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Calculus 2 : Integrals

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Example Questions

Example Question #92 : Definite Integrals

Evaluate.

$\int_{0}^{1} \cos(6x) \ dx$

Possible Answers:

1.59

-2.17

0.36

Answer not listed.

-0.047

Correct answer:

-0.047

Explanation:

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

In this case, $f(x) = \cos(6x)$ .

The antiderivative is $F(x) = \frac{\sin(6x)}{6}$ .

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

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

$= \frac{\sin(6(1))}{6} - \left ( \frac{\sin(6(0))}{6} \right )$

= -0.047

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Example Question #461 : Integrals

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

Possible Answers:

1.5

2.5

Correct answer:

Explanation:

First, integrate the expression. Remember to raise the exponent by 1 and then put that result on the denominator as well: $\frac{2x^4}{4}-\frac{4x^2}{2}+x=\frac{x^4}{2}-2x^2+x$ . Evaluate at 2 and then 0. Subtract the results. $(\frac{2^4}{2}-8+2)-0=2$ .

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Example Question #462 : Integrals

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

Possible Answers:

$\frac{140}{4}$

$\frac{140}{3}$

$\frac{131}{3}$

$\frac{137}{2}$

Correct answer:

$\frac{140}{3}$

Explanation:

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

Therefore, after integrating, it should look like:

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

Then, first evaluate at 4 and then 0.

Subtract the results:

$(\frac{2(64)}{3}-8+12)-0=\frac{128}{3}+4=\frac{140}{3}$ .

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Example Question #463 : Integrals

$\int_{0}^{2}9x^2-3x+1dx$

Possible Answers:

Correct answer:

Explanation:

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

$9(\frac{x^3}{3})-\frac{3x^2}{2}+x=3x^3-\frac{3x^2}{2}+x$ .

Then, evaluate at 2 and then 0.

Subtract the two:

3(2^3)-3(2)+2=24-6+2=20.

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Example Question #102 : Definite Integrals

What is the area underneath the curve f(x)=x^2-3x+1 from x=4 to x=6 and bounded by the axis?

Possible Answers:

$\frac{68}{5}$

$\frac{67}{3}$

$\frac{68}{3}$

Correct answer:

$\frac{68}{3}$

Explanation:

First set up the integral expression:

$\int_{4}^{6}x^2-3x+1dx$ .

Then, integrate each term separately. Remember to raise the exponent by 1 and then put that result on the denominator as well:

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

Then evaluate at 6 and then 4. Subtract the results.

$(\frac{216}{3}-\frac{108}{2}+6)-(\frac{64}{3}-\frac{48}{2}+4)$ .

Simplify to get your answer:

$(72-54+6)-(\frac{64}{3}-24+4)=24-(\frac{64}{3}-20)=\frac{68}{3}$ .

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Example Question #464 : Integrals

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

Possible Answers:

2.0578

3.0794

1.0794

Correct answer:

3.0794

Explanation:

First, chop up the fraction into two separate terms:

$\int \frac{3}{x}+1dx$ .

Then, integrate those to get

$3ln\left | x \right |+x$ .

Then evaluate at 2 and then 1. Subtract the results:

(3ln2+2)-(3ln1+1)=4.0794-1=3.0794 .

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Example Question #465 : Integrals

Calculate the value of the definite integral
$\int_0^{\pi/4} sec^2x\, dx$

Possible Answers:

$-\pi$

Correct answer:

Explanation:

The antiderivative of sec^2x is $tan\,x$ .

Using the corollary to the Fundamental Theorem of Calculus we have

$\int_0^{\pi/4} sec^2x\, dx=tan\,x|_0^{\pi/4}=tan(\pi/4)-tan(0)=1-0=1$

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Example Question #105 : Definite Integrals

Calculate the value of the definite integral
$\int_4^7 \frac{2x}{1+x^2} \,dx$

Possible Answers:

$\frac{2}{3}$

$ln(\frac{7}{4})$

$=ln(\frac{50}{17})$

$\frac{7}{4}$

Correct answer:

$=ln(\frac{50}{17})$

Explanation:

To solve this definite integral, we use u-substitution.

u=1+x^2 and $du=2x\,dx$

We then solve the integral

$\int_{17}^{50} \frac{du}{u}=(ln|u||_{17}^{50}$

=ln50-ln17

$=ln(\frac{50}{17})$

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Example Question #106 : Definite Integrals

Evaluate.

$\int_{20}^{21} \frac{1}{15x- 250} \ dx$

Possible Answers:

0.06

0.04

Answer not listed.

0.02

0.12

Correct answer:

0.02

Explanation:

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

In this case, $f(x) = \frac{1}{15x- 250}$ .

The antiderivative is $F(x) =\frac{\ln(15x-250)}{15}$ .

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

$\int_{20}^{21} \frac{1}{15x- 250} \ dx = \left ( \frac{\ln(15x-250)}{15} \right )_{20}^{21}$

$= \frac{\ln(15(21)-250)}{15} - \left ( \frac{\ln(15(20)-250)}{15} \right )$

=0.02

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Example Question #107 : Definite Integrals

Evaluate.

$\int_{2}^{3}\sin (3x - 7) \ dx$

Possible Answers:

0.32

Answer not listed.

1.98

-1.24

0.53

Correct answer:

0.32

Explanation:

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

In this case, $f(x) = \sin (3x - 7)$ .

The antiderivative is $F(x) =-\frac{\cos(3x-7)}{3}$ .

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

$\int_{2}^{3}\sin (3x - 7) \ dx = \left ( -\frac{\cos(3x-7)}{3} \right )_{2}^{3}$

$= -\frac{\cos(3(3)-7)}{3} - \left ( -\frac{\cos(3(2)-7)}{3} \right )$

=0.32

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #92 : Definite Integrals

Example Question #461 : Integrals

Example Question #462 : Integrals

Example Question #463 : Integrals

Example Question #102 : Definite Integrals

Example Question #464 : Integrals

Example Question #465 : Integrals

Example Question #105 : Definite Integrals

Example Question #106 : Definite Integrals

Example Question #107 : Definite Integrals

All Calculus 2 Resources

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