(a) State both parts of the Fundamental Theorem of Calculus.
(b) Evaluate the integral

(c) Compute

Solution:
(a)
| Step 1:
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| The Fundamental Theorem of Calculus has two parts.
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| The Fundamental Theorem of Calculus, Part 1
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Let be continuous on and let
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Then, is a differentiable function on and
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| Step 2:
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| The Fundamental Theorem of Calculus, Part 2
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Let be continuous on and let be any antiderivative of
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Then,
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(b)
| Step 1:
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| The Fundamental Theorem of Calculus Part 2 says that
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| 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_0^1 \frac{d}{dx}\bigg(e^{\arctan(x)}\bigg)~dx=F(1)-F(0)}
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| where 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 F(x)}
is any antiderivative of 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}\bigg(e^{\arctan(x)}\bigg).}
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| Thus, we can take
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| 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 F(x)=e^{\arctan(x)}}
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| since then 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 F'(x)=\frac{d}{dx}\bigg(e^{\arctan(x)}\bigg).}
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| Step 2:
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| Now, we have
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| 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_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}}
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(c)
| Step 1:
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| Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, we have
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| 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)\frac{d}{dx}\bigg(\frac{1}{x}\bigg).}
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|
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| Step 2:
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| Hence, we have
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| 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).}
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| Final Answer:
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| (a) See above
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| (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}
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| (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)}
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