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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 #431 : Integrals

Evaluate.

$\int_{1}^{2}5x^4 + 12x \ dx$

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

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

In this case, f(x) =5x^4 + 12x .

The antiderivative is F(x) = x^5 + 6x^2 .

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

$\int_{1}^{2}5x^4 + 12x \ dx = \left ( x^5 + 6x^2 \right )_{1}^{2}$

$= (2)^5 + 6(2)^2 - \left ( (1)^5 + 6(1)^2 \right )$

= 49

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Example Question #432 : Integrals

Evaluate.

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

Possible Answers:

5.42

3.25

8.83

2.77

Answer not listed

Correct answer:

2.77

Explanation:

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

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

The antiderivative is $F(x) = 4 \ln(x-1)$ .

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

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

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

= 2.77

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Example Question #433 : Integrals

Evaluate.

$\int_{1}^{2}12 \ dx$

Possible Answers:

Answer not listed

144

Correct answer:

Explanation:

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

In this case, f(x) = 12 .

The antiderivative is F(x) = 12x .

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

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

$= 12(2) - \left ( 12(1) \right )$

= 12

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Example Question #434 : Integrals

Evaluate.

$\int_{0}^{1}-\sin(6x) +3 \ dx$

Round to the nearest whole number.

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

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

In this case, $f(x) = -\sin(6x) +3$ .

The antiderivative is $F(x) = \frac{\cos(6x)}{6} + 3x$ .

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

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

$= \frac{\cos(6(1))}{6} + 3(1) - \left ( \frac{\cos(6(0))}{6} + 3(0) \right )$

= 3

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Example Question #71 : Definite Integrals

Evaluate.

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

Possible Answers:

4.25

-2.37

0.32

2.77

Answer not listed.

Correct answer:

2.77

Explanation:

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

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

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

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

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

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

= 2.77

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Example Question #436 : Integrals

Evaluate.

$\int_{0}^{1} e^{7x} - 7 \ dx$

Possible Answers:

100.35

149.52

Answer not listed.

42.63

74.79

Correct answer:

149.52

Explanation:

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

In this case, $f(x) = e^{7x} + 7$ .

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

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

$\int_{0}^{1} e^{7x} - 7 \ dx = \left ( \frac{e^{7x}}{7} - 7x \right )_{0}^{1}$

$= \frac{e^{7(1)}}{7} - 7(1) - \left ( \frac{e^{7(0)}}{7} - 7(0) \right )$

= 149.52

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Example Question #437 : Integrals

Evaluate.

$\int_{1}^{2} \sin(x) + 6 - \frac{1}{5x}\ dx$

Possible Answers:

Answer not listed.

4.27

6.82

5.79

8.36

Correct answer:

6.82

Explanation:

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

In this case, $f(x) = \sin(x) + 6 - \frac{1}{5x}$ .

The antiderivative is $F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$ .

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

$\int_{1}^{2} \sin(x) + 6 - \frac{1}{5x}\ dx = \left ( - \cos(x) + 6x - \frac{\ln (x)}{5} \right )_{1}^{2}$

$= - \cos(2) + 6(2) - \frac{\ln (2)}{5} - \left ( - \cos(1) + 6(1) - \frac{\ln (1)}{5} \right )$

= 6.82

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Example Question #438 : Integrals

Evaluate.

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

Possible Answers:

12.7

20.2

6.1

89.3

Answer not listed.

Correct answer:

20.2

Explanation:

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

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

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

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

$= \frac{3^4}{4} - \frac{\sin (2(3))}{2} - \left ( \frac{0^4}{4} - \frac{\sin (2(0))}{2} \right )$

= 20.2

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Example Question #439 : Integrals

Evaluate.

$\int_{0}^{1} 5e^{5x} \ dx$

Possible Answers:

67.2

12.5

Answer not listed.

102.4

147.4

Correct answer:

147.4

Explanation:

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

The antiderivative is $F(x) = e^{5x}$ .

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

$\int_{0}^{1} 5e^{5x} \ dx = \left ( e^{5x}\right )_{0}^{1}$

$= e^{5(1)} - \left ( e^{5(0)} \right )$

= 147.4

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Example Question #440 : Integrals

Evaluate.

$\int_{0}^{1} \frac{1}{6x+5} \ dx$

Possible Answers:

0.98

4.64

0.13

Answer not listed.

1.35

Correct answer:

0.13

Explanation:

In this case, $f(x) = \frac{1}{6x+5}$ .

The antiderivative is $F(x) = \frac{\ln(6x+5)}{6}$ .

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

$\int_{0}^{1} \frac{1}{6x+5} \ dx = \left ( \frac{\ln(6x+5)}{6} \right )_{0}^{1}$

$= \frac{\ln(6(1)+5)}{6} - \left ( \frac{\ln(6(0)+5)}{6} \right )$

= 0.13

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #431 : Integrals

Example Question #432 : Integrals

Example Question #433 : Integrals

Example Question #434 : Integrals

Example Question #71 : Definite Integrals

Example Question #436 : Integrals

Example Question #437 : Integrals

Example Question #438 : Integrals

Example Question #439 : Integrals

Example Question #440 : Integrals

All Calculus 2 Resources

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