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Calculus 2 : Calculus II

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

Example Questions

Example Question #75 : Definite Integrals

Evaluate.

$\int_{0}^{1} 3x - 5 \ dx$

Possible Answers:

-3.5

2.5

Answer not listed.

-4.5

-1.5

Correct answer:

-3.5

Explanation:

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

The antiderivative is $F(x) = \frac{3x^2}{2}-5x$ .

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

$\int_{0}^{1} 3x - 5 \ dx = \left ( \frac{3x^2}{2}-5x \right )_{0}^{1}$

$= \frac{3(1)^2}{2}-5(1) - \left ( \frac{3(0)^2}{2}-5(0) \right )$

=-3.5

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Example Question #91 : Finding Integrals

Evaluate.

$\int_{0}^{1} e^{8x} + \cos(4x) \ dx$

Possible Answers:

Answer not listed.

17.9

205.1

56.3

372.6

Correct answer:

372.6

Explanation:

In this case, $f(x) = e^{8x} + \cos(4x)$ .

The antiderivative is $F(x) =\frac{ \sin(4x)}{4} + \frac{e^{8x}}{8}$ .

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

$\int_{0}^{1} e^{8x} + \cos(4x) \ dx = \left ( \frac{ \sin(4x)}{4} + \frac{e^{8x}}{8} \right )_{0}^{1}$

$=\frac{ \sin(4(1))}{4} + \frac{e^{8(1)}}{8} - \left ( \frac{ \sin(4(0))}{4} + \frac{e^{8(0)}}{8} \right )$

= 372.6

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Example Question #72 : Definite Integrals

Evaluate.

$\int_{0}^{1} e^{9 - 2x}\ dx$

Possible Answers:

Answer not listed.

3503.2

126.5

352.4

43.7

Correct answer:

3503.2

Explanation:

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

In this case, $f(x) = e^{9 - 2x}$ .

The antiderivative is $F(x) =-\frac{ e^{9 - 2x}}{2}$ .

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

$\int_{0}^{1} e^{9 - 2x}\ dx = \left ( -\frac{ e^{9 - 2x}}{2} \right )_{0}^{1}$

$= -\frac{ e^{9 - 2(1)}}{2} - \left ( -\frac{ e^{9 - 2(0)}}{2} \right )$

= 3503.2

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Example Question #81 : Definite Integrals

Evaluate.

$\int_{1}^{2} x^6 - \frac{1}{3} x^{2} \ dx$

Possible Answers:

17.37

102.65

Answer not listed.

2.44

87.93

Correct answer:

17.37

Explanation:

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

In this case, $f(x) = x^6 - \frac{1}{3} x^{2}$ .

The antiderivative is $F(x) =\frac{x^7}{7} - \frac{x^3}{9}$ .

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

$\int_{1}^{2} x^6 - \frac{1}{3} x^{2} \ dx = \left ( \frac{x^7}{7} - \frac{x^3}{9} \right )_{1}^{2}$

$= \frac{2^7}{7} - \frac{2^3}{9} - \left ( \frac{1^7}{7} - \frac{1^3}{9} \right )$

= 17.37

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Example Question #442 : Integrals

Evaluate.

$\int_{1}^{2} 13,525 \ dx$

Possible Answers:

13,525

Answer not listed.

27,050

13,525x

Correct answer:

13,525

Explanation:

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

In this case, f(x) = 13,525 .

The antiderivative is F(x) = 13,525x .

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

$\int_{1}^{2} 13,525 \ dx = \left ( 13,525x \right )_{1}^{2}$

$= 13,525(2) - \left ( 13,525(1) \right )$

= 13,525

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Example Question #443 : Integrals

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

Possible Answers:

$\frac{69}{2}$

$\frac{63}{2}$

$\frac{71}{2}$

Correct answer:

$\frac{69}{2}$

Explanation:

First, integrate the function. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator. The first step should look like this: $3(\frac{x^3}{3})-\frac{x^2}{2}+4x=x^3-\frac{x^2}{2}+4x$ . Then, evaluate the function at 3 and subtract from the result when you plug in 0. $(27-\frac{9}{2}+12)-0=39-\frac{9}{2}=\frac{69}{2}$ .

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Example Question #2192 : Calculus Ii

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

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

To integrate, remember to raise the exponent by 1 and then put that result on the denominator: $\frac{x^4}{4}-\frac{3x^3}{3}+\frac{x^2}{2}=\frac{x^4}{4}-x^3+\frac{x^2}{2}$ . Then, evaluate at 3 and then 1. Subtract the two results. $(\frac{81}{4}-27+\frac{9}{2})-(\frac{1}{4}-1+\frac{1}{2})=(\frac{99}{4}-27)-(\frac{3}{4}-1)=-2$ .

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Example Question #91 : Finding Integrals

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

Possible Answers:

$\frac{4}{3}$

$-\frac{8}{3}$

$-\frac{4}{3}$

$-\frac{11}{3}$

$-\frac{2}{3}$

Correct answer:

$-\frac{4}{3}$

Explanation:

First, integrate the expression, remembering to add one to the exponent and then putting that result on the denominator: $\frac{x^3}{3}-\frac{3x^2}{2}+x$ . Then evaluate at 3 and then 1. Subtract the results: $(9-\frac{27}{2}+3)-(\frac{1}{3}-\frac{3}{2}+1)=(12-\frac{27}{2})-(-\frac{7}{6}+1)=11-\frac{74}{6}=\frac{-4}{3}$ .

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Example Question #2191 : Calculus Ii

Evaluate the following definite integral:

$\int^2_1\frac{3x^4-x^2+7}{x^2}dx$

Possible Answers:

$\frac{19}{2}$

$\frac{17}{2}$

$\frac{19}{4}$

$\frac{29}{6}$

$\frac{13}{2}$

Correct answer:

$\frac{19}{2}$

Explanation:

$\int^2_1\frac{3x^4-x^2+7}{x^2}dx=\int^2_1(3x^2-1+7x^{-2})dx=(x^3-x-\frac{7}{x})\left.\begin{matrix} \\ \end{matrix}\right| ^2_1=\frac{19}{2}$

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Example Question #82 : Definite Integrals

Evaluate.

$\int_{1}^{2}-\cos(3x) + \frac{1}{3x} \ dx$

Possible Answers:

3.26

0.46

0.37

Answer not listed.

12.74

Correct answer:

0.37

Explanation:

In this case, $f(x) = -\cos(3x) + \frac{1}{3x}$ .

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

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

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

$= \frac{\ln(2) - \sin(3(2))}{3} - \left ( \frac{\ln(1) - \sin(3(1))}{3} \right )$

= 0.37

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #75 : Definite Integrals

Example Question #91 : Finding Integrals

Example Question #72 : Definite Integrals

Example Question #81 : Definite Integrals

Example Question #442 : Integrals

Example Question #443 : Integrals

Example Question #2192 : Calculus Ii

Example Question #91 : Finding Integrals

Example Question #2191 : Calculus Ii

Example Question #82 : Definite Integrals

All Calculus 2 Resources

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