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Calculus 2 : Definite Integrals

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Example Question #51 : Definite Integrals

Evaluate.

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

Possible Answers:

4.12

Answer not listed

5.63

2.46

3.24

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 #53 : Definite 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 #57 : Definite Integrals

Evaluate.

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

Possible Answers:

$- \frac{1}{2}$

$\frac{1}{6}$

$\frac{3}{5}$

Answer not listed

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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Example Question #58 : Definite Integrals

Evaluate.

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

Possible Answers:

-0.2i

1.24

0.86

0.34

Answer not listed.

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 #59 : Definite Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

-1.05

1.23

0.98

0.42

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 #60 : Definite Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

1.35

0.43

8.57

0.21

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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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Definite Integrals

Example Question #52 : Definite Integrals

Example Question #53 : Definite Integrals

Example Question #54 : Definite Integrals

Example Question #55 : Definite Integrals

Example Question #56 : Definite Integrals

Example Question #57 : Definite Integrals

Example Question #58 : Definite Integrals

Example Question #59 : Definite Integrals

Example Question #60 : Definite Integrals

All Calculus 2 Resources

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