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Calculus 2 : Integral Applications

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Example Questions

Example Question #71 : Integral Applications

Find the area under the curve $y=6x^{2}-2x$ between $0\leq x\leq 4$ .

Possible Answers:

114

112

110

116

118

Correct answer:

112

Explanation:

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space. Thus, given $y=6x^{2}-2x$ , then the area over $0\leq x\leq 4$ is $\int_{0}^{4}(6x^{2}-2x)dx$ .

Using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$\int_{0}^{4}(6x^{2}-2x)dx$

$=[2x^{3}-x^{2}]_{0}^{4}\textrm{}$

$=[2(4)^{3}-(4)^{2}] -[2(0)^{3}-(0)^{2}]$

=[2(64)-16]

=128-16

=112

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Example Question #72 : Integral Applications

Find the area under the curve $y=\frac{1}{2}x^{2}-3x$ between $0\leq x\leq 2$ .

Possible Answers:

$\frac{28}{6}$

$\frac{6}{28}$

$-\frac{14}{3}$

$-\frac{6}{28}$

None of the above

Correct answer:

$-\frac{14}{3}$

Explanation:

We can define the area underneath a curve provided by a function as the definite integral of the function over a given space.

Thus, given $y=\frac{1}{2}x^{2}-3x$ , then the area over $0\leq x\leq 2$ is $\int_{0}^{2}\left(\frac{1}{2}x^{2}-3x\right)dx$ .

Using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$\int_{0}^{2}\left(\frac{1}{2}x^{2}-3x\right)dx$

$=\left[\frac{1}{6}x^{3}-\frac{3}{2}x^{2}\right]_{0}^{2}\textrm{}$

$=\left[\frac{1}{6}(2)^{3}-\frac{3}{2}(2)^{2}\right]-\left[\frac{1}{6}(0)^{3}-\frac{3}{2}(0)^{2}\right]$

$=\left[\frac{8}{6}-\frac{12}{2}\right]$

$=\left[\frac{8}{6}-6\right]$

$=\frac{8}{6}-\frac{36}{6}$

$=-\frac{28}{6}=-\frac{14}{3}$

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Example Question #71 : Integral Applications

Find the area under the curve for f(x)=2-cos(x) from $x=-\pi/2$ to $x=\pi/2$

Possible Answers:

$\pi-1$

$2\pi-1$

$2\pi-2$

$\pi-2$

Correct answer:

$2\pi-2$

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:

$\int_{-\pi/2}^{\pi/2}2-cos(x)dx$

Solution:

$\int2-cos(x)dx = 2x-sin(x)$

$2\pi/2 - sin(\pi/2)-(-2\pi/2-sin(-\pi/2) =$

$2\pi-2$

Note:

$sin(\pi/2) = 1$

$\int cos(x) = sin(x)$

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Example Question #71 : Integral Applications

Find the area under the curve for f(x)=3x+2 from x=3 to x=6 , 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:

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

Solution:

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

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

$=\frac{3}{2}\cdot 6^2+2\cdot6 -(\frac{3}{2}\cdot 3^2+2\cdot 3)$

$=\frac{3}{2}\cdot36+12-\frac{3}{2}\cdot9-6$

$=\frac{93}{2} = 46.5$

after rounding

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Example Question #12 : Area Under A Curve

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

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:

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

Solution:

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

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

$=\frac{2}{3}\cdot 4^3+4\cdot4 -(\frac{2}{3}1^3+4)$

$=\frac{2}{3}\cdot 64+16-\frac{2}{3}-4=54$

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Example Question #76 : Integral Applications

Find the area under the curve for f(x)=4x^4+10 from x=2 to x=4 , rounded to the nearest integer.

Possible Answers:

847

932

756

814

Correct answer:

814

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:

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

Solution:

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

$=\frac{4}{5}x^5+10x\Big |_{2}^{4}$

$=\frac{4}{5}\cdot4^5+10\cdot4 -(\frac{4}{5}\cdot 2^5+10\cdot2)$

$=\frac{4}{5}\cdot1024+40 -\frac{4}{5}\cdot32-20$

$\frac{4068}{5}$ = 813.6

= 814 after rounding

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Example Question #11 : Area Under A Curve

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

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

Solution:

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

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

$=\frac{3}{4}\cdot3^4+\frac{4}{3}\cdot3^3+3\cdot3 -(\frac{3}{4}\cdot2^4+\frac{4}{3}\cdot2^3+3\cdot2)$

$=\frac{3}{4}\cdot81+\frac{4}{3}\cdot27+9 -\frac{3}{4}\cdot16-\frac{4}{3}\cdot8-6$

$\frac{317}{12}$ = 26.4

= 26 after rounding

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Example Question #78 : Integral Applications

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

Possible Answers:

126

136

112

120

Correct answer:

120

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:

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

Solution:

This function is negative for the entire region, so multiply the integral by -1 to drop the absolute value signs.

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

$=10x-\frac{1}{4}x^4\Big|_{-4}^{2}$

$=10\cdot2-\frac{1}{4}\cdot2^4-(10\cdot(-4)-\frac{1}{4}\cdot(-4)^4)$

$=20-\frac{1}{4}\cdot16+40+\frac{1}{4}\cdot256$

=20-4+40+64

=120

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Example Question #79 : Integral Applications

Find the area under the curve for f(x)=|x^3-x| from x=-1 to x=3

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:

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

Solution:

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

In other words, find this sum.

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

First piece:

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

$=0-(\frac{1}{4} - \frac{1}{2}) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Second piece:

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

$=\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Third piece:

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

$=\frac{1}{4}\cdot 3^4 - \frac{1}{2}\cdot 3^2 - (\frac{1}{4} - \frac{1}{2})$

$=\frac{1}{4}\cdot 81 - \frac{1}{2}\cdot 9 -\frac{1}{4}+\frac{1}{2} = 16$

Sum:

$16 + \frac{1}{4} + \frac{1}{4} = 16.5$ , when rounded, is

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Example Question #80 : Integral Applications

Find the area under the curve for f(x)=|2x^2-10x| from x=-2 to x=5

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:

$\int_{-2}^{5}|2x^2-10x|dx$

Solution:

This function is negative from x=[0,5] , and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,5] region by , and sum everything up.

In other words, sum up these two integrals.

$\int_{-2}^{0}2x^2-10xdx + \int_{0}^{5}10x-2x^2dx$

First piece:

$\int_{-2}^{0}2x^2-10xdx = \frac{2}{3}x^3-5x^2\Big|_{-2}^{0}$

$=0-(-\frac{2}{3}\cdot8-5\cdot4) = \frac{16}{3}+20$

Second piece:

$\int_{0}^{5}10x-2x^2dx = 5x^2-\frac{2}{3}x^3\Big|_{0}^{5}$

$=125-\frac{2}{3}\cdot125$

Sum:

$\frac{16}{3}+20 + 125-\frac{2}{3}\cdot125 = 67$

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 #71 : Integral Applications

Example Question #72 : Integral Applications

Example Question #71 : Integral Applications

Example Question #71 : Integral Applications

Example Question #12 : Area Under A Curve

Example Question #76 : Integral Applications

Example Question #11 : Area Under A Curve

Example Question #78 : Integral Applications

Example Question #79 : Integral Applications

Example Question #80 : Integral Applications

All Calculus 2 Resources

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