I am having some trouble with a math problem ( attached document). It is probably something easy, but I just can't see it. Thank you in advance!

2. On part (a), did you try directly computing both sides and solving the resulting system of equations for $\omega_0, \omega_1$?

3. I did. I get w0 + w1 =2. I was not sure that fully prove that w0=w1=1. Any ideas for b) ?

4. You're right. Your single equation doesn't prove part (a) yet. Let me just review the situation for you:

We want to compute $\int_{-1}^1 f(x)\text{ dx}=\omega_0 f(-\alpha)+\omega_1 f(\alpha)$, where $f(x)=mx+b$, for $m\ne 0$. Taking the integral of the left hand side and evaluating the right hand side gives

$2b=\omega_0(-m\alpha+b)+\omega_1(m\alpha+b)$

Expanding the terms:

$2b=(\omega_0-\omega_1)(m\alpha)+(\omega_0+\omega_1)b$

Now, we have two systems of equations (by comparing the coefficients of each $m\alpha$ and $b$ on both sides):

$\omega_0-\omega_1=0$
$\omega_0+\omega_1=2$

Solving the system shows that $\omega_0=\omega_1=1$.

For part (b), I might apply the same strategy and see what happens.

5. Any luck for b) ?

6. Did you try it?

7. I did. I applied the same strategy for b). However, I did no get the answer. To be honest, this happened probably because I am not exactly sure what I am supposed to get as an answer and whether I should put w0=w1=1 (the result from a) ). Any thoughts?

8. I agree it is not very clear but to get a single value of $\alpha$ in (b) you must use the values from (a): $\omega_0= \omega_1= 1$.

For a quadratic function, $a_0+ a_1x+ a_2x^2$, the formula becomes $\int_{-1}^1 a_0+ a_1x+ a_2x^2 dx= f(-\alpha)+ f(\alpha)$.

It is easy to calculate that
$\int_{-1}^1 a_0+ a_1x+ a_2x^2 dx= 2a_0+ \frac{2}{3}a_2$ while $\omega_0f(-\alpha)+ \omega_1f(\alpha)=$ $a_0- a_1\alpha+ a_2\alpha^2+ a_0+ a_1\alpha+ a_2\alpha^2= 2a_0+ 2a_2\alpha^2$.

9. So, as a result we get that alpha equals +- the square root of 1/3. Is that correct?

10. Yes, I believe it is.