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Calculus 2 : Finding Integrals

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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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Example Question #61 : Definite Integrals

Evaluate.

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

Possible Answers:

-1.18

-4.35

6.22

Answer not listed.

1.35

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 #62 : Definite 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 #63 : Definite Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

392

456

4102

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 #64 : Definite Integrals

Evaluate.

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

Possible Answers:

2.34

6.42

-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 #65 : Definite Integrals

Evaluate.

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

Possible Answers:

0.93

1.67

1.04

Answer not listed.

0.26

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 #66 : Definite Integrals

Evaluate.

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

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{7}$

$\frac{2}{3}$

$\frac{6}{7}$

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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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #57 : Definite Integrals

Example Question #58 : Definite Integrals

Example Question #59 : Definite Integrals

Example Question #60 : Definite Integrals

Example Question #61 : Definite Integrals

Example Question #62 : Definite Integrals

Example Question #63 : Definite Integrals

Example Question #64 : Definite Integrals

Example Question #65 : Definite Integrals

Example Question #66 : Definite Integrals

All Calculus 2 Resources

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