009B Sample Final 2, Problem 1

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

(b) Evaluate the integral

$\int _{0}^{1}{\frac {d}{dx}}{\bigg (}e^{\tan ^{-1}(x)}{\bigg )}dx$

(c) Compute

${\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt$

Foundations:

Solution:

(a)

Step 1:
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
Let  $f$  be continuous on  $[a,b]$  and let  $F(x)=\int _{a}^{x}f(t)~dt.$
Then,  $F$  is a differentiable function on  $(a,b)$  and  $F'(x)=f(x).$

Step 2:
The Fundamental Theorem of Calculus, Part 2
Let  $f$  be continuous on  $[a,b]$  and let  $F$  be any antiderivative of  $f.$
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 \int_a^b f(x)~dx=F(b)-F(a).}

(b)

Step 1:
The Fundamental Theorem of Calculus Part 2 says that
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}(e^{\arctan(x)})~dx=F(1)-F(0)}
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}(e^{\arctan(x)}).}
Thus, we can take
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)}}
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}(e^{\arctan(x)}).}

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {d}{dx}}(e^{\arctan(x)})~dx}&=&\displaystyle {F(1)-F(0)}\\&&\\&=&\displaystyle {e^{\arctan(1}-e^{\arctan(0)}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-e^{0}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-1.}\end{array}}$

(c)

Step 1:
Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, 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)\frac{d}{dx}\bigg(\frac{1}{x}\bigg).}

Step 2:
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:
(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)}

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