# Difference between revisions of "009B Sample Midterm 2, Problem 1"

This problem has three parts:

(a) State the Fundamental Theorem of Calculus.

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

(c) Evaluate  ${\displaystyle \int _{0}^{\pi /4}\sec ^{2}x~dx.}$

Foundations:
1. 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

${\displaystyle {\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt=\sin(x).}$
2. 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  ${\displaystyle \sec ^{2}x.}$

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:
Let  ${\displaystyle F(x)=\int _{0}^{\cos(x)}\sin(t)~dt.}$
The problem is asking us to find  ${\displaystyle F'(x).}$
Let  ${\displaystyle g(x)=\cos(x)}$  and  ${\displaystyle G(x)=\int _{0}^{x}\sin(t)~dt.}$
Then,
${\displaystyle F(x)=G(g(x)).}$
Step 2:
If we take the derivative of both sides of the last equation,
we get
${\displaystyle F'(x)=G'(g(x))g'(x)}$
by the Chain Rule.
Step 3:
Now,  ${\displaystyle g'(x)=-\sin(x)}$  and  ${\displaystyle G'(x)=\sin(x)}$
by the Fundamental Theorem of Calculus, Part 1.
Since
${\displaystyle G'(g(x))=\sin(g(x))=\sin(\cos(x)),}$
we have
${\displaystyle F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot (-\sin(x)).}$

(c)

Step 1:
Using the Fundamental Theorem of Calculus, Part 2, we have
${\displaystyle \int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan(x){\biggr |}_{0}^{\pi /4}.}$
Step 2:
So, we get
${\displaystyle \int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}-\tan(0)=1.}$

(b)     ${\displaystyle \sin(\cos(x))\cdot (-\sin(x))}$
(c)     ${\displaystyle 1}$