009B Sample Midterm 1, Problem 3
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Evaluate the indefinite and definite integrals.
- a)
- b)
| Foundations: |
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| Review integration by parts. |
Solution:
(a)
| Step 1: |
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| 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: |
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| Now, we need to use integration by parts again. Let and . Then, and . |
| Building on the previous step, we have |
| . |
(b)
| Step 1: |
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| 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: |
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| 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: |
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| (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}} |