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

Part 2 of the Fundamental Theorem of Calculus says that

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

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

(a)

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

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