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

Final Answer:
(a)    $f(x)=-\cos(x^{2})+\cos(1)$
(b)    $f'(x)=\sin(x^{2})2x$
(c)    See above
(d)    $\sin(x^{2})2x$

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