AP Calculus BC - Double Integration over General Regions

Calculate the following Integral.

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

$-\frac{3}{120}$

$-\frac{1}{120}$

$\frac{1}{120}$

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.

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

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

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

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

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

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

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

Evaluate the following integral on the region specified:

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

Where R is the region defined by the conditions:

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

$2\sqrt{2}$

$4\sqrt{2}$

Explanation

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

Calculate the following Integral.

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

$-\frac{3}{120}$

$-\frac{1}{120}$

$\frac{1}{120}$

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.

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

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

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

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

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

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

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

Evaluate the double integral

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

Explanation

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

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

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

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

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

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

Explanation

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 $\frac{1}{2}(\frac{3^4}{4}-\frac{1}{4})=10$ .

Calculate the definite integral of the function , given below as

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

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}+\frac{1}{8}\cos(2x)\sin(4y)+C$

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

Cannot be solved.

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)$

Explanation

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

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

Evaluate the following integral.

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

$\frac{5}{2}$

Explanation

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 $\frac{5}{2}$ .

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

Explanation

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 .

Evaluate the following iterated integrals:

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

$\frac{28}{3}$

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 ; hence, we evaluate the double integrals with respect to first, and then integrate the result with respect to , as shown: