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 $f(x)=\int _{-1}^{x}\sin(t^{2})2tdt$.

b) Find $f'(x)$.

c) State the Fundamental Theorem of Calculus.

d) Use the Fundamental Theorem of Calculus to compute  ${\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2tdt{\bigg )}$  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, $du=2tdt$.
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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $f(x)=-\cos(x^{2})+\cos(1)$.

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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b f(x)~dx=F(b)-F(a)} .
(d) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(x^2)2x}

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