# Week 1 Assignment
## Advanced Level
1. Provide a clear description of the method of u-substitution for definite integrals (chapter 5.6, called “change of variables formula”) along with a proof of its validity, and a description of its relation to the chain rule.
$u$ subsitution is a method that is used to try to make an equation easier to solve by replacing a set of numbers with a $u$. Below is the change of variables formula. $$\int^b_af(u(x))u'(x)dx=\int^{u(b)}_{u(a)}(u)du$$
Here is an example of a problem that we can use $u$ subsitution to help solve it.
$$\int^2_0 \frac{x}{(1+x^2)^3}\cdot dx$$ To make this easier I will set $u=1+x^2$ now he equation can be rewritten to look like $\int^2_0\frac{x}{u^3}\cdot dx$.
Now to get the rest of the equation in terms of $u$ we must find $\frac{du}{dx}$, which is $2x$.
The equation can now become $\int^2_0\frac{du}{2u^3}$. It is now time to take the antiderivative of that equation. $\int^2_0\frac{1}{-4}u^2$ . It is time to plug $1+x^2$ back into the equation. $$\int^2_0\frac{1}{-4(1+x^2)^3}$$ Now the intergral can be solved with ease. After pluging in the $x$ values the answer should come out to be $\frac{-2}{9}$
Demonstrate its use by computing the integral:
$$\int^π_0\frac {cos(x)}{sin(x)^2+1}dx$$
You solution must include a drawing of the function and shading of the relevant integral area using Desmos, and in such a way that the 0 and π endpoints can be moved around.

First let $u = sin(x)$
Then you can find that $dx=\frac{du}{cos(x)}$
Now substitute $u$ into the equation $$\int^π_0\frac {cos(x)}{u^2+1}\cdot dx$$
After that plug $\frac{du}{cos(x)}$in for $dx$ $$\int^π_0\frac {cos(x)}{u^2+1}\cdot \frac{du}{cos(x)}$$
$$\int^π_0\frac 1{u^2+1}\cdot du$$
$\frac{d}{dx}\cdot tan^-(x)=\frac{1}{x^2+1}$ Can be appiled to this equation to yeild the following result
$$\int^π_0 tan^-(u)$$ After this step you can substitute $sin(x)$ back in place of $u$$$\int^π_0 tan^-(sinx)$$
$$tan^-(sin(0))-tan^-(sin(π))=0-0$$
$$0-0=0$$
Therefore, $$\int^π_0\frac {cos(x)}{sin(x)^2+1}dx=0$$
<iframe src="https://www.desmos.com/calculator/wjhl8rp5o2?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>
2.State the two forms of the fundamental theorem of calculus and discuss the kinds of problems that each form helps us solve. Include at least one example for each form. Also show the relation between the two forms: Show that if one form is true then we can use it to prove the other form is also true, and vice versa.

2. The first FTC is, if G antiderivative of $f$ on $[a,b]$ $(G'(x)=f(x))$, then$\int^b_a f(x)dx=G(b)-G(a)$
It would be used to intregrate integrals, and in turn would give the area under the curve of a given function.
Ex.
$$\int^4_2x^3dx$$
$$x^3dx=0.25x^4$$
$$\int^4_2x^3dx=0.25(4)^4-0.25(2)^4=60$$

The second FTC is $A(x)=\int^x_af(t)dt$, if $f(x)$ continued on $[a,b]$, then $A'(x)=f(x)$
This one is used in situations when you do not really know what range you are working with yet, so you can plug in a number for x accordingly.
Ex.
$\frac{d}{dx}[\int^x_0\sqrt{x+4}\cdot d(t)]$
$f(t)=\sqrt{x+4}$
$\frac{d}{dx}[\int^x_0f(t)d(t)]$=$\frac{d}{dx}[F(t)|^x_0]$
$\frac{d}{dx}[F(x)-F(0)]=f(x)-0$
$$f(x)=\sqrt{x+4}$$
(I'm sorry I couldn't find out how to do something like this in desmos)
The two theorems are linked by the following proof:
$F(x)=\int^x_af(t)\cdot dt$, where $a≤x≤b$
From that we can say that $F'(x)= \frac{lim}{Δx->0}\frac{F(x+Δx)-F(x)}{Δx}$ (I could not find out how to do the limit without the fraction)
That can be put into an intergal as shown below.
$$\frac{lim}{Δx->0}\frac{1}{Δx}(\int^{x+Δx}_af(t)dt-\int^x_af(t)dt)$$
This can be simplified into
$$\frac{lim}{Δx->0}\int^{x+Δx}_xf(t)dt$$ Now it can be said that $\frac{lim}{Δx->0}\int^{x+Δx}_xf(t)dt$ $=F'(x)$
So using the MVT of a definet ingegral we can say there is a $c$ in $x≤c≤(x+Δx)$
To find the area of something we would usually take the length times the width and for $c$ it would normally look like $f(c)\cdot Δx$. The next step in the proof would then be as follows
$$f(c)\cdot Δx= \int^{x+Δx}_xf(t)dt$$ This simplifies into $$f(c)=\frac{\int^{x+Δx}_xf(t)dt}{Δx}$$
We can now use subsitution and get that $f(c)=F'(x)$
This is an important step, but not the last one. Now we must prove that $f(x)=F'(x)$.
We have to remember that $x≤c≤(x+Δx)$ and from this we can determine that $\frac{lim}{Δx->0}x=x$ and $\frac{lim}{Δx->0}x+Δx=x$ from these two things we can say that $\frac{lim}{Δx->0}c(Δx)=x$
Therefore we can say that as $x$ approches 0 $c$ approches $x$, which would then mean that $f(c)$ is getting closer to $f(x)$ as $x$ approches 0.
When this all comes together we get that $\frac{lim}{Δx->0}f(c)=F'(x)$
This allows us to turn integrals into action and get real numbers from them.
3. $f'(a)=\frac{lim}{h->0}\frac{f(a+h)-f(a)}{h}$
$$f(x+h)=\frac{1}{(x+h)^2}=\frac{1}{x^2+2xh+^2}$$
$f(x)=x^{-2}$
$f'(a)=\frac{(x^2+2xh+h^2)^--x^{-2}}{h}$= $-\frac{(2x+h}{x^2(x^2+2xh+h^2)}$
Plug 0 in as h
$-2x^{-3}$= $f'(x)$
It is graphed as shown below.
<iframe src="https://www.desmos.com/calculator/8bsi5icout?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

The second method is $f'(a)=\frac{lim}{x->a}\frac{f(x)-f(a)}{x-a}$
$$-2x^{-3}=\frac{lim}{x->a}\frac{f(x)-f(a)}{x-a}$$
$$-2x^{-3}(x-a)=f(x)-f(a)$$
$$-2x^{-2}+2x^{-3}a=x^{-2}-f(a)
$$f(a)=2x^{-2}-2x^{-3}a+x^2$$
$x=a$ $3x^{-2}-2x^{-2}=f(a)$
$x^{-2}=f(a)$
I know this was not done the way you asked, but I could not find out the correct way and this still proves that the equation works because $f(x)=f(a)=x^{-2}$
4.
I did not realize how long it would take to program all of this and that's my fault. So, I am willing to use a token to correct my mistake next week. Sorry.