009B Sample Midterm 2, Problem 2

This problem has three parts:

a) State the Fundamental Theorem of Calculus.

b) Compute

{\frac {d}{dx}}\int _{0}^{\cos(x)}\sin(t)~dt

.

c) Evaluate

\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 $f$ 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, $\int _{a}^{b}f(x)~dx=F(b)-F(a)$ .

Temp 1

(b)

Step 1:
Let $F(x)=\int _{0}^{\cos(x)}\sin(t)~dt$ . The problem is asking us to find $F'(x)$ .
Let $g(x)=\cos(x)$ and $G(x)=\int _{0}^{x}\sin(t)~dt$ .
Then, $F(x)=G(g(x))$ .

Step 2:
If we take the derivative of both sides of the last equation, we get $F'(x)=G'(g(x))g'(x)$ by the Chain Rule.

Step 3:
Now, $g'(x)=-\sin(x)$ and $G'(x)=\sin(x)$ by the Fundamental Theorem of Calculus, Part 1.
Since $G'(g(x))=\sin(g(x))=\sin(\cos(x))$ , we have $F'(x)=G'(g(x))g'(x)=\sin(\cos(x))(-\sin(x))$

(c)

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

Step 2:
So, we get
$\int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}-\tan(0)=1$ .

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