009B Sample Midterm 2, Problem 1

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

Final Answer:
(a)     See solution above.
(b)     $\sin(\cos(x))\cdot (-\sin(x))$
(c)     $1$

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