Prove that $\int_{0}^{1} x e^{x} d x = 1$.

(N/A) To evaluate the integral $\int_{0}^{1} x e^{x} d x$,we use the method of integration by parts.
The formula for integration by parts is $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = e^{x} dx$.
Then,$du = dx$ and $v = \int e^{x} dx = e^{x}$.
Applying the formula:
$\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} = e^{x}(x - 1)$.
Now,we apply the limits from $0$ to $1$:
$\int_{0}^{1} x e^{x} dx = [e^{x}(x - 1)]_{0}^{1}$.
Evaluating at the upper limit $(x = 1)$:
$e^{1}(1 - 1) = e(0) = 0$.
Evaluating at the lower limit $(x = 0)$:
$e^{0}(0 - 1) = 1(-1) = -1$.
Subtracting the lower limit value from the upper limit value:
$0 - (-1) = 1$.
Thus,$\int_{0}^{1} x e^{x} dx = 1$.