009B Sample Final 1, Problem 2

We would like to evaluate

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

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:
$u$-substitution

Solution:

(a)

Step 1:
We proceed using $u$-substitution. Let $u=t^{2}$. Then, $du=2tdt$.
Since this is a definite integral, we need to change the bounds of integration.
Plugging in 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
$f(x)=\int _{-1}^{x}\sin(t^{2})2t~dt=\int _{1}^{x^{2}}\sin(u)du=\left.-\cos(u)\right|_{1}^{x^{2}}=-\cos(x^{2})+\cos(1)$.

(b)

Step 1:
From part (a), we have $f(x)=-\cos(x^{2})+\cos(1)$.

Step 2:
If we take the derivative, we get $f'(x)=\sin(x^{2})2x$.

(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)

Step 1:
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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