009B Sample Midterm 2, Problem 3
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Evaluate:
- a)
- b)
| Foundations: |
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| Review -substitution |
Solution:
(a)
| Step 1: |
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| We multiply the product inside the integral to get |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\int _{1}^{2}{\bigg (}8t^{3}-10+12-{\frac {15}{t^{3}}}{\bigg )}~dt=\int _{1}^{2}(8t^{3}+2-15t^{-3})~dt} . |
| Step 2: |
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| We integrate to get |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\left.2t^{4}+2t+{\frac {15}{2}}t^{-2}\right|_{1}^{2}} . |
| We now evaluate to get |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=2(2)^{4}+2(2)+{\frac {15}{2(2)^{2}}}-{\bigg (}2+2+{\frac {15}{2}}{\bigg )}=36+{\frac {15}{8}}-4-{\frac {15}{2}}={\frac {211}{8}}} . |
(b)
| Step 1: |
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| We use -substitution. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x^{4}+2x^{2}+4} . Then, and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{4}}=(x^{3}+x)dx} . Also, we need to change the bounds of integration. |
| Plugging in our values into the equation Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x^{4}+2x^{2}+4} , we get and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=2^{4}+2(2)^{2}+4=28} . |
| Therefore, the integral becomes Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du} . |
| Step 2: |
|---|
| We now have: |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du=\left.{\frac {1}{6}}u^{\frac {3}{2}}\right|_{4}^{28}={\frac {1}{6}}(28^{\frac {3}{2}}-4^{\frac {3}{2}})={\frac {1}{6}}(({\sqrt {28}})^{3}-({\sqrt {4}})^{3})={\frac {1}{6}}((2{\sqrt {7}})^{3}-2^{3})} . |
| So, 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 \int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}} . |
| 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 \frac{211}{8}} |
| (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{28\sqrt{7}-4}{3}} |