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)  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, $\int _{a}^{b}f(x)~dx=F(b)-F(a)$.
(d)  $\sin(x^{2})2x$

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