∫limits_0^1xe^x dx is equal to

The correct answer is Option (2) → 1

$\displaystyle \int_{0}^{1} x e^{x}\,dx$

Using integration by parts:

Let $u = x,\; dv = e^{x}dx$

$\Rightarrow du = dx,\; v = e^{x}$

$\int x e^{x}dx = x e^{x} - \int e^{x}dx = e^{x}(x - 1) + C$

Now apply limits 0 to 1:

$\left[e^{x}(x - 1)\right]_{0}^{1} = e(1 - 1) - (1)(0 - 1) = 0 - (-1) = 1 $

$\displaystyle \int_{0}^{1} x e^{x}dx = 1 $