Question

The value of the integral ${\int }_0^2{\int }_0^x{e}^{x+y}dydx$ is

• A. $\frac{1}{2}(e-1)$
• B. $\frac{1}{2}(e^2-1{)}^2$
• C. $\frac{1}{2}(e^2-e)$
• D. $\frac{1}{2}{(e-\frac{1}{e})}^2$

Answer 1

The correct option is B $\frac{1}{2}(e^2-1{)}^2$

Given intergal is
$I={\int }_0^2{\int }_0^x{e}^{x-y}dydx$
$={\int }_0^2{e}^x.{\left[e^y\right]}_0^{x}dx$
$={\int }_0^2{e}^x(e^x-1)dx$
$={[\frac{e^{2x}}{2}-e^x]}_0^2$
$=\frac{e^4}{2}-e^2+\frac{1}{2}$
$I=\frac{1}{2}{(e^2-1)}^2$