Problem source

Sketch the graph of 
\[
y=\frac{x^{2}\mathrm{e}^{-x}}{1+x},
\]
for $-\infty< x< \infty.$ 
Show that the value of 
\[
\int_{0}^{\infty}\frac{x^{2}\mathrm{e}^{-x}}{1+x}\,\mathrm{d}x
\]
lies between $0$ and $1$.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[->] (-3, 0) -- (3,0) node [right] {$x$};
        \draw[->] (0, -3) -- (0,3) node [above] {$y$};

        \draw[domain = -0.7:3, samples=50, variable = \x, blue, thick]  plot ({\x},{\x*\x*exp(-\x)/(1+\x)});
        \draw[domain = -2.7:-1.05, samples=50, variable = \x, blue, thick]  plot ({\x},{\x*\x*exp(-\x)/(1+\x)/exp(3)});
        \draw[dashed] (-1,-3) -- (-1,3);
    \end{tikzpicture}
\end{center}

First notice the integrand is always positive over the range we are integrating, so the integral is greater than $0$. Since  $\frac{x}{1+x} \leq 1$ for $x \geq 0$ we can note that:

\begin{align*}
\int_0^{\infty} \frac{x^2e^{-x}}{1+x} \d x &=\int_0^{\infty} \frac{x}{1+x}xe^{-x} \d x \\
&< \int_0^\infty xe^{-x} \d x \\
&= \left [ -xe^{-x} \right]_0^{\infty} + \int_0^{\infty} e^{-x} \d x \\
&= 0 + 1 \\
&= 1
\end{align*}

and so we are done.