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Calculus 2 : Integrals

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

Example Questions

Example Question #44 : Definite Integrals

Evaluate.

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

Possible Answers:

-124.36

141.02

23.54

Answer not listed

65.21

Correct answer:

141.02

Explanation:

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

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

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

= 141.02

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Example Question #45 : Definite Integrals

Evaluate.

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

Possible Answers:

8.75

3.69

15.67

Answer not listed

2.14

Correct answer:

3.69

Explanation:

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

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

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

= 3.69

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Example Question #2163 : Calculus Ii

Evaluate.

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

Possible Answers:

-0.35

1.47

1.23

2.94

Answer not listed

Correct answer:

-0.35

Explanation:

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

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

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

= -0.35

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Example Question #2164 : Calculus Ii

Evaluate.

$\int_{0}^{1} 6x^2 + \sin(x) \ dx$

Possible Answers:

Answer not listed

5.63

4.12

3.24

2.46

Correct answer:

2.46

Explanation:

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

In this case, $f(x) = 6x^2 + \sin(x)$ .

The antiderivative is $F(x) = 2x^3 - \cos(x)$ .

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

$\int_{0}^{1} 6x^2 + \sin(x) \ dx = \left ( 2x^3 - \cos(x) \right )_{0}^{1}$

$= 2(1)^3 - \cos(1) - \left ( 2(0)^3 - \cos(0) \right )$

= 2.46

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Example Question #52 : Definite Integrals

Evaluate.

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

Possible Answers:

Answer not listed

0.31

0.47

1.06

1.23

Correct answer:

0.31

Explanation:

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

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

The antiderivative is $F(x) =\frac{\ln(3x-1)}{3}$ .

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

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

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

= 0.31

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Example Question #411 : Integrals

Evaluate.

$\int_{0}^{1} 12x^5 - 3x^2 \ dx$

Possible Answers:

Answer not listed

Correct answer:

Explanation:

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

In this case, f(x) = 12x^5 - 3x^2 .

The antiderivative is F(x) =2x^6 - x^3 .

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

$\int_{0}^{1} 12x^5 - 3x^2 \ dx = \left ( 2x^6 - x^3 \right )_{0}^{1}$

$= 2(1)^6 - (1)^3 - \left ( 2(0)^6 - (0)^3 \right )$

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Example Question #54 : Definite Integrals

Evaluate.

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

Possible Answers:

-0.98

0.63

0.42

-1.35

Answer not listed

Correct answer:

-0.98

Explanation:

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

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

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

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

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

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

=-0.98

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Example Question #55 : Definite Integrals

Evaluate.

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

Possible Answers:

$\frac{4}{5}$

$\frac{1}{3}$

$\frac{2}{5}$

Answer not listed

$\frac{1}{2}$

Correct answer:

$\frac{1}{3}$

Explanation:

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

In this case, f(x) = (x-1)^2 .

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

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

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

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

$=\frac{1}{3}$

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Example Question #56 : Definite Integrals

Evaluate.

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

Possible Answers:

0.02

0.06

0.12

Answer not listed

0.08

Correct answer:

0.06

Explanation:

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

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

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

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

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

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

= 0.06

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Example Question #51 : Definite Integrals

Evaluate.

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

Possible Answers:

Answer not listed

$- \frac{1}{2}$

$\frac{3}{5}$

$\frac{1}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

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

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

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

The antiderivative is $F(x) = \frac{1}{4}x^2 - \frac{1}{9} x^3$ .

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

$\int_{0}^{1} \frac{1}{2}x - \frac{1}{3} x^2 \ dx = \left ( \frac{1}{4}x^2 - \frac{1}{9} x^3 \right )_{0}^{1}$

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

$= \frac{1}{6}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #44 : Definite Integrals

Example Question #45 : Definite Integrals

Example Question #2163 : Calculus Ii

Example Question #2164 : Calculus Ii

Example Question #52 : Definite Integrals

Example Question #411 : Integrals

Example Question #54 : Definite Integrals

Example Question #55 : Definite Integrals

Example Question #56 : Definite Integrals

Example Question #51 : Definite Integrals

All Calculus 2 Resources

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