How to rewrite integral of Brownian motion $\int^1_0 \left( W(r)-\int_0^1W(s)ds\right)^2dr$

Question

How would I go to rewrite

$$\int^1_0 \left( W(r)-\int_0^1W(s)ds\right)^2dr$$

into $$\int_0^1 W(r)^2dr -\left( \int_0^1W(r)dr \right)^2$$

where $W(r)$ is a standard Brownian Motion.

Answer 1

Note,

$$\begin{align} & \int^1_0 \left( W(r)-\int_0^1W(s)ds\right)^2dr \\ & =\int^1_0 \left( W^2(r)-2W(r)\int_0^1W(s)ds+\left( \int_0^1W(s)ds \right)^2\right)dr \\ & =\int^1_0 W^2(r)dr-2\int^1_0 W(r)dr \int_0^1W(s)ds+\int^1_0 dr\left( \int_0^1 W(s)ds \right)^2 \\ & =\int^1_0 W^2(r)dr-2\left( \int_0^1 W(r)dr \right)^2+ \left( \int_0^1 W(s)ds \right)^2 \\ & =\int_0^1 W(r)^2dr -\left( \int_0^1W(r)dr \right)^2 \\ \end{align}$$