For real numbers $\alpha_1$ & $\alpha_2$, if the formula $\int_1^{-1}f(x) \ dx=\alpha_1f(1/2)+\alpha_2f(-1/2)$ is exact for all polynomials of degree $\le1$ then what is the value of $2\alpha_1+3\alpha_2$ ?
My trial: assume $f(x)=ax+b$ then we have $$\int_1^{-1}(ax+b)\ dx=\alpha_1\left(\frac{a}{2}+b\right)+\alpha_2\left(-\frac{a}{2}+b\right)$$ $$0+b(1+1)=a\left(\frac{\alpha_1-\alpha_2}{2}\right)+b\left(\alpha_1+\alpha_2\right)$$ $$0\cdot a+2b=a\left(\frac{\alpha_1-\alpha_2}{2}\right)+b\left(\alpha_1+\alpha_2\right)$$ by comparing coefficients of a & b on both the sides, we get $$\alpha_1-\alpha_2=0\tag 1$$ $$\alpha_1+\alpha_2=2\tag 2$$ solving (1), (2) gives us $\alpha_1= \alpha_2=1$ hence $$2\alpha_1+3\alpha_2=2\cdot1+3\cdot 1=5$$ But I am not sure if i am correct. please explain if i am wrong. thanks
The integral of ax+b is equal to: $$ (a/2)x^{2}+bx+C$$ Now from -1 to 1 it is going to be: $$a/2+b+C-a/2+b-C=2b$$ Now we have $$2b=α_{1}f(1/2)+α_{2}f(-1/2)$$ and we now have: $$f(1/2)=1/2a+b$$ and $$f(-1/2)=-1/2a+b$$ So we have $$2b=α_{1}(1/2a+b)+α_{2}(-1/2a+b)$$ And so $$α_{1}=α_{2}=1$$