Question

Obtain the double integral of the function, f (x, y) = 2xy + 5exponential x over the region
bounded by the triangle formed by the y-axis and the lines 5y=x and y=z

Answer 1

A sketch the region D is follow:

.$5y=x \rArr y=\frac{x}{5}$

Point of intersection:

$\frac{x}{5}=z \rArr x=5z$

The region D: $\lbrace (x, y): 0 \le x \le 5z, \;\frac{x}{5} \le y \le z \rbrace$

$I=\iint_D(2xy+5e^x)dA=\int^{5z}_0(\int^{z}_{\frac{x}{5}}(2xy+5e^x)dy)dx=\int^{5z}_0((xy^2+5ye^x)\vert ^{z}_{\frac{x}{5}})dx=\int^{5z}_0(xz^2+5ze^x-\frac{x^3}{25}-xe^x)dx=I_{1}+I_{2}+I_{3}+I_{4}$

$I_{1}=z^2\int^{5z}_0xdx=z^2(\frac{x^2}{2})\vert ^{5z}_{0}=\frac{25z^4}{2}$

$I_{2}=5z\int^{5z}_0e^xdx=5z(e^x)\vert ^{5z}_{0}=5ze^{5z}-5z$

$I_{3}=-\frac{1}{25}\int^{5z}_0x^3dx=-\frac{1}{25}(\frac{x^4}{4})\vert ^{5z}_{0}=\frac{625z^4}{100}=\frac{25z^4}{4}$

$I_{4}=-\int^{5z}_0xe^xdx=\begin{Bmatrix} u=x & du=dx \\ dv=e^xdx & v=e^x \end{Bmatrix}=-(xe^x)\vert^{5z}_0+\int^{5z}_0e^xdx=-5ze^{5z}+(e^x)\vert^{5z}_0=-5ze^{5z}+e^{5z}-1$

$I=\frac{25z^4}{2}+5ze^{5z}-5z+\frac{25z^4}{4}-5ze^{5z}+e^{5z}-1=\frac{50z^4}{4}+\cancel{5ze^{5z}}-5z+\frac{25z^4}{4}-\cancel{5ze^{5z}}+e^{5z}-1=e^{5z}+\frac{75z^4}{4}-5z-1$