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Calculus 2 : Integrals

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Example Questions

Example Question #118 : Definite Integrals

Evaluate.

$\int_{3}^{4}e^{2x-5} \ dx$

Possible Answers:

9.3

4.1

5.8

2.4

8.7

Correct answer:

8.7

Explanation:

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

In this case, $f(x) = e^{2x-5}$ .

The antiderivative is $F(x) = \frac{e^{2x-5} }{2}$ .

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

$\int_{3}^{4}e^{2x-5} \ dx= \left ( \frac{e^{2x-5} }{2} \right )_{3}^{4}$

$=\frac{e^{2(4)-5} }{2} - \left ( \frac{e^{2(3)-5} }{2} \right )$

= 8.7

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Example Question #119 : Definite Integrals

Evaluate.

$\int_{1}^{2} \sin (5x-3) \ dx$

Possible Answers:

1.023

2.421

-0.545

0.935

0.001

Correct answer:

0.001

Explanation:

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

In this case, $f(x) = \sin (5x-3)$ .

The antiderivative is $F(x) = -\frac{ \cos(5x-3) }{5}$ .

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

$\int_{1}^{2} \sin (5x-3) \ dx = \left ( -\frac{ \cos(5x-3) }{5} \right )_{1}^{2}$

$=-\frac{ \cos(5(2)-3) }{5} - \left ( -\frac{ \cos(5(1)-3) }{5} \right )$

= 0.001

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Example Question #481 : Integrals

Evaluate.

$\int_{0}^{1}\sec^2(3x) \ dx$

Possible Answers:

-0.05

Answer not listed.

1.09

-0.35

1.24

Correct answer:

-0.05

Explanation:

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

In this case, $f(x) = \sec^2(3x)$ .

The antiderivative is $F(x) = \frac{ \tan(3x) }{3}$ .

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

$\int_{0}^{1}\sec^2(3x) \ dx = \left ( \frac{ \tan(3x) }{3} \right )_{0}^{1}$

$= \frac{ \tan(3(1)) }{3} - \left ( \frac{ \tan(3(0)) }{3} \right )$

= -0.05

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Example Question #121 : Definite Integrals

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

Possible Answers:

$\frac{58}{3}$

$\frac{5}{3}$

$\frac{32}{3}$

$\frac{59}{3}$

$\frac{7}{3}$

Correct answer:

$\frac{32}{3}$

Explanation:

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

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

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

$(\frac{81}{4}-2(9)+12)-(\frac{1}{4}-\frac{2}{3}+4)$

Simplify to get your answer:

$(\frac{81}{4}+\frac{5}{12}-6-4)=\frac{32}{3}$ .

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Example Question #122 : Definite Integrals

$\int_{0}^{4}x^{\frac{1}{3}}dx$

Possible Answers:

$3\sqrt[3]{4}$

$3\sqrt{2}$

$\sqrt[3]{4}$

$3\sqrt{4}$

Correct answer:

$3\sqrt[3]{4}$

Explanation:

First, integrate this expression, remembering to raise the exponent by 1 and then put that result on the denominator:

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

Simplify to get:

$(\frac{3}{4})x^{\frac{4}{3}}$

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

$\frac{3}{4}(4^{\frac{4}{3}})-0=\frac{3}{4}(4\sqrt[3]{4})=3\sqrt[3]{4}$

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Example Question #123 : Definite Integrals

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

Possible Answers:

54.75

54.25

54.87

53.75

52.75

Correct answer:

54.75

Explanation:

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

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

Now, evaluate at 4 and then 1. Subtract the results.

$((\frac{4^4}{4})-16+8)-(\frac{1}{4}-1+2)=56-1-\frac{1}{4}=54.75$ .

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Example Question #121 : Definite Integrals

$\int_{1}^{2}2t^2-4t+3dt$

Possible Answers:

$\frac{5}{3}$

$\frac{11}{3}$

$\frac{1}{3}$

$\frac{5}{2}$

$\frac{4}{3}$

Correct answer:

$\frac{5}{3}$

Explanation:

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

$2(\frac{t^3}{3})-4(\frac{t^2}{2})+3t=\frac{2}{3}t^3-2t^2+3t$

Then, evaluate at 2 and then 1. Subtract the results:

$((\frac{2}{3})8-8+6)-(\frac{2}{3}-2+3)=(\frac{16}{3}-2)-(\frac{2}{3}+1)=\frac{5}{3}$ .

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Example Question #125 : Definite Integrals

$\int_{0}^{4}t^2-t+1dt$

Possible Answers:

$\frac{64}{3}$

$\frac{52}{3}$

$\frac{1}{3}$

$\frac{50}{3}$

$\frac{49}{3}$

Correct answer:

$\frac{52}{3}$

Explanation:

First, integrate this expression, remembering to raise the exponent by 1 and then also putting that result on the denominator:

$\frac{t^3}{3}-\frac{t^2}{2}+t$

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

$(\frac{64}{3}-8+4)-0=\frac{64}{3}-4=\frac{52}{3}$

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Example Question #126 : Definite Integrals

$\int_{0}^{3}t^3+5t^2+tdt$

Possible Answers:

$\frac{243}{4}$

$\frac{207}{4}$

$\frac{199}{4}$

$\frac{201}{4}$

$\frac{241}{4}$

Correct answer:

$\frac{243}{4}$

Explanation:

First, integrate the expression, remembering to add one to the exponent and then put that result on the denominator:

$\frac{t^4}{4}+\frac{5t^3}{3}+\frac{t^2}{2}$

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

$(\frac{81}{4}+\frac{135}{3}+\frac{9}{2})-0=\frac{63}{4}+45=\frac{243}{4}$

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Example Question #127 : Definite Integrals

$\int_{0}^{2}t^{\frac{3}{2}}dt$

Possible Answers:

$\frac{2}{5}(2^{\frac{5}{2}})+C$

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

$\frac{2}{5}(2^{\frac{2}{5}})$

$\frac{1}{5}(2^{\frac{2}{5}})+C$

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

Correct answer:

$\frac{2}{5}(2^{\frac{5}{2}})+C$

Explanation:

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

$\frac{t^{\frac{5}{2}}}{\frac{5}{2}}$

Simplify to get:

$\frac{2}{5}t^{\frac{5}{2}}$

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

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

Now, add a C because it is an indefinite integral:
$\frac{2}{5}(2^{\frac{2}{5}})+C$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #118 : Definite Integrals

Example Question #119 : Definite Integrals

Example Question #481 : Integrals

Example Question #121 : Definite Integrals

Example Question #122 : Definite Integrals

Example Question #123 : Definite Integrals

Example Question #121 : Definite Integrals

Example Question #125 : Definite Integrals

Example Question #126 : Definite Integrals

Example Question #127 : Definite Integrals

All Calculus 2 Resources

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