$\int\limits_0^1xe^x\,dx$ is equal to |
0 1 $e$ $e^2$ |
1 |
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 $ |