Lagrange Interpolation Polynomial word problem proofs I can't even read this, I have no idea what I am being asked. I just need some help intrepetitng this mess of a question. 
Suppose we are given three points (x, y) (x1, y1) (x2,y2) and we want to find a unique parabola P(x) that passes through these three points. This is given the Lagrange Interpolation Polynomial
$$P(x) = y_0 \frac{(x - x_ 1)(x- x_2)}{(x_0 - x_1)(x_0 - x_2)} y_1 \frac{(x - x_ 1)(x- x_2)}{(x_0 - x_1)(x_0 - x_2)} y_2 \frac{(x - x_ 1)(x- x_2)}{(x_0 - x_1)(x_0 - x_2)}$$
a) Confirm that $P(x_i) = y_i$ for $i = 0, 1 ,2$.
I have no idea what is being asked here at all. What does i reference? 
b) Show that $\int_{x_0}^{x_2} P(x) dx = \frac{\Delta x}{3}(P(x_0) + 4P(x_1) + P(x_2))$
What does all of that mean? This problem is dirty.
Then I am suppose to derive Simpsons rule...
 A: The Lagrange Interpolation formula you wrote down is not right. That may be the source of your problems. We will copy your formula, and make the (several) necessary changes.
$$P(x) = y_0 \frac{(x - x_ 1)(x- x_2)}{(x_0 - x_1)(x_0 - x_2)}+ y_1 \frac{(x - x_ 0)(x- x_2)}{(x_1 - x_0)(x_1 - x_2)} +y_2 \frac{(x - x_0)(x- x_1)}{(x_2 - x_0)(x_2 - x_1)}.$$
The first thing you are asked to do is to show that this formula "works." To do that, you need to verify that $P(x_0)=y_0$, and $P(x_1)=y_1$, and $P(x_2)=y_2$.
(That's what the "$i$" stuff is about, you need to verify three things, though the verifications are essentially identical.)
That is straightforward. Plug in $x_0$ for $x$ in the formula. Notice how the last two terms die, and  in the first term we get 
$y_0\frac{(x_0-x_1)(x_0-x_2)}{(x_0-x_1)(x_0-x_2)}$, which is a fancy way of writing $y_0$.
By the way, we don't necessarily get a parabola. We do get a polynomial of degree $\le 2$, but if your given points $(x_0,y_0)$, $(x_1,y_1)$, $(x_2,y_2)$ lie on a line, then the "$x^2$" term might be $0$. And in the degenerate case where the $y_i$ are all the same, there will not be an $x$ term either. 
The (b) part is slightly unpleasant, but fairly important, since it is connected with Simpson's Rule for numerical integration.  The notation is undoubtedly explained in the problem, or in the text above it. There will be probably a very specific choice of $x_1$, halfway between $x_0$ and $x_2$. And $\Delta x$ will be half the distance between $x_0$ and $x_2$. So the terms in the Lagrange Interpolation Formula are not "general." The three denominators are $2(\Delta x)^2$, $(\Delta x)^2$, and $2(\Delta x)^2$.  It will be simplest if you use $x_1-\Delta x$ for $x_0$, and $x_1+\Delta x$ for $x_2$. But do check the specific instructions in your book. The integration is not as messy as it looks, since you are integrating a quadratic. 
Remark: We tend to think of subscripts as necessary. But Euler happily did without, using $a,b,c,d\dots,z$. And I hear he did OK. 
A: You probably meant:
$$P(x_i) = y_i$$
Meaning: "show that this equation actually does pass through the three points $(x_0,y_0)\ldots$ You can show this by plugging in $x_0, x_1, x_2$ instead of $x$, simplifying and seeing that you indeed get $y_i$.
