# 009B Sample Final 2, Problem 1

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

(b) Evaluate the integral

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

(c) Compute

${\displaystyle {\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt}$
Foundations:
1. What does Part 2 of the Fundamental Theorem of Calculus say about  ${\displaystyle \int _{a}^{b}\sec ^{2}x~dx}$  where  ${\displaystyle a,b}$  are constants?

Part 2 of the Fundamental Theorem of Calculus says that

${\displaystyle \int _{a}^{b}\sec ^{2}x~dx=F(b)-F(a)}$  where  ${\displaystyle F}$  is any antiderivative of  ${\displaystyle \sec ^{2}x.}$
2. What does Part 1 of the Fundamental Theorem of Calculus say about  ${\displaystyle {\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt?}$

Part 1 of the Fundamental Theorem of Calculus says that

${\displaystyle {\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt=\sin(x).}$

Solution:

(a)

Step 1:
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
Let  ${\displaystyle f}$  be continuous on  ${\displaystyle [a,b]}$  and let  ${\displaystyle F(x)=\int _{a}^{x}f(t)~dt.}$
Then,  ${\displaystyle F}$  is a differentiable function on  ${\displaystyle (a,b)}$  and  ${\displaystyle F'(x)=f(x).}$
Step 2:
The Fundamental Theorem of Calculus, Part 2
Let  ${\displaystyle f}$  be continuous on  ${\displaystyle [a,b]}$  and let  ${\displaystyle F}$  be any antiderivative of  ${\displaystyle f.}$
Then,  ${\displaystyle \int _{a}^{b}f(x)~dx=F(b)-F(a).}$

(b)

Step 1:
The Fundamental Theorem of Calculus Part 2 says that
${\displaystyle \int _{0}^{1}{\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}~dx=F(1)-F(0)}$
where  ${\displaystyle F(x)}$  is any antiderivative of  ${\displaystyle {\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}.}$
Thus, we can take
${\displaystyle F(x)=e^{\arctan(x)}}$
since then ${\displaystyle F'(x)={\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}.}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {d}{dx}}{\bigg (}e^{\arctan(x)}{\bigg )}~dx}&=&\displaystyle {F(1)-F(0)}\\&&\\&=&\displaystyle {e^{\arctan(1)}-e^{\arctan(0)}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-e^{0}}\\&&\\&=&\displaystyle {e^{\frac {\pi }{4}}-1.}\end{array}}}$

(c)

Step 1:
Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, we have
${\displaystyle {\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt=\sin {\bigg (}{\frac {1}{x}}{\bigg )}{\frac {d}{dx}}{\bigg (}{\frac {1}{x}}{\bigg )}.}$
Step 2:
Hence, we have
${\displaystyle {\frac {d}{dx}}\int _{1}^{\frac {1}{x}}\sin t~dt=\sin {\bigg (}{\frac {1}{x}}{\bigg )}{\bigg (}-{\frac {1}{x^{2}}}{\bigg )}.}$
(b)    ${\displaystyle e^{\frac {\pi }{4}}-1}$
(c)    ${\displaystyle \sin {\bigg (}{\frac {1}{x}}{\bigg )}{\bigg (}-{\frac {1}{x^{2}}}{\bigg )}}$