# 009B Sample Final 1, Problem 2

We would like to evaluate

${\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t\,dt{\bigg )}.$ (a) Compute  $f(x)=\int _{-1}^{x}\sin(t^{2})2t\,dt.$ (b) Find  $f'(x).$ (c) State the Fundamental Theorem of Calculus.

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

Foundations:
How would you integrate  $\int e^{x^{2}}2x~dx?$ You could use  $u$ -substitution.

Let  $u=x^{2}.$ Then,  $du=2xdx.$ So, we get  $\int e^{u}~du=e^{u}+C=e^{x^{2}}+C.$ Solution:

(a)

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

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

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

(c)

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,  $\int _{a}^{b}f(x)~dx=F(b)-F(a).$ (d)
By the Fundamental Theorem of Calculus, Part 1,

${\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t~dt{\bigg )}\,=\,\sin(x^{2})2x.$ (a)    $f(x)=-\cos(x^{2})+\cos(1)$ (b)    $f'(x)=\sin(x^{2})2x$ (d)    $\sin(x^{2})2x$ 