Question

Given the shaded triangular region $P$ shown in the figure. What is $\int \int xydxdy$?

• A. $\frac{1}{6}$
• B. $\frac{2}{9}$
• C. $\frac{7}{16}$
• D. $1$

Answer 1

The correct option is A $\frac{1}{6}$

The equation of straight line is
$\frac{x}{2}+\frac{y}{1}=1$
$\Rightarrow y=1-\frac{x}{2}$

Now,
$\int \int xy.dxdy={\int }_0^2{\int }_{y=0}^{1-\frac{x}{2}}\left(xy\right)dydx$
$={\int }_0^{2}x{\left[\frac{y^2}{2}\right]}_0^{1-\frac{x}{2}}dx$
$={\int }_0^2\frac{x}{2}{(1-\frac{x}{2})}^{2}dx$
$={\int }_0^2\frac{x}{2}(1+\frac{x^2}{4}-x)dx$
$=\frac{1}{2}{\int }_0^2(x+\frac{x^3}{4}-x^2)dx$
$=\frac{1}{2}{[\frac{x^2}{2}+\frac{x^4}{16}-\frac{x^3}{3}]}_0^2$
$=\frac{1}{2}[2+1-\frac{8}{3}]=\frac{1}{6}$