- #36

etotheipi

$$\pi^{\bot}(v) = \frac{\langle v, x - \frac{1}{2} \rangle}{|x - \frac{1}{2}|^2} (x - \frac{1}{2}) + \langle v , 1 \rangle$$then$$\begin{align*}

\int_0^1 (x-\frac{1}{2}) e^x dx &= \frac{3-e}{2} \\

\int_0^1 (x-\frac{1}{2})^2 dx &= \frac{1}{12} \\

\int_0^1 e^x dx = e-1

\end{align*}$$so you just get$$\pi^{\bot}(e^x) = 6(3-e)(x-\frac{1}{2}) + (e-1)$$lol idk if that's right, but it's 3am so cba to check atm