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Calculus 3 : Multiple Integration

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Example Questions

Example Question #4 : Double Integration Over General Regions

Compute the following integral:

$\int_{1}^{2}\int_{0}^{2}(x^3+y^2)dydx$

Possible Answers:

$\frac{120}{15}$

$\frac{130}{12}$

$\frac{122}{12}$

Correct answer:

$\frac{122}{12}$

Explanation:

First, you must evaluate the integral with respect to y and solving within the bounds.

In doing so, you get $x^3y+\frac{y^3}{3}$ and you evaluate for y from 0 to 2.

This gets you

$\int_{1}^{2}(2x^3+\frac{8}{3})dx$ .

This time evaluating the integral with respect to x gets you

$\frac{2x^4}{4}+\frac{8}{3}x$ .

Evaluating for x from 1 to 2 gets you

$\frac{122}{12}$ .

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Example Question #7 : Double Integration Over General Regions

Evaluate the double integral.

$\int_{0}^{7} \int_{0}^{5} 2x+7\,dxdy$

Possible Answers:

710

125

365

420

Correct answer:

420

Explanation:

When solving double integrals, we compute the integral on the inside first.

$\int_{0}^{7} \int_{0}^{5} 2x+7\,dxdy=\int_0^7x^2+7x|_{x=0}^{x=5}\,dy$

$=\int_0^75^2+7(5)-0^2+7(0)\,dy$

$=\int_0^760\,dy$

$=60y|_{y=0}^{y=7}$

=420-0

=420

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Example Question #1 : Double Integration Over General Regions

Evaluate the double integral.

$\int_{0}^{6} \int_{0}^{4} 6y+8\,dxdy$

Possible Answers:

624

315

540

312

Correct answer:

624

Explanation:

When solving double integrals, we compute the integral on the inside first.

$\int_{0}^{6} \int_{0}^{4} 6y+8\,dxdy=\int_0^66xy+8x|_{x=0}^{x=4}\,dy$

$=\int_0^6 6(4)y+8(4)-6(0)y-8(0)\,dy$

$=\int_0^624y+32\,dy$

$=12y^2+32y|_{y=0}^{y=6}$

=12(6)^2+32(6)-[12(0)^2+32(0)]

=624

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Example Question #6 : Double Integration Over General Regions

Evaluate the double integral

$\int_{0}^{3} \int_{0}^{4} 3x+6\,dydx$

Possible Answers:

324

256

126

200

Correct answer:

126

Explanation:

When solving double integrals, we compute the integral on the inside first.

$\int_{0}^{3} \int_{0}^{4} 3x+6\,dydx=\int_0^33xy+6y|_{y=0}^{y=4}\,dx$

$=\int_0^33x(4)+6(4)-[3x(0)+6(0)]\,dx$

$=\int_0^312x+24\,dx$

$=6x^2+24x|_{x=0}^{x=3}$

=6(3)^2+24(3)-6(0)^2-24(0)

=126

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Example Question #7 : Double Integration Over General Regions

Evaluate the integral $\int_{1}^{3}\int_{0}^{2}(xy^3)dxdy$

Possible Answers:

Correct answer:

Explanation:

First, you must evaluate the integral with respect to x. This gets you $y^3\frac{x^2}{2}dy$ evaluated from x=0 to x=2 . This becomes $\int_{1}^{3}2y^3dy$ . Solving this integral with respect to y gets you $\frac{1}{2}y^4$ . Evaluating from y=1 to y=3 , you get $\frac{1}{2}(\frac{3^4}{4}-\frac{1}{4})=10$ .

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Example Question #671 : Multiple Integration

Evaluate the following integral: $\int_{0}^{1}\int_{3}^{6}(6x^2z)dzdx$

Possible Answers:

Correct answer:

Explanation:

First, you must evaluate the integral with respect to z. Using the rules for integration, we get 3x^2z^2dx evaluated from z=3 to z=6 . The result is $\int_{0}^{1}54x^2dx$ . This becomes 27x^3 , evaluated from x=0 to x=1 . The final answer is .

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Example Question #671 : Multiple Integration

Evaluate:

$\int_{1}^{3}\int_{y}^{2y}dxdy$

Possible Answers:

Correct answer:

Explanation:

To evaluate the iterated integral, we start with the innermost integral, evaluated with respect to x:

$\int_{y}^{2y}dx=x\left.\begin{matrix} 2y\\y \end{matrix}\right|=2y-y=y$

The integral was found using the following rule:

$\int dx=x+C$

Now, we evaluate the last remaining integral, using our answer above from the previous integral as our integrand:

$\int_{1}^{3}ydy=\frac{y^2}{2}\left.\begin{matrix} 3\\1 \end{matrix}\right|=\frac{9}{2}-\frac{1}{2}=4$

The integral was found using the following rule:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

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Example Question #11 : Double Integration Over General Regions

Evaluate the double integral

$\int_{0}^{1} \int_{0}^{2}x+y\,dx\,dy$

Possible Answers:

Correct answer:

Explanation:

To evaluate the double integral, compute the inside integral first.

$\int_{0}^{1} \int_{0}^{2}x+y\,dx\,dy=\int_0^1 (\frac{1}{2}x^2+xy|_{x=0}^{x=2}dy$

$=\int_0^12+2y\,dy$

$=2y+y^2|_{y=0}^{y=1}$

=2(1)+1^2-[2(0)+0^2]

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Example Question #191 : Double Integrals

Evaluate the double integral

$\int_0^2 \int_0^1 y-2x\,dx\,dy$

Possible Answers:

Correct answer:

Explanation:

aTo evaluate the double integral, compute the inside integral first.

$\int_{0}^{2} \int_{0}^{1}y-2x\,dx\,dy=\int_0^2 (xy-x^2|_{x=0}^{x=1}dy$

$=\int_0^2 y-1\,dy$

$=\frac{1}{2}y^2-y|_{y=0}^{y=2}$

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

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Example Question #672 : Multiple Integration

Evaluate the double integral

$\int_0^3 \int_0^4 2x-2y \,dy\,dx$

Possible Answers:

Correct answer:

Explanation:

To evaluate the double integral, compute the inside integral first.

$\int_{0}^{3} \int_{0}^{4}2x-2y\,dy\,dx=\int_0^3 2xy-y^2|_{y=0}^{y=4}dx$

$=\int_0^3 8x-16\,dx$

$=4x^2-16x|_{x=0}^{x=3}$

=4(3)^2-16(3)-[4(0)^2-16(0)]

=-12

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Calculus 3 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #4 : Double Integration Over General Regions

Example Question #7 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

Example Question #6 : Double Integration Over General Regions

Example Question #7 : Double Integration Over General Regions

Example Question #671 : Multiple Integration

Example Question #671 : Multiple Integration

Example Question #11 : Double Integration Over General Regions

Example Question #191 : Double Integrals

Example Question #672 : Multiple Integration

All Calculus 3 Resources

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