Question

The value of integral $\underset{-1}{\overset{1}{\int }}\underset{-1}{\overset{1}{\int }}|x+y|dxdy$ is

• A. $\frac{4}{3}$
• B. 2
• C. $\frac{8}{3}$
• D. 4

Answer 1

The correct option is C $\frac{8}{3}$



$|x+y|=\left\{\begin{array}{cc}(x+y)& x+y\ge 0\\ -(x+y)& x+y<0\end{array}\right\}$
Changing th order of integration,
$2\underset{-1}{\overset{1}{\int }}\underset{-y}{\overset{1}{\int }}(x+y)dxdy$
$\Rightarrow 2\underset{-1}{\overset{1}{\int }}{(\frac{x^2}{2}+xy)}_{-y}^{1}dy$
$\Rightarrow 2\underset{-1}{\overset{1}{\int }}(\frac{1}{2}+y-\frac{y^2}{2}+y^2)dy$
$\Rightarrow 2\underset{-1}{\overset{1}{\int }}(\frac{y^2}{2}+y+\frac{1}{2})dy$
$\Rightarrow 2{[\frac{y^3}{6}+\frac{y^2}{2}+\frac{y}{2}]}_{-1}^1$
$\Rightarrow 2[(\frac{1}{6}+\frac{1}{2}+\frac{1}{2})-(\frac{-1}{6}+\frac{1}{2}-\frac{1}{2})]$
$\Rightarrow 2[\frac{7}{6}+\frac{1}{6}]=\frac{8}{3}$