009B Sample Midterm 2, Problem 1 Detailed Solution

This problem has three parts:

(a) State both parts of 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:
1. 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
${\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt=\sin(x).$
2. 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 $\sec ^{2}x.$

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

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

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