009B Sample Final 2, Problem 1
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(a) State both parts of the Fundamental Theorem of Calculus.
(b) Evaluate the integral
(c) Compute
| Foundations: |
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| 1. What does Part 2 of the Fundamental Theorem of Calculus say about where are constants? |
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Part 2 of the Fundamental Theorem of Calculus says that |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{a}^{b}\sec ^{2}x~dx=F(b)-F(a)} where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F} is any antiderivative of |
| 2. What does Part 1 of the Fundamental Theorem of Calculus say about |
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Part 1 of the Fundamental Theorem of Calculus says that |
Solution:
(a)
| Step 1: |
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| The Fundamental Theorem of Calculus has two parts. |
| The Fundamental Theorem of Calculus, Part 1 |
| Let be continuous on and let |
| Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F} is a differentiable function on and |
| Step 2: |
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| The Fundamental Theorem of Calculus, Part 2 |
| Let be continuous on and let be any antiderivative of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f.} |
| Then, |
(b)
| Step 1: |
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| The Fundamental Theorem of Calculus Part 2 says that |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{1}{\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}~dx=F(1)-F(0)} |
| where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F(x)} is any antiderivative of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}.} |
| Thus, we can take |
| since then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F'(x)={\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}.} |
| Step 2: |
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| Now, we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}~dx}&=&\displaystyle {F(1)-F(0)}\\&&\\&=&\displaystyle {e^{\arctan(1)}-e^{\arctan(0)}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-e^{0}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-1.}\end{array}}} |
(c)
| Step 1: |
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| Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, we have |
| Step 2: |
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| Hence, 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 \frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt=\sin\bigg(\frac{1}{x}\bigg)\bigg(-\frac{1}{x^2}\bigg).} |
| Final Answer: |
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| (a) See above |
| (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 e^{\frac{\pi}{4}}-1} |
| (c) 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 \sin\bigg(\frac{1}{x}\bigg)\bigg(-\frac{1}{x^2}\bigg)} |