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,  $\int _{a}^{b}f(x)~dx=F(b)-F(a).$

(b)

Step 1:
The Fundamental Theorem of Calculus Part 2 says that
$\int _{0}^{1}{\frac {d}{dx}}(e^{\arctan(x)})~dx=F(1)-F(0)$
where  $F(x)$  is any antiderivative of  ${\frac {d}{dx}}(e^{\arctan(x)}).$
Thus, we can take
$F(x)=e^{\arctan(x)}$
since then $F'(x)={\frac {d}{dx}}(e^{\arctan(x)}).$

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
${\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt=\sin {\bigg (}{\frac {1}{x}}{\bigg )}{\frac {d}{dx}}{\bigg (}{\frac {1}{x}}{\bigg )}.$

Step 2:
Hence, we have
${\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt=\sin {\bigg (}{\frac {1}{x}}{\bigg )}{\bigg (}-{\frac {1}{x^{2}}}{\bigg )}.$

Final Answer:
(a)    See above
(b)    $e^{\frac {\pi }{4}}-1$
(c)    $\sin {\bigg (}{\frac {1}{x}}{\bigg )}{\bigg (}-{\frac {1}{x^{2}}}{\bigg )}$

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