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Calculus 2 : Integrals

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Example Questions

Example Question #261 : Integrals

Find the area under the curve for f(x)=14x^3+5x^2 from x=-2 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}14x^3+5x^2dx$

Solution:

$\int14x^3+5x^2dx$

$\frac{14}{4}x^4+\frac{5}{3}x^3\Big|_{-2}^{2}$

$(\frac{7}{2}2^4 + \frac{5}{3}2^3)-(\frac{7}{2}(-2)^4+\frac{5}{3}(-2)^3)$

$(\frac{7}{2}\cdot16+\frac{5}{3}\cdot8)-(\frac{7}{2}\cdot16-\frac{5}{3}\cdot8)$

$(56+\frac{40}{3})-(56-\frac{40}{3}) = \frac{80}{3} = 26.67$

When rounded, the area under the curve is

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Example Question #2011 : Calculus Ii

Find the area under the curve for f(x)=3x^2+e^x from x=0 to x=2

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{2}3x^2+e^xdx$

Solution:

$\int3x^2+e^xdx$

$\frac{3}{3}x^3+e^x\Big|_{0}^{2}$

$x^3+e^x\Big|_{0}^{2}$

(2^3+e^2)-(e^0)

8+e^2-1 = e^2 + 7 = 14.389

When rounded to the nearest integer, the area under the curve is

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Example Question #88 : Integral Applications

Find the area under the curve for $f(x)=7x+e^{2x}$ from x=0 to x=2 , when rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{2}7x+e^{2x}dx$

Solution:

$\int7x+e^{2x}dx$

$\frac{7}{2}x^2+\frac{1}{2}e^{2x}\Big|_{0}^{2}$

$(\frac{7}{2}\cdot2^2+\frac{1}{2}e^{2\cdot2})-(\frac{1}{2})$

$14+\frac{1}{2}e^4-\frac{1}{2} = \frac{27}{2}+\frac{1}{2}e^4 = 40.799$

When rounded, the area under the curve is

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Example Question #2012 : Calculus Ii

Find the area under the curve for f(x)=|4x+x^2| from x=-2 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

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

Solution:

This function is negative from x=[-2,0], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[-2,0] region by -1, and sum everything up.

1st Piece:

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

$-2x^2-\frac{1}{3}x^3\Big|_{-2}^{0}$

$0-(-2\cdot4-\frac{1}{3}\cdot(-8)) = -(-8+\frac{8}{3}) = 8 - \frac{8}{3} = \frac{16}{3}$

2nd piece:

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

$2x^2+\frac{1}{3}x^3\Big|_{0}^{2}$

$2\cdot4+\frac{8}{3} = 8+\frac{8}{3} = \frac{32}{3}$

Sum:

$\frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16$

The area under the curve is

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Example Question #89 : Integral Applications

Find the area under the curve for f(x)=|-6x+2x^2| from x=-3 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

inding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}|-6x+2x^2|dx$

Solution:

This function is negative from x=[0,2], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,2] region by -1, and sum everything up.

1st piece:

$\int_{-3}^{0}|-6x+2x^2|dx$

$-3x^2+\frac{2}{3}x^3\Big|_{-3}^{0}$

$0-(-3\cdot(-3)^2+\frac{2}{3}\cdot(-3)^3)$

$9+\frac{2}{3}\cdot27 = 9+18=27$

2nd piece:

$\int_{0}^{2}|-6x+2x^2|dx$

$3x^2-\frac{2}{3}x^3\Big|_{0}^{2}$

$3*4-\frac{2}{3}\cdot8 = 12-\frac{16}{3}$

sum:

$27+12-\frac{16}{3} = \frac{155}{3} = 51.667$

When rounded to the nearest integer, the area under the curve is

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Example Question #33 : Area Under A Curve

Find the area under the curve for f(x)=|-3x+2x^2+x^3| from x=-1 to x=1 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

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

Solution:

This function is negative from x=[0,1] , and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,1] region by , and sum everything up.

First piece:

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

$-\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{4}x^4\Big|_{-1}^{0}$

$0-(-\frac{3}{2}-\frac{2}{3}+\frac{1}{4}) = \frac{3}{2}+\frac{2}{3}-\frac{1}{4} = \frac{23}{12}$

Second piece:

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

$\frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4\Big|_{0}^{1}$

$\frac{3}{2}-\frac{2}{3}-\frac{1}{4} = \frac{7}{12}$

Sum:

$\frac{23}{12}+\frac{7}{12} = \frac{30}{12} = \frac{5}{2}=2.5$

When rounded to the nearest integer, the area under the curve is

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Example Question #91 : Integral Applications

Find the area under the curve for f(x)=sin(x)+1 from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sin(x)+1dx$

Solution:

$\int sin(x)+1dx$

$x-cos(x)\Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

$({\frac{\pi}{2}}-cos({\frac{\pi}{2}}))-({\frac{\pi}{4}} - cos({\frac{\pi}{4}}))$

$\frac{\pi}{4} + \frac{1}{\sqrt2} = 1.4925$

Rounded to the nearest integer, the area under the curve is

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Example Question #91 : Integral Applications

Find the area under the curve for f(x)=2sin(x)+x from x=0 to $x=\pi$

Possible Answers:

$\frac{1}{2}\pi^2 + 6$

$\pi^2 + 2$

$\frac{1}{2}\pi + 2$

$\frac{1}{2}\pi^2 + 4$

Correct answer:

$\frac{1}{2}\pi^2 + 4$

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{\pi}2sin(x)+xdx$

Solution:

$\int2sin(x)+xdx$

$\frac{1}{2}x^2-2cos(x)\Big|_{0}^{\pi}$

$\frac{1}{2}\pi^2-2\cdot cos(\pi)-(-2cos(0)) = \frac{1}{2}\pi^2-2\cdot cos(\pi) + 2cos(0)$

$\frac{1}{2}\pi^2-2\cdot-1 + 2 = \frac{1}{2}\pi^2 + 4$

The area under the curve is $\frac{1}{2}\pi^2 + 4$

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Example Question #36 : Area Under A Curve

Find the area under the curve for f(x)=|3x^2-4x| from x=-2 to x=2

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

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

Solution:

This function is negative from $x=[0,\frac{4}{3}],$ and positve everywhere else. Split this integral up into 3 pieces, multiplying x=[0,\frac{4}{3}] region by -1, and sum everything up.

1st piece:

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

$3\cdot\frac{1}{3}x^3-\frac{4}{2}x^2\Big|_{-2}^{0}$

$x^3-2x^2\Big|_{-2}^{0}$

$0-((-2)^3-2\cdot4) = 8+8 = 16$

2nd piece:

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

$2x^2 - x^3\Big|_{0}^{\frac{4}{3}}$

$2\cdot(\frac{4}{3})^2 - (\frac{4}{3})^3 = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

3rd piece:

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

$x^3 - 2x^2\Big|_{\frac{4}{3}}^{2}$

$(8-2\cdot4)-((\frac{4}{3})^3-2\cdot(\frac{4}{3})^2) = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

Sum:

$16+\frac{32}{27} + \frac{32}{27} = \frac{496}{27} = 18.37$

When rounded to the nearest integer, the area under the curve is

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Example Question #37 : Area Under A Curve

Find the area under the curve for f(x)=(x+6)^2-36 from x=1 to x=4

Possible Answers:

111

123

107

127

Correct answer:

111

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{1}^{4} (x+6)^2-36dx$

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify the function

(x+6)^2-36

x^2+36+12x-36

x^2+12x

2. Evaluate the integral

$\int_{1}^{4} x^2+12xdx$

$\frac{1}{3}x^3+6x^2\Big|_{1}^{4}$

$\frac{64}{3}+6\cdot16-(\frac{1}{3}+6) = 111$

The area under the curve is 111

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #261 : Integrals

Example Question #2011 : Calculus Ii

Example Question #88 : Integral Applications

Example Question #2012 : Calculus Ii

Example Question #89 : Integral Applications

Example Question #33 : Area Under A Curve

Example Question #91 : Integral Applications

Example Question #91 : Integral Applications

Example Question #36 : Area Under A Curve

Example Question #37 : Area Under A Curve

All Calculus 2 Resources

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