Question

Prove: $∫_0^1x.e^x\ dx=1$

Answer 1

Let $I= ∫_0^1x.e^x\ dx$

Integrating by parts, we obtain

$I=x∫_0^1e^xdx - ∫_0^1[{(\frac {d}{dx}(x))∫e^xdx}]dx$

$I$ = $[xe^x]_0^1$ - $∫_0^1e^xdx$

$I$= $[xe^x]_0^1$ - $[e^x]_0^1$

$I$= $e-e+1$

$I$=$1$

Hence, the given result is proved.