I have $f:[-3,3]\to \mathbb{C}$ and $\langle f,g\rangle=\int_{-3}^{3}f(x)\overline{g(x)}dx$
So $\langle x^2,x+x^3\rangle=\int_{-3}^{3}=x^2\cdot (x-x^3)dx?$
I have $f:[-3,3]\to \mathbb{C}$ and $\langle f,g\rangle=\int_{-3}^{3}f(x)\overline{g(x)}dx$
So $\langle x^2,x+x^3\rangle=\int_{-3}^{3}=x^2\cdot (x-x^3)dx?$
No. It's $\int_{-3}^3x^2(x+x^3)\,\mathrm dx$.