009B Sample Midterm 1, Problem 3

Evaluate the indefinite and definite integrals.

(a)   $\int x^{2}e^{x}~dx$

(b)   $\int _{1}^{e}x^{3}\ln x~dx$

Foundations:
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v={\frac {x^{4}}{4}}.}
Therefore, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
${\begin{array}{rcl}\displaystyle {\int _{1}^{e}x^{3}\ln x~dx}&=&\displaystyle {{\bigg (}(\ln e){\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}{\bigg )}-{\bigg (}(\ln 1){\frac {1^{4}}{4}}-{\frac {1^{4}}{16}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}+{\frac {1}{16}}}\\&&\\&=&\displaystyle {{\frac {3e^{4}+1}{16}}.}\end{array}}$

Final Answer:
(a)     $x^{2}e^{x}-2xe^{x}+2e^{x}+C$
(b)     ${\frac {3e^{4}+1}{16}}$

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