Answer
$$\operatorname{proj}_{{g}} {f}
=0.$$
Work Step by Step
Let $f(x)=x, \quad g(x)=1, \quad C[-1,1] , \quad\langle f, g\rangle=\int_{-1}^1 f(x)g(x) d x.$
We have
$$\langle f, g\rangle=\int_{-1}^1 x d x=\left[\frac{ 1}{2 } x^2\right]_{-1}^1=0,$$
$$\langle g,g\rangle=\int_{-1}^{1} 1 d x= \left[x\right]_{-1}^{1}=2.$$
Now, the orthogonal projection of $f$ onto $g$ is given by
$$\operatorname{proj}_{{g}} {f}
=\frac{\langle{f}, {g}\rangle}{\langle{g}, {g}\rangle} {g}=0.$$
