{3}$

Explanation:

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

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

Simplify to get:

$\frac{x^3}{3}-2x^2$

Now, evaluate at 3 and then at 1. Subtract the results:
$(9+18)-(\frac{1}{3}+2)=27-(2\frac{1}{3})=\frac{74}{3}$

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Example Question #151 : Finding Integrals

$\int_{0}^{3}x^{\frac{4}{5}}dx$

Possible Answers:

3.0137

4.0137

4.0297

4.0283

4.0172

Correct answer:

4.0137

Explanation:

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

$\frac{x^{\frac{9}{5}}}{\frac{9}{5}}$

Simplify to get:

$\frac{5}{9}x^{\frac{9}{5}}$

Now, evaluate at 3 and then 0. Subtract the results:

$(\frac{5}{9}(3)^{\frac{9}{5}})-(\frac{5}{9}(0)^{\frac{9}{5}})=4.0137$

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Example Question #2252 : Calculus Ii

$\int_{1}^{2}3x^2+4x-1dx$

Possible Answers:

Correct answer:

Explanation:

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

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

Now, evaluate at 2 and then 1:

(8+8-2)-(1+2-1)=14-2=12 .

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Example Question #152 : Finding Integrals

$\int_{0}^{1}x^{\frac{1}{7}}dx$

Possible Answers:

$\frac{7}{8}$

$\frac{2}{3}$

$\frac{9}{8}$

$\frac{1}{8}$

$\frac{1}{4}$

Correct answer:

$\frac{7}{8}$

Explanation:

Recall that when integrating, you must raise the exponent by 1 and also put that result on the denominator:

$\frac{x^{\frac{8}{7}}}{\frac{8}{7}}$

Simplify to get:
$\frac{7}{8}x^{\frac{8}{7}}$

Now, evaluate at 1 and then at 0:

$(\frac{7}{8}1^{\frac{8}{7}})-(\frac{7}{8}0^{\frac{8}{7}})=\frac{7}{8}$

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Example Question #142 : Definite Integrals

$\int_{1}^{2}x^3+4x^2+xdx$

Possible Answers:

$\frac{75}{12}$

$\frac{175}{12}$

$\frac{179}{12}$

$\frac{175}{8}$

$\frac{173}{12}$

Correct answer:

$\frac{175}{12}$

Explanation:

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

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

Next, evaluate at 2 and then 1. Subtract the results:
$(4+\frac{32}{3}+2)-(\frac{1}{4}+\frac{4}{3}+\frac{1}{2})$

Simplify:

$(6+\frac{32}{3})-(\frac{50}{24})=\frac{175}{12}$

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Example Question #162 : Finding Integrals

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

Possible Answers:

1.9992

1.5452

1.5430

1.5474

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 #512 : 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 #142 : Definite Integrals

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

Possible Answers:

$\frac{1}{4}$

$\frac{37}{4}$

$\frac{39}{4}$

$\frac{35}{3}$

$\frac{35}{4}$

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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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #141 : Definite Integrals

Example Question #142 : Definite Integrals

Example Question #143 : Definite Integrals

Example Question #151 : Finding Integrals

Example Question #2252 : Calculus Ii

Example Question #152 : Finding Integrals

Example Question #142 : Definite Integrals

Example Question #162 : Finding Integrals

Example Question #512 : Integrals

Example Question #142 : Definite Integrals

All Calculus 2 Resources

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