# 007B 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  $\int _{0}^{\pi /4}\sec ^{2}x~dx.$ Background Information:
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
$\int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}-\tan(0)=1.$ (b)     $\sin(\cos(x))\cdot (-\sin(x))$ (c)     $1$ 