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 $F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot (-\sin(x))$.

Temp 2

(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) $\int _{0}^{\pi /4}\sec ^{2}x~dx\,=\,1$.

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