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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #71 : Integral Applications

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

Possible Answers:

120

126

112

136

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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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 #251 : Integrals

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 #251 : Integrals

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

Possible Answers:

126

132

107

119

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #76 : Integral Applications

Example Question #11 : Area Under A Curve

Example Question #71 : Integral Applications

Example Question #79 : Integral Applications

Example Question #80 : Integral Applications

Example Question #81 : Integral Applications

Example Question #251 : Integrals

Example Question #251 : Integrals

Example Question #251 : Integrals

Example Question #2002 : Calculus Ii

All Calculus 2 Resources

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