# 009B Sample Midterm 3, Problem 2 Detailed Solution

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

$F(x)=\int _{\cos(x)}^{5}{\frac {1}{1+u^{10}}}~du.$ Background Information:
What does Part 1 of the Fundamental Theorem of Calculus
say is the derivative of  $G(x)=\int _{x}^{5}{\frac {1}{1+u^{10}}}~du?$ First, we need to switch the bounds of integration.

So, we have  $G(x)=-\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.$ By Part 1 of the Fundamental Theorem of Calculus,  $G'(x)=-{\frac {1}{1+x^{10}}}.$ Solution:

Step 1:
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,

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

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

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

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