Question

The values of integrals ${\int }_0^1\left({\int }_0^1\frac{x-y}{(x+y{)}^3}dy\right)dx$ and ${\int }_0^1\left({\int }_0^1\frac{x-y}{(x+y{)}^3}dx\right)dy$ are

• A. same and equal to $0.5$
• B. same and equal to $-0.5$
• C. $0.5$ and $-0.5,$ respectively
• D. $-0.5$ and $0.5,$ respectively

Answer 1

The correct option is C $0.5$ and $-0.5,$ respectively

$\left(i\right){\int }_0^1\left({\int }_0^1\frac{x-y}{(x+y{)}^3}dy\right)dx$
Let $x+y=t\therefore dy=dt$
$y=(t-x)$
So after chaning the limits we get,
$={\int }_0^1{\int }_x^{x+1}\left(\frac{2x-t}{\left(t^3\right)}dt\right)dx$
$={\int }_0^1\left({\int }_x^{x+1}(\frac{2x}{t^3}-\frac{1}{t^2})dt\right)dx$
$={\int }_0^1{(\frac{-x}{t^2}+\frac{1}{t})}_x^{x+1}dx$
$={\int }_0^1\left[\frac{1}{(x+1{)}^2}\right]dx=0.5$

$\left(ii\right){\int }_0^1\left({\int }_0^1\frac{x-y}{(x+y{)}^3}dx\right)dy$
Let $x+y=t$
$dx=dt$
$x=(t-y)$
So after changing the limits we get,
$={\int }_0^1\left({\int }_y^{1+y}\frac{(t-2y)}{t^3}dt\right)dy$
$={\int }_0^1\left\{{\int }_y^{1+y}(\frac{1}{t^2}-\frac{2y}{t^3})dt\right\}dy$
$={\int }_0^1{[\frac{-1}{t}+\frac{y}{t^2}]}_y^{y+1}dy$
$={\int }_0^1\frac{-1}{(y+1{)}^2}dy=-0.5$