Prove that $L:P_5(\Bbb R) \mapsto \Bbb R^5$ defined by $L(f(x))=(f(-2),f(-1),f(0),f(1),f(2))$ is an isomorphism, then show that for any $f(x)$ there exists unique $a_1,a_2$ such that $$\int_{-1}^1 f(x)dx=2(1-a_1-a_2)f(0)+a_1(f(1)+f(-1))+a_2(f(2)+f(-2))$$
I have shown that the linear map is an isomorphism, but I'm stuck at the second part. Suppose $L^{-1}$ is the inverse map, $L_0,\dots,L_4$ are linear combination of the values $(a,b,c,d,e)\in \Bbb R^5$, then we have
$$f(x)=L^{-1}(a,b,c,d,e)=L_0(a,b,c,d,e)+L_1(a,b,c,d,e)x+L_2(a,b,c,d,e)x^2+L_3(a,b,c,d,e)x^3+L_4(a,b,c,d,e)x^4$$
$$\int_{-1}^1f(x)dx=2L_0+\frac 23 L_2 +\frac 25 L_4$$
How do I complete the proof?
Using your equation where you write $f(x) = L^{-1}(a,b,c,d,e)$, we get: $$\int_{-1}^1f(x)dx = 2(1-a_1-a_2)f(0)+a_1(f(1)+f(-1))+a_2(f(2)+f(-2))$$$$ = 2(1-a_1-a_2)L_0+a_1(2L_0+2L_2+2L_4)+a_2(2L_0+8L_2+32L_4) $$$$= 2L_0+(2a_1+8a_2)L_2+(2a_1+32a_2)L_4$$
You also have the equation you've already obtained, $$2L_0+\frac{2}{3}L_2+\frac{2}{5}L_4.$$
All that's left is to equate coefficients, leaving you to solve a system of two equations with two variables.