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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #14 : Double Integration Over General Regions

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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Example Question #2201 : Calculus 3

Integrate:

$\int_{0}^{1} \int_{0}^{5x}4ydydx$

Possible Answers:

$\frac{25}{3}$

$\frac{50}{3}$

Correct answer:

$\frac{50}{3}$

Explanation:

To perform the iterated integration, we must work from inside outwards. To start we perform the following integration:

$\int_{0}^{5x}4ydy=2y^2 \Big|^{5x}_{0}= 2(5x)^2=50x^2$

This becomes the integrand for the outermost integral:

$\int_{0}^{1}50x^2dx=\frac{50x^3}{3}\Big|^{1}_{0}= \frac{50}{3}$

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Example Question #2202 : Calculus 3

$\int_{0}^{2}\int_{y}^{3y}2xdxdy$

Possible Answers:

$\frac{64}{3}$

$\frac{32}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{64}{3}$

Explanation:

To perform the iterated integral, we work from inside outwards.

The first integral we perform is

$\int_{y}^{3y}2xdx=x^2\left.\begin{matrix} 3y\\y \end{matrix}\right|=9y^2-y^2=8y^2$

This becomes the integrand for the outermost integral,

$\int_{0}^{2}8y^2dy=\frac{8y^3}{3}\left.\begin{matrix} 2\\ 0 \end{matrix}\right|=\frac{64}{3}$

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Example Question #13 : Double Integration Over General Regions

Solve:

$\int_{1}^{2}\int_{0}^{x^2}ydydx$

Possible Answers:

$\frac{31}{2}$

$\frac{31}{10}$

$\frac{32}{10}$

Correct answer:

$\frac{31}{10}$

Explanation:

To evaluate the iterated integral, we must work from inside outward.

The first integral we evaluate is

$\int_{0}^{x^2}ydy= \frac{y^2}{2}\left.\begin{matrix} x^2\\0 \end{matrix}\right|= \frac{x^4}{2}$

This becomes the integrand for the outermost integral.

The final integral we evaluate is

$\int_{1}^{2}\frac{x^4}{2}dx= \frac{x^5}{10}\left.\begin{matrix} 2\\1 \end{matrix}\right|= \frac{32}{10}-\frac{1}{10}=\frac{31}{10}$

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Example Question #11 : Double Integration Over General Regions

Let f(x) be any continuous, one-to-one function over the interval I=[a,b] , with f(x)>0 on and 0<a<b . Select all integrals which correctly define the indicated area under the curve.

Question 5 calc 3

$1)\: \: \: \: \: \: \: \: A=\int_a^bf(x)dx$

$2)\: \: \: \: \:\, \, \: A=\int_0^{f(x)}\int_{a}^bdydx$

$3)\: \: \: \: \: \: \: \: A = \int_a^b\int_0^{f(x)}dydx$

$4) \: \: \: \: \: \: \: A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy$

Possible Answers:

1 3 and 4

3 and 4

1 and 3

1 and 4

Correct answer:

1 and 3

Explanation:

$1)\: \: \: \: \: \: \: \: A=\int_a^bf(x)dx$

No explanation necessary. This is the familiar integral for computing the area under the curve y=f(x) over [a,b] .

_____________________________________________________________

$2)\: \: \: \: \:\, \, \: A=\int_0^{f(x)}\int_{a}^bdydx$

This integral does not give the area, in fact, it will result in a function of which certainly cannot be interpreted as the area of corresponding to a specific interval I=[a,b] .

$A = \int _0^{f(x)}\int_a^bdydx=\int_0^{f(x)}[y]_{y=a}^{y=b}dx=\int_0^{f(x)}(b-a)dx$

Now carry out the integration with respect to we obtain,

$\int_0^{f(x)}(b-a)dx=\left[ (a-b)x\right ]_{x=0}^{x=f(x)}=(a-b)f(x)$

Certainly this is not the area of the area under f(x) . One obvious problem is that the integral gave back a function of and no real number.

_____________________________________________________________

$3)\: \: \: \: \: \: \: \: A = \int_a^b\int_0^{f(x)}dydx$

This equation will reduce down to the first option, option , after we carry out the integration with respect to .

$A = \int_a^b\int_0^{f(x)}dydx=\int_a^b \left[ \: \: y \, \, \right]_{y=o}^{y=f(x)} dx =\int_a^bf(x)dx$

Therefore, option is a valid representation for the area.

___________________________________________________________

$4) \: \: \: \: \: \: \: A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy$

The problem with this choice can be seen immediately. Clearly the first integration in will simply be the area under the curve on , but the second will now multiply the area by f(b)-f(a) .

$A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy = \int_{f(a)}^{f(b)}Ady = [Ay]_{y=f(a)}^{y=f(b)}=A[f(b)-f(a)]$

A = A[f(b)-f(a)] could only be true if f(b)-f(a)=1 . But because none of the listed characteristics of our function imply this, we cannot assume it. In fact, the problem stated that f(x) is any function that satisfies the stated conditions. Certainly is not true for just any function.

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Example Question #12 : Double Integration Over General Regions

Evaluate the following iterated integrals:

$\int_-_1^3 \int_0^\frac{\pi}{2} y^2 \cos(x) \ dxdy$

Possible Answers:

$\frac{28}{3}$

Correct answer:

$\frac{28}{3}$

Explanation:

When evaluating double integrals, work from the inside out: that is, evaluate the integrand with respect to the first variable listed by the differential operators, and then evaluate the result with respect to the second variable listed by the differential operators.

Here, we have the order of integration specified by dxdy ; hence, we evaluate the double integrals with respect to first, and then integrate the result with respect to , as shown:

$\int_-_1^3 \int_0^\frac{\pi}{2} y^2 \cos(x) \ dxdy$

$=\int_-_1^3 \left [\int_0^\frac{\pi}{2} y^2 \cos(x) \ dx \right ] dy$

$=\int_-_1^3 \left [y^2 \sin(x) \Big|_0^\frac{\pi}{2} \right ] dy$

$=\int_-_1^3 \left [y^2 (1)-y^2(0) \right ] dy$

$=\int_-_1^3 y^2 \ dy$

$= \frac{y^3}{3} \Big|_-_1^3$

$=\frac{27}{3}-\frac{-1}{3}$

$=\frac{28}{3}$

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Example Question #21 : Double Integration Over General Regions

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

Possible Answers:

$-\frac{3}{120}$

$-\frac{1}{120}$

120

$\frac{1}{120}$

Correct answer:

$-\frac{3}{120}$

Explanation:

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

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

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

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Example Question #21 : Double Integration Over General Regions

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

Possible Answers:

$\frac{1}{120}$

$-\frac{3}{120}$

$-\frac{1}{120}$

120

Correct answer:

$-\frac{3}{120}$

Explanation:

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

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

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

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Example Question #1 : Vector Addition

Let $\vec{a}=(1,2,3)$ , $\vec{b}=(10,-10,3)$ , and $\vec{c}=(0,0,1)$ ,

find $\vec{a}+\vec{b}+\vec{c}$

Possible Answers:

(11, 0, 1)

(7, -8,11)

(0,0,1)

(11, -8, 7)

(11, 7, -8)

Correct answer:

(11, -8, 7)

Explanation:

In order to find $\vec{a}+\vec{b}+\vec{c}$ , we need to add like components. $\vec{a}+\vec{b}+\vec{c}=(1+10+0,2-10+0,3+3+1)=(11,-8,7)$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #14 : Double Integration Over General Regions

Example Question #672 : Multiple Integration

Example Question #2201 : Calculus 3

Example Question #2202 : Calculus 3

Example Question #13 : Double Integration Over General Regions

Example Question #11 : Double Integration Over General Regions

Example Question #12 : Double Integration Over General Regions

Example Question #21 : Double Integration Over General Regions

Example Question #21 : Double Integration Over General Regions

Example Question #1 : Vector Addition

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