Question : Evaluate double integral x = 0 to 3 integral y =x to 3 of (3+xy) dydx
Solution :
We shall proceed as follows to Evaluate double integral x = 0 to 3 integral y =x to 3 of (3+xy) dydx
Let double integral as \[I=\int_{x=0}^{3}{\int_{y=x}^{3}\left(3+xy\right)dydx}\] \[I=\int_{x=0}^{3}\left(\int_{y=x}^{3}\left(3+xy\right)dy\right)dx\] First we shall evaluate inner integral \[I=\int_{x=0}^{3}{\left(3y+x\frac{y^2}{2}\right)_x^3dy}\] And Evaluate it at the y limits \[I=\int_{x=0}^{3}\left(3\ast3+x\frac{3^2}{2}-3x-x\frac{x^2}{2}\right)dx\] We get \[I=\int_{x=0}^{3}\left(9+\frac{3}{2}\ x-\frac{x^3}{2}\right)dx\] Now we shall evaluate outer integral \[I=\left(9x+\frac{3}{2}\frac{x^2}{2}-\frac{x^4}{8}\right)_0^3\] And evaluating it at the x limits \[I=27+\frac{27}{4}-\frac{81}{8}-0+0-0=\frac{189}{8}\] Thus the value of the double integral is \[\int_{x=0}^{3}{\int_{y=x}^{3}\left(3+xy\right)dxdy}=\frac{189}{8}\] We get the value of the double integral x = 0 to 3 integral y =x to 3 of (3+xy) dydx as 189/8