# 009B Sample Midterm 2, Problem 2

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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:
Review the Fundamental Theorem of Calculus.

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
${\displaystyle \int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}-\tan(0)=1}$.
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 ${\displaystyle F'(x)=f(x)}$.
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)   ${\displaystyle {\frac {d}{dx}}\int _{0}^{\cos(x)}\sin(t)~dt\,=\,\sin(\cos(x))\cdot (-\sin(x))}$.
(c) ${\displaystyle \int _{0}^{\pi /4}\sec ^{2}x~dx\,=\,1}$.