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Calculus 2 : Finding Integrals

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Example Questions

Example Question #2195 : Calculus Ii

Evaluate.

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

Possible Answers:

12.74

Answer not listed.

3.26

0.37

0.46

Correct answer:

0.37

Explanation:

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

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

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

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

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

= 0.37

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Example Question #2201 : Calculus Ii

Evaluate.

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

Possible Answers:

Answer not listed

12.35

133.16

93.66

83.52

Correct answer:

Answer not listed

Explanation:

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

In this case, $f(x) = \frac{3}{12x-5}$ .

The antiderivative is $F(x) =\frac{\ln (12x-5)}{4}$ .

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

$\int_{0}^{1} \frac{3}{12x-5} \ dx = \left ( \frac{\ln (12x-5)}{4} \right )_{0}^{1}$

$= \frac{\ln (12(1)-5)}{4} - \left ( \frac{\ln (12(0)-5)}{4} \right )$

In this step it should be pointed out that natural log cannot be evaluated for values less than 1 thus there is no solution to this problem.

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Example Question #101 : Finding Integrals

Evaluate.

$\int_{0}^{1} \cos (2x-9) + 4 \ dx$

Possible Answers:

64.44

-24.39

-0.75

Answer not listed.

-4.12

Correct answer:

-4.12

Explanation:

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

In this case, $f(x) = \cos (2x-9) + 4$ .

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

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

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

$= \frac{\sin (2(1)-9)}{2} - 4(1) - \left ( \frac{\sin (2(0)-9)}{2} - 4(0) \right )$

= -4.12

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Example Question #101 : Finding Integrals

Evaluate.

$\int_{0}^{1} \sec^2(5x-7) \ dx$

Possible Answers:

0.36

1.35

0.61

-5.26

Answer not listed.

Correct answer:

0.61

Explanation:

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

In this case, $f(x) = \sec^2(5x-7)$ .

The antiderivative is $F(x) = \frac{\tan (5x-7)}{5}$ .

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

$\int_{0}^{1} \sec^2(5x-7) \ dx = \left ( \frac{\tan (5x-7)}{5} \right )_{0}^{1}$

$= \frac{\tan (5(1)-7)}{5} - \left ( \frac{\tan (5(0)-7)}{5} \right )$

= 0.61

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Example Question #102 : Finding Integrals

If $f{}'(x)=3x^2-1$ and f(1)=2 , what is f(x) ?

Possible Answers:

f(x)=x^2-2x+2

f(x)=x^2-x-2

f(x)=-x^2-x+2

f(x)=x^2-x+2

f(x)=x^2-x

Correct answer:

f(x)=x^2-x+2

Explanation:

First off, you're going to want to integrate the derivative to get the original function.

When integrating, raise the exponent by 1 and put that result on the denominator.

Therefore, the integral is

$\frac{3x^3}{3}-x=x^3-x+C$ .

Remember to add a C because it is an indefinite integral at this point.

To find C, plug in 1 for x and set the integral equal to 2 from your initial conditions:

1^3-3+c=2; c=2 .

Plug 2 in for C to get your function:

f(x)=x^2-x+2 .

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Example Question #103 : Finding Integrals

If a(t)=4t and v(1)=1, what is v(t) ?

Possible Answers:

v(t)=2t^2

v(t)=t^2-1

v(t)=2t^3-1

v(t)=2t^2+1

v(t)=2t^2-1

Correct answer:

v(t)=2t^2-1

Explanation:

Recall that the integral of acceleration is velocity.

Therefore, integrate the acceleration function first. Recall that you raise the exponent by 1 and then put that result on the denominator.

Therefore,

$\int 4t=\frac{4t^2}{2}=2t^2+C$ .

Remember to add a +C because it is an indefinite integral.

Then, plug in your initial conditions to find C:

2(1)+C=1; C=-1 .

Plug your value in for C to get an answer of

v(t)=2t^2-1 .

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Example Question #104 : Finding Integrals

Evaluate.

$\int_{1}^{2} \frac{1}{2}x - 30x^{14} \ dx$

Possible Answers:

- 65,533.25

- 658,149

Answer not listed.

12,572

91,498

Correct answer:

- 65,533.25

Explanation:

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

In this case, $f(x) = \frac{1}{2}x - 30x^{14}$ .

The antiderivative is $F(x) = \frac{x^2}{4} - 2x^{15}$ .

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

$\int_{1}^{2} \frac{1}{2}x - 30x^{14} \ dx = \left ( \frac{x^2}{4} - 2x^{15} \right )_{1}^{2}$

$= \frac{2^2}{4} - 2(2)^{15} - \left ( \frac{1^2}{4} - 2(1)^{15} \right )$

=- 65,533.25

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Example Question #105 : Finding Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

$\frac{25}{2}$

$\frac{13}{2}$

$\frac{1}{2}$

$\frac{27}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

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

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

The antiderivative is $F(x) = 12x + \frac{x^2}{2}$ .

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

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

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

$= \frac{27}{2}$

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Example Question #106 : Finding Integrals

Evaluate.

$\int_{6}^7{} \frac{3}{x-5} \ dx$

Possible Answers:

9.42

Answer not listed.

4.17

3.26

2.08

Correct answer:

2.08

Explanation:

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

In this case, $f(x) = \frac{3}{x-5}$ .

The antiderivative is $F(x) = 3\ln(x-5)$ .

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

$\int_{6}^7{} \frac{3}{x-5} \ dx = \left ( 3\ln(x-5) \right )_{6}^{7}$

$= 3\ln(7-5) - \left ( 3\ln(6-5) \right )$

= 2.08

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Example Question #107 : Finding Integrals

Evaluate.

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

Possible Answers:

156.5

Answer not listed.

65.2

23.4

-188.7

Correct answer:

156.5

Explanation:

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

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

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

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

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

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

= 156.5

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2195 : Calculus Ii

Example Question #2201 : Calculus Ii

Example Question #101 : Finding Integrals

Example Question #101 : Finding Integrals

Example Question #102 : Finding Integrals

Example Question #103 : Finding Integrals

Example Question #104 : Finding Integrals

Example Question #105 : Finding Integrals

Example Question #106 : Finding Integrals

Example Question #107 : Finding Integrals

All Calculus 2 Resources

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