# 009B Sample Final 1, Problem 2

We would like to evaluate

${\displaystyle {\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t\,dt{\bigg )}.}$

(a) Compute  ${\displaystyle f(x)=\int _{-1}^{x}\sin(t^{2})2t\,dt.}$

(b) Find  ${\displaystyle f'(x).}$

(c) State the Fundamental Theorem of Calculus.

(d) Use the Fundamental Theorem of Calculus to compute  ${\displaystyle {\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t\,dt{\bigg )}}$  without first computing the integral.

Foundations:
How would you integrate  ${\displaystyle \int e^{x^{2}}2x~dx?}$

You could use  ${\displaystyle u}$-substitution.

Let  ${\displaystyle u=x^{2}.}$  Then,  ${\displaystyle du=2xdx.}$

So, we get  ${\displaystyle \int e^{u}~du=e^{u}+C=e^{x^{2}}+C.}$

Solution:

(a)

Step 1:
We proceed using  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=t^{2}.}$  Then,  ${\displaystyle du=2t\,dt.}$
Since this is a definite integral, we need to change the bounds of integration.
Plugging our values into the equation  ${\displaystyle u=t^{2},}$  we get
${\displaystyle u_{1}=(-1)^{2}=1}$  and  ${\displaystyle u_{2}=x^{2}.}$
Step 2:
So, we have

${\displaystyle {\begin{array}{rcl}f(x)&=&\displaystyle {\int _{-1}^{x}\sin(t^{2})2t~dt}\\&&\\&=&\displaystyle {\int _{1}^{x^{2}}\sin(u)~du}\\&&\\&=&\displaystyle {-\cos(u){\bigg |}_{1}^{x^{2}}}\\&&\\&=&\displaystyle {-\cos(x^{2})+\cos(1)}.\\\end{array}}}$

(b)

Step 1:
From part (a), we have  ${\displaystyle f(x)=-\cos(x^{2})+\cos(1).}$
Step 2:
If we take the derivative, we get  ${\displaystyle f'(x)=\sin(x^{2})2x,}$  since  ${\displaystyle \cos(1)}$  is a constant.

(c)

Step 1:
The Fundamental Theorem of Calculus has two parts.
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  ${\displaystyle F'(x)=f(x).}$
Step 2:
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,  ${\displaystyle \int _{a}^{b}f(x)~dx=F(b)-F(a).}$
(d)
By the Fundamental Theorem of Calculus, Part 1,

${\displaystyle {\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t~dt{\bigg )}\,=\,\sin(x^{2})2x.}$

(a)    ${\displaystyle f(x)=-\cos(x^{2})+\cos(1)}$
(b)    ${\displaystyle f'(x)=\sin(x^{2})2x}$
(d)    ${\displaystyle \sin(x^{2})2x}$