Double Integration over General Regions - AP Calculus BC

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Question

Calculate the following Integral.

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

Answer

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

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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Question

Calculate the definite integral of the function , given below as

$f(x,y)= 3x^3\sqrt{y}+ \sin(2x)\cos(4y)$

Answer

Because there are no nested terms containing both and , we can rewrite the integral as

$\iint f(x,y)dxdy=\int3x^3dx \int \sqrt{y}dy+\int\sin(2x)dx \int \cos(4y)dy$

This enables us to evaluate the double integral and the product of two independent single integrals. From the integration rules from single-variable calculus, we should arrive at the result

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{12}\cos(2x)\sin(4y)+C$ .

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Question

Evaluate the following integral on the region specified:

$\int\int_R(x^3y):dxdy$

Where R is the region defined by the conditions:

$0\leq x \leq 2, ; 1\leq y \leq2$

Answer

$\int^2_1\int^2_0(x^3y):dxdy=\int^2_1(\frac{x^4y}{4}|^2_0)dy\int^2_14y:dy =2y^2|^2_1=2(4-1)=2(3)=6$

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Question

Evaluate:

$\int \int x^2dxdy$

Answer

Because the x and y terms in the integrand are independent of one another, we can move them to their respective integrals:

$\int dy \cdot \int x^2dx= \frac{x^3y}{3}+C$

We used the following rules for integration:

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

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Question

Evaluate the double integral

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

Answer

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

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

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

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

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

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Question

Evaluate the double integral.

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

Answer

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

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

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

$=\int_0^760,dy$

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

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Question

Evaluate the double integral.

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

Answer

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

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

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

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

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

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Question

Compute the following integral:

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

Answer

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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Question

Evaluate the following integral.

$\int_{0}^{1}\int_{2}^{3}\left (5x^4y \right )dydx$

Answer

First, you must evaluate the integral with respect to y (because of the notation ).

Using the rules of integration, this gets us

$5x^4\frac{y^2}{2}dx$ .

Evaluated from y=2 to y=3, we get

$5x^4(\frac{9}{2}-\frac{4}{2})dx=\frac{25}{2}x^4dx$ .

Integrating this with respect to x gets us $\frac{5}{2}x^5$ , and evaluating from x=0 to x=1, you get $\frac{5}{2}$ .

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Question

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

Answer

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

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Question

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

Answer

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

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Question

Evaluate:

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

Answer

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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Question

Evaluate the double integral

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

Answer

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

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

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

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

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Question

Evaluate the double integral

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

Answer

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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Question

Evaluate the double integral

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

Answer

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

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

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

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

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Question

Integrate:

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

Answer

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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Question

Evaluate the following iterated integrals:

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

Answer

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 ; 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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Question

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

Answer

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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Question

Solve:

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

Answer

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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Question

.

Let be any continuous, one-to-one function over the interval , with on and . Select all integrals which correctly define the indicated area under the curve.

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

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

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

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

Answer

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

No explanation necessary. This is the familiar integral for computing the area under the curve over .

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

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 .

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

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 . One obvious problem is that the integral gave back a function of and no real number.

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

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.

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

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 .

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

could only be true if . But because none of the listed characteristics of our function imply this, we cannot assume it. In fact, the problem stated that is any function that satisfies the stated conditions. Certainly is not true for just any function.

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