Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 20

Evaluate $\displaystyle \int^1_0 \left( x^2 + 1 \right) e^{-x} dx$
$\displaystyle \int^1_0 \left( x^2 + 1 \right) e^{-x} dx = \int^1_0 \left( x^2 e^{-x} \right) dx + \int^1_0 e^{-x} dx $

To evaluate $\displaystyle \int^1_0\left( x^2 e^{-x} \right) dx$, we must use integration by parts, so if
we let $u = x^2$ and $dv = e^{-x} dx$, then
$du = 2x dx$ and $\displaystyle v = \int e^{-x} dx = -e^{-x}$

So,

$
\begin{equation}
\begin{aligned}
\int^1_0 \left( x^2 e^{-x} \right) dx = uv - \int vdu &= -x^2 e^{-x} -\int -e^{-x} (2x dx)\\
\\
&= -x^2 e^{-x} + 2 \int e^{-x} (xdx)
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int xe^{-x} dx$, we must use integration by parts once more, so if we let $u_1 = x$ and $dv_1 = e^{-x} dx$, then $du_1 = dx$ and $\displaystyle v_1 = \int e^{-x} dx = - e^{-x}$

So,

$
\begin{equation}
\begin{aligned}
\int xe^{-x} dx = uv - \int vdu &= -xe^{-x} - \int - e^{-x} dx\\
\\
&= - x e^{-x} + \int e^{-x} dx\\
\\
&= -x e^{-x} - e^{-x}
\end{aligned}
\end{equation}
$


Therefore,

$
\begin{equation}
\begin{aligned}
\int^1_0 \left( x^2 + 1 \right) e^{-x} dx &= \int^1_0 \left( x^2 e^{-x} \right) dx + \int^1_0 e^{-x} dx\\
\\
&= -x^2 e^{-x} + 2 \int x e^{-x} dx + \int^1_0 e^{-x} dx\\
\\
&= -x^2 e^{-x} + 2 \left[ -x e^{-x} - e^{-x} \right] + \int^1_0 e^{-x} dx\\
\\
&= -x^2 e^{-x} - 2xe^{-x} - 2e^{-x} + \left[ -e^{-x} \right]^1_0\\
\\
&= \left[ -x^2 e^{-x} - 2x e^{-x} - 3e^{-x} \right]^1_0\\
\\
&= \left[ -1^2 e^{-1} -2(1)e^{-1} - 3e^{-1} - \left( -0^2e^{-0} - 2 (0) e^{-0} - 3e^{-0}\right) \right]\\
\\
&= -\frac{1}{e} - \frac{2}{e} - \frac{3}{e} + 0 + 0 + \frac{3}{e^0}\\
\\
&= 3 - \frac{6}{e}
\end{aligned}
\end{equation}
$