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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 #52 : Definite Integrals

Evaluate.

$\int_{0}^1{} e^{3x- 2} \ dx$

Possible Answers:

0.86

1.24

-0.2i

Answer not listed.

0.34

Correct answer:

0.86

Explanation:

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

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

In this case, $f(x) = e^{3x- 2}$ .

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

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

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

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

= 0.86

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Example Question #53 : Definite Integrals

Evaluate.

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

Possible Answers:

0.42

Answer not listed.

0.98

1.23

-1.05

Correct answer:

0.42

Explanation:

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

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

In this case, $f(x) = \cos (2x-1)$ .

The antiderivative is $F(x) = \frac{\sin (2x-1)}{2}$ .

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

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

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

= 0.42

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Example Question #421 : Integrals

Evaluate.

$\int_{2}^{3}\frac{1}{4x-5} \ dx$

Possible Answers:

8.57

0.21

0.43

Answer not listed.

1.35

Correct answer:

0.21

Explanation:

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

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

In this case, $f(x) = \frac{1}{4x-5}$ .

The antiderivative is $F(x) = \frac{ \ln (4x-5)}{4}$ .

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

$\int_{2}^{3}\frac{1}{4x-5} \ dx = \left ( \frac{ \ln (4x-5)}{4} \right )_{2}^{3}$

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

= 0.21

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Example Question #71 : Finding Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

-1.18

1.35

6.22

-4.35

Correct answer:

-1.18

Explanation:

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

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

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

The antiderivative is $F(x) = \frac{ \sin(2x)}{2} - \frac{\ln(2x)}{2}$ .

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

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

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

= -1.18

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Example Question #72 : Finding Integrals

Evaluate.

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

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) = 2x + 3x^2 + 4x^3 .

The antiderivative is F(x) =x^2 + x^3 + x^4 .

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

$\int_{1}^{2} 2x + 3x^2 + 4x^3 \ dx = \left ( x^2 + x^3 + x^4 \right )_{1}^{2}$

$=(2)^2 + (2)^3 + (2)^4 - \left ( (1)^2 + (1)^3 + (1)^4 \right )$

= 25

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Example Question #71 : Finding Integrals

Evaluate.

$\int_{1}^{2} 3x^2 + 12x^{11} \ dx$

Possible Answers:

Answer not listed.

456

4102

392

Correct answer:

4102

Explanation:

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

In this case, $f(x) = 3x^2 + 12x^{11}$ .

The antiderivative is $F(x) = x^3 + x^{12}$ .

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

$\int_{1}^{2} 3x^2 + 12x^{11} \ dx = \left ( x^3 + x^{12} \right )_{1}^{2}$

$= (2)^3 + (2)^{12} - \left ( (1)^3 + (1)^{12} \right )$

= 4102

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Example Question #61 : Definite Integrals

Evaluate.

$\int_{0}^{1} e^{4x- 2} \ dx$

Possible Answers:

6.42

2.34

-3.26

1.81

Answer not listed.

Correct answer:

1.81

Explanation:

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

In this case, $f(x) = e^{4x- 2}$ .

The antiderivative is $F(x) = \frac{e^{4x- 2} }{4}$ .

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

$\int_{0}^{1} e^{4x- 2} \ dx = \left ( \frac{e^{4x- 2} }{4} \right )_{0}^{1}$

$= \frac{e^{4(1)- 2} }{4} - \left ( \frac{e^{4(0)- 2} }{4} \right )$

= 1.81

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Example Question #73 : Finding Integrals

Evaluate.

$\int_{1}^{2}\frac{1}{9x-8} \ dx$

Possible Answers:

1.04

Answer not listed.

1.67

0.26

0.93

Correct answer:

0.26

Explanation:

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

In this case, $f(x) = \frac{1}{9x-8}$ .

The antiderivative is $F(x) = \frac{\ln (9x-8)}{9}$ .

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

$\int_{1}^{2}\frac{1}{9x-8} \ dx = \left ( \frac{\ln (9x-8)}{9} \right )_{1}^{2}$

$= \frac{\ln (9(2)-8)}{9} - \left ( \frac{\ln (9(1)-8)}{9} \right )$

= 0.26

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Example Question #63 : Definite Integrals

Evaluate.

$\int_{0}^{\frac{\pi}{2}} -\cos(7x) \ dx$

Possible Answers:

$\frac{1}{2}$

$\frac{6}{7}$

$\frac{2}{3}$

$-\frac{1}{7}$

Answer not listed.

Correct answer:

$-\frac{1}{7}$

Explanation:

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

In this case, $f(x) =-\cos(7x)$ .

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

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

$\int_{0}^{\frac{\pi}{2}} -\cos(7x) \ dx = \left ( - \frac{\sin(7x)}{7} \right )_{0}^{\frac{\pi}{2}}$

$= - \frac{\sin(7(\frac{\pi}{2}))}{7} - \left ( - \frac{\sin(7(0))}{7} \right )$

$= -\frac{1}{7}$

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Example Question #421 : Integrals

Evaluate.

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

Possible Answers:

7.92

6.42

9.31

2.64

Answer not listed.

Correct answer:

7.92

Explanation:

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

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

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

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

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

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

= 7.92

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #52 : Definite Integrals

Example Question #53 : Definite Integrals

Example Question #421 : Integrals

Example Question #71 : Finding Integrals

Example Question #72 : Finding Integrals

Example Question #71 : Finding Integrals

Example Question #61 : Definite Integrals

Example Question #73 : Finding Integrals

Example Question #63 : Definite Integrals

Example Question #421 : Integrals

All Calculus 2 Resources

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