Question

Prove the following question.

${\int }_0^{1}x{e}^{x}dx=1$

Answer 1

${\int }_0^{1}x{e}^{x}dx=1$

$\text{Applying rule of integration by parts taking x as the first function and}{e}^x\text{as the second function, we get}$

$={[x\int e^{x}dx-\int 1.e^{x}dx]}_0^1={[x{e}^x-e^x]}_0^1={\left[e^x\right(x-1\left)\right]}_0^1=e(1-1)-e^0(0-1)=0+1=1.Henceproved.$