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Calculus 2 : Integrals

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Example Questions

Example Question #128 : Definite Integrals

If f{}'(x)=2x+1 and f(1)=3 , what is the original f(x) function?

Possible Answers:

f(x)=x^2+x+1

f(x)=x^2+x

f(x)=x^2-x+1

f(x)=2x^2+x+1

f(x)=x^2+x-1

Correct answer:

f(x)=x^2+x+1

Explanation:

First, set up the integral expression:

$\int 2x+1dx$

Now, integrate. Remember to raise the exponent by 1 and then put that result on the denominator:

$\frac{2x^2}{2}+x=x^2+x+C$

Plug in your initial conditions to find C:

1^2+1+C=3; C=1

Now plug back in to get your initial f(x) function:

f(x)=x^2+x+1

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Example Question #491 : Integrals

Evaluate.

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

Possible Answers:

Answer not listed.

$\frac{2}{3}$

$\frac{1}{5}$

$\frac{5}{7}$

$\frac{9}{8}$

Correct answer:

Answer not listed.

Explanation:

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

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

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

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

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

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

$= \frac{3}{4}$

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Example Question #492 : Integrals

Evaluate.

$\int_{12}^{14} e^ {x -7}\ dx$

Possible Answers:

573.8

1,579.4

948.2

Answer not listed.

232.4

Correct answer:

948.2

Explanation:

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

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

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

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

$\int_{12}^{14} e^ {x -7}\ dx = \left (e^ {x -7} \right )_{12}^{14}$

$= e^ {14 -7} - \left ( e^ {12 -7} \right )$

= 948.2

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Example Question #493 : Integrals

Evaluate.

$\int_{6}^{7} \frac{x-4}{x-5} \ dx$

Possible Answers:

45.2

Answer not listed.

3.5

18.3

16.7

Correct answer:

Answer not listed.

Explanation:

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

In this case, $f(x) =\frac{x-4}{x-5}$ .

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

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

$\int_{6}^{7} \frac{x-4}{x-5} \ dx = \left (\ln(x-5) + x \right )_{6}^{7}$

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

= 13.7

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Example Question #494 : Integrals

Evaluate.

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

Possible Answers:

0.02

3.8

-4.2

1.54

Answer not listed.

Correct answer:

0.02

Explanation:

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

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

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

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

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

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

= 0.02

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Example Question #495 : Integrals

Evaluate.

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

Possible Answers:

-0.06

2.04

Answer not listed.

1.38

-2.03

Correct answer:

-0.06

Explanation:

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

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

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

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

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

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

= -0.06

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Example Question #134 : Definite Integrals

Evaluate.

$\int_{2}^{3}\sec^2 (12x-1)\ dx$

Possible Answers:

4.26

2.74

0.94

0.02

Answer not listed.

Correct answer:

0.02

Explanation:

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

In this case, $f(x) =\sec^2 (12x-1)$ .

The antiderivative is $F(x) = \frac{\tan(12x-1)}{12}$ .

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

$\int_{2}^{3}\sec^2 (12x-1)\ dx = \left ( \frac{\tan(12x-1)}{12} \right )_{2}^{3}$

$= \frac{\tan(12(3)-1)}{12} - \left ( \frac{\tan(12(2)-1)}{12} \right )$

= 0.02

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Example Question #135 : Definite Integrals

$\int_{0}^{2}x^3+4x^2-x+1dx$

Possible Answers:

$\frac{44}{3}$

$\frac{41}{3}$

$\frac{43}{3}$

$\frac{38}{3}$

$\frac{50}{3}$

Correct answer:

$\frac{44}{3}$

Explanation:

First, integrate this expression. Remember to raise the exponent by 1 and then put that result on the denominator:

$\frac{x^4}{4}+\frac{4x^3}{3}-\frac{x^2}{2}+x$

Now, evaluate at 2 and then 0. Subtract the results:
$(\frac{16}{4}+\frac{32}{3}-\frac{4}{2}+2)-0=4+\frac{32}{3}-2+2=\frac{44}{3}$

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Example Question #496 : Integrals

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

Possible Answers:

2.3456

1.2950

3.5069

1.3506

1.8058

Correct answer:

1.3506

Explanation:

First, integrate this expression. Remember to add one to the exponent and then also put that result on the denominator:

$\frac{x^{\frac{7}{4}}}{\frac{7}{4}}$

Simplify:

$\frac{4}{7}(x^{\frac{7}{4}})$

Evaluate at 2 and then 1. Subtract the results:

$(\frac{4}{7}(2^{\frac{7}{4}}))-(\frac{4}{7}(1^{\frac{7}{4}}))$

Round to four places:

1.9220-.5714=1.3506

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Example Question #137 : Definite Integrals

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

Possible Answers:

14.4328

14.4315

14.4372

9.4315

12.4315

Correct answer:

14.4315

Explanation:

First, chop up the fraction into two separate terms:

$\int \frac{5}{x}+xdx$

Now, integrate:

$5ln\left | x \right |+\frac{x^2}{2}$

Evaluate at 4 and then at 1. Subtract the results:

$(5ln4+8)-(5ln1+\frac{1}{2})$

Round to four places:
14.9315-.5=14.4315

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #128 : Definite Integrals

Example Question #491 : Integrals

Example Question #492 : Integrals

Example Question #493 : Integrals

Example Question #494 : Integrals

Example Question #495 : Integrals

Example Question #134 : Definite Integrals

Example Question #135 : Definite Integrals

Example Question #496 : Integrals

Example Question #137 : Definite Integrals

All Calculus 2 Resources

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