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:
Review integration by parts.

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
$\int x^{2}e^{x}~dx=x^{2}e^{x}-\int 2x~dx$.

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
$\int x^{2}e^{x}~dx=x^{2}e^{x}-{\bigg (}2xe^{x}-\int 2e^{x}~dx{\bigg )}=x^{2}e^{x}-2xe^{x}+2e^{x}+C$.

(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
$\int _{1}^{e}x^{3}\ln x~dx=\left.\ln x{\frac {x^{4}}{4}}\right|_{1}^{e}-\int _{1}^{e}{\frac {x^{3}}{4}}~dx=\left.\ln x{\frac {x^{4}}{4}}-{\frac {x^{4}}{16}}\right|_{1}^{e}$.

Step 2:
Now, we evaluate to get
$\int _{1}^{e}x^{3}\ln x~dx={\bigg (}\ln e{\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}{\bigg )}-{\bigg (}\ln 1{\frac {1^{4}}{4}}-{\frac {1^{4}}{16}}{\bigg )}={\frac {e^{4}}{4}}-{\frac {e^{4}}{16}}+{\frac {1}{16}}={\frac {3e^{4}+1}{16}}$.

Final Answer:
(a)   $x^{2}e^{x}-2xe^{x}+2e^{x}+C$
(b)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3e^4+1}{16}}

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