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Calculus 2 : Area Under a Curve

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Example Question #61 : Integral Applications

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

Possible Answers:

Correct answer:

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

Using the Power Rule for Integrals

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

we can determine that:

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

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

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

=27-3

=24

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Example Question #11 : Area Under A Curve

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

Possible Answers:

$\frac{3}{10}$

$-\frac{3}{10}$

None of the above

$\frac{10}{3}$

$-\frac{10}{3}$

Correct answer:

$\frac{10}{3}$

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 $y=\frac{1}{2}x^{2}+x$ , then the area over $0\leq x\leq 2$ is $\int_{0}^{2}(\frac{1}{2}x^{2}+x)dx$ .

Using the Power Rule for Integrals

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

we can determine that:

$\int_{0}^{2}(\frac{1}{2}x^{2}+x)dx$

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

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

$=[\frac{8}{6}+2]$

$=[\frac{8}{6}+\frac{12}{6}]$

$=\frac{20}{6}$

$=\frac{10}{3}$

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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 #73 : Integral Applications

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

Possible Answers:

$2\pi-1$

$\pi-2$

$\pi-1$

$2\pi-2$

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 #11 : Area Under A Curve

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 #75 : Integral Applications

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 #77 : Integral Applications

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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Calculus 2 : Area Under a Curve

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Integral Applications

Example Question #11 : Area Under A Curve

Example Question #71 : Integral Applications

Example Question #72 : Integral Applications

Example Question #73 : Integral Applications

Example Question #11 : Area Under A Curve

Example Question #75 : Integral Applications

Example Question #76 : Integral Applications

Example Question #77 : Integral Applications

Example Question #78 : Integral Applications

All Calculus 2 Resources

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