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Calculus 2 : Integral Applications

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Example Questions

Example Question #81 : Integral Applications

Find the area under the curve for f(x)=|2x^3-3x-10| from x=-2 to x=4

Possible Answers:

138

114

122

Correct answer:

122

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:

$\int_{-2}^{4}|2x^3-3x-10|dx$

Solution:

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

In other words, find the sum of these two integrals.

$\int_{-2}^{2}10+3x-2x^3dx + \int_{2}^{4}2x^3-3x-10dx$

First piece:

$\int_{-2}^{2}10+3x-2x^3dx = 10x+\frac{3}{2}x^2-\frac{2}{4}x^4\Big|_{-2}^{2}$

$=10\cdot2+\frac{3}{2}2^2-\frac{2}{4}2^4 - (10\cdot-2+\frac{3}{2}(-2)^2-\frac{2}{4}(-2)^4)$

=20+6-8-(-20+6-8) = 40

Second piece:

$\int_{2}^{4}2x^3-3x-10dx = \frac{2}{4}x^4-\frac{3}{2}x^2-10x\Big|_{2}^{4}$

$\frac{2}{4}\cdot256-\frac{3}{2}\cdot16 - 40 - (8-6-20) = 128-24-40+18 = 82$

Sum:

Add the 2 integrals together.

40+82 = 122

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Example Question #251 : Integrals

Find the area under the curve for f(x)=sin(x)+x^3 from x=0 to $x=\pi/2$

Possible Answers:

$\frac{\pi^4}{64}+1$

$\frac{\pi^2}{32}$

$\frac{\pi^2}{64}-1$

$\frac{\pi^4}{32}+1$

Correct answer:

$\frac{\pi^4}{64}+1$

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:

$\int_{0}^{\pi/2}sin(x)+x^3dx$

Solution:

$\int sin(x)+x^3dx$

$-cos(x)+\frac{1}{4}x^4\Big|_{0}^{\frac{\pi}{2}}$

$\frac{1}{4}\cdot\frac{\pi^4}{16}+cos(0)$

Answer:

$\frac{\pi^4}{64}+1$

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Example Question #81 : Integral Applications

Find the area under the curve for f(x)=(x-3)^2+6 from x=0 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:

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

Solution:

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

$=\int_{0}^{2} x^2 + 9 -6x + 6 dx$

$=\frac{1}{3}x^3 -3x^2 + 15x \Big|_{0}^{2}$

$=\frac{8}{3} - 3\cdot4 + 15\cdot2$

$=\frac{8}{3} + 18 = \frac{62}{3} = 20.6666$

When rounded, it is equal to

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Example Question #2003 : Calculus Ii

Find the area under the curve for f(x)=(x-7)^2+3x^2 from x=1 to x=4

Possible Answers:

107

132

119

126

Correct answer:

126

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:

$\int_{1}^{4}(x-7)^2+3x^2dx$

Solution:

$\int_{1}^{4}(x-7)^2+3x^2dx$

$=\int_{1}^{4} x^2-14x+49+3x^2 dx$

$=\int_{1}^{4} 4x^2 - 14x + 49 dx$

$=\frac{4}{3} x^3 - 7x^2 + 49x\Big|_{1}^{4}$

$=\frac{4}{3} 4^3 - 7\cdot4^2 + 49\cdot4 - (\frac{4}{3} - 7 + 49)$

After simplifying, the answer is 126

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Example Question #2002 : Calculus Ii

Find the area under the curve for f(x)=21x^2-40x from x=-2 to x=1 , rounded to the nearest integer.

Possible Answers:

107

115

123

132

Correct answer:

123

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:

$\int_{-2}^{1}21x^2-40xdx$

Solution:

$\int21x^2-40xdx$

$\frac{21}{3}x^3-\frac{40}{2}x^2\Big|_{-2}^{1}$

$(7\cdot1^3-20\cdot1^2)-(7\cdot(-2)^3-20\cdot(-2)^2)$

$(7-20)-(7\cdot-8-(20\cdot4))$

-13-(-56-80)

-13-(-136)=-13+136=123

The area under the curve is 123

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Example Question #82 : Integral Applications

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 #87 : Integral Applications

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 #82 : Integral Applications

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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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Integral Applications

Example Question #251 : Integrals

Example Question #81 : Integral Applications

Example Question #2003 : Calculus Ii

Example Question #2002 : Calculus Ii

Example Question #82 : Integral Applications

Example Question #87 : Integral Applications

Example Question #88 : Integral Applications

Example Question #82 : Integral Applications

Example Question #89 : Integral Applications

All Calculus 2 Resources

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