009B Sample Midterm 1, Problem 3

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Evaluate the indefinite and definite integrals.

a)
b)


Foundations:  
Review integration by parts.

Solution:

(a)

Step 1:  
We proceed using integration by parts. Let and . Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2xdx} and .
Therefore, we have
   .
Step 2:  
Now, we need to use integration by parts again. Let and . Then, and .
Building on the previous step, we have
   .

(b)

Step 1:  
We proceed using integration by parts. Let and . Then, and .
Therefore, we have
   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{e}x^{3}\ln x~dx=\left.\ln x{\bigg (}{\frac {x^{4}}{4}}{\bigg )}\right|_{1}^{e}-\int _{1}^{e}{\frac {x^{3}}{4}}~dx=\left.\ln x{\bigg (}{\frac {x^{4}}{4}}{\bigg )}-{\frac {x^{4}}{16}}\right|_{1}^{e}} .
Step 2:  
Now, we evaluate to get
   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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)   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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