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