# 009B Sample Midterm 3, Problem 2 Detailed Solution

State the fundamental theorem of calculus, and use this theorem to find the derivative of

${\displaystyle F(x)=\int _{\cos(x)}^{5}{\frac {1}{1+u^{10}}}~du.}$

Background Information:
What does Part 1 of the Fundamental Theorem of Calculus
say is the derivative of  ${\displaystyle G(x)=\int _{x}^{5}{\frac {1}{1+u^{10}}}~du?}$

First, we need to switch the bounds of integration.

So, we have  ${\displaystyle G(x)=-\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.}$

By Part 1 of the Fundamental Theorem of Calculus,  ${\displaystyle G'(x)=-{\frac {1}{1+x^{10}}}.}$

Solution:

Step 1:
The Fundamental Theorem of Calculus, Part 1

Let  ${\displaystyle f}$  be continuous on  ${\displaystyle [a,b]}$  and let  ${\displaystyle F(x)=\int _{a}^{x}f(t)~dt.}$

Then,  ${\displaystyle F}$  is a differentiable function on  ${\displaystyle (a,b)}$  and  ${\displaystyle F'(x)=f(x).}$

The Fundamental Theorem of Calculus, Part 2

Let  ${\displaystyle f}$  be continuous on  ${\displaystyle [a,b]}$  and let  ${\displaystyle F}$  be any antiderivative of  ${\displaystyle f.}$ Then,

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

Step 2:
First,
${\displaystyle F(x)=-\int _{5}^{\cos(x)}{\frac {1}{1+u^{10}}}~du.}$
Now, let  ${\displaystyle g(x)=\cos(x)}$  and  ${\displaystyle G(x)=\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.}$
Therefore,

${\displaystyle F(x)=-G(g(x)).}$

Hence,
${\displaystyle F'(x)=-G'(g(x))g'(x)}$
by the Chain Rule.
Step 3:
Now,
${\displaystyle g'(x)=-\sin(x).}$
By the Fundamental Theorem of Calculus,

${\displaystyle G'(x)={\frac {1}{1+x^{10}}}.}$

Hence,

${\displaystyle {\begin{array}{rcl}\displaystyle {F'(x)}&=&\displaystyle {-{\frac {1}{1+\cos ^{10}x}}(-\sin(x))}\\&&\\&=&\displaystyle {{\frac {\sin(x)}{1+\cos ^{10}x}}.}\\\end{array}}}$

${\displaystyle F'(x)={\frac {\sin(x)}{1+\cos ^{10}x}}}$