show that $L_{0}(x)=1$ ,$L_{1}(x)=1-x$ and $L_{2}(x)=1-2x +\frac{x^{2}}{2}$ are mutually orthogonal on the interval $(0,\infty )$ with respect to the weighting function $r(x)=e^{-x}$
I have trouble advancing with this problem, though it seemed quite straight forward at glance.
I've done this, but I don't know if it's okay
$\left \langle L_{0},L_{1} \right \rangle=\int_{0}^{\infty }(1-x)e^{-x}dx$
$\left \langle L_{0},L_{2} \right \rangle=\int_{0}^{\infty }(1-2x +\frac{x^{2}}{2})e^{-x}dx$
$\left \langle L_{1},L_{2} \right \rangle=\int_{0}^{\infty }(1-x)(1-2x +\frac{x^{2}}{2})e^{-x}dx$
Yes, your approach is right. Now all you need is to compute the integrals. Note that you can simplify your calculations. After you show $\left \langle L_{0},L_{1} \right \rangle=0$, you rewrite $\left \langle L_{0},L_{2} \right \rangle$ as $$\left \langle L_{0},L_{2} \right \rangle=\int_0^\infty(1-x)e^{-x}dx+\int_0^\infty(-x+\frac{x^2}{2})e^{-x}dx$$ The first term is just $\left \langle L_{0},L_{1} \right \rangle=0$. Similarly, for the last integral you write $$\left \langle L_{1},L_{2} \right \rangle=\int_0^\infty(1-2x+\frac{x^2}{2})e^{-x}dx+\int_0^\infty(-x+2x^2-\frac{x^3}{2})e^{-x}dx\\=\left \langle L_{0},L_{2} \right \rangle+\int_0^\infty(-x+\frac{x^2}{2}+\frac{3x^2}{2}-\frac{x^3}{2})e^{-x}dx$$ Notice that the last integral you already computed the integral of the first two terms.