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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=x~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^{2}}{2}}.} Thus,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
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 \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
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 \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)   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 x^2e^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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