# 009B Sample Midterm 1, Problem 4 Detailed Solution

Evaluate the indefinite and definite integrals.

(a)   $\int x^{2}e^{x}~dx$ (b)   $\int _{1}^{e}x^{3}\ln x~dx$ Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$ 2. How would you integrate  $\int x\ln x~dx?$ You could use integration by parts.

Let  $u=\ln x$ and  $dv=x~dx.$ Then,  $du={\frac {1}{x}}dx$ and  $v={\frac {x^{2}}{2}}.$ ${\begin{array}{rcl}\displaystyle {\int x\ln x~dx}&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-\int {\frac {x}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-{\frac {x^{2}}{4}}+C.}\end{array}}$ Solution:

(a)

Step 1:
We proceed using integration by parts.
Let  $u=x^{2}$ and  $dv=e^{x}dx.$ Then,  $du=2xdx$ and  $v=e^{x}.$ Therefore, we have
$\int x^{2}e^{x}~dx=x^{2}e^{x}-\int 2xe^{x}~dx.$ Step 2:
Now, we need to use integration by parts again.
Let  $u=2x$ and  $dv=e^{x}dx.$ Then,  $du=2dx$ and  $v=e^{x}.$ Building on the previous step, we have
${\begin{array}{rcl}\displaystyle {\int x^{2}e^{x}~dx}&=&\displaystyle {x^{2}e^{x}-{\bigg (}2xe^{x}-\int 2e^{x}~dx{\bigg )}}\\&&\\&=&\displaystyle {x^{2}e^{x}-2xe^{x}+2e^{x}+C.}\end{array}}$ (b)

Step 1:
We proceed using integration by parts.
Let  $u=\ln x$ and  $dv=x^{3}dx.$ Then,  $du={\frac {1}{x}}dx$ and  $v={\frac {x^{4}}{4}}.$ Therefore, we have

${\begin{array}{rcl}\displaystyle {\int _{1}^{e}x^{3}\ln x~dx}&=&\displaystyle {\left.\ln x{\bigg (}{\frac {x^{4}}{4}}{\bigg )}\right|_{1}^{e}-\int _{1}^{e}{\frac {x^{3}}{4}}~dx}\\&&\\&=&\displaystyle {\left.\ln x{\bigg (}{\frac {x^{4}}{4}}{\bigg )}-{\frac {x^{4}}{16}}\right|_{1}^{e}.}\end{array}}$ Step 2:
Now, we evaluate to get
${\begin{array}{rcl}\displaystyle {\int _{1}^{e}x^{3}\ln x~dx}&=&\displaystyle {{\bigg (}(\ln e){\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}{\bigg )}-{\bigg (}(\ln 1){\frac {1^{4}}{4}}-{\frac {1^{4}}{16}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}+{\frac {1}{16}}}\\&&\\&=&\displaystyle {{\frac {3e^{4}+1}{16}}.}\end{array}}$ (a)     $x^{2}e^{x}-2xe^{x}+2e^{x}+C$ (b)     ${\frac {3e^{4}+1}{16}}$ 