Help with basic Double Integral

Question

Given $$ I = \int_t^T(X(t,s)\int_t^sX(t,u)du)ds, \;\;\text{how do I get }\;I = \frac{(\int_t^TX(t,u)du)^2}{2}. $$

Many thanks for helping! And pardon me if this is straightforward.

Answer 1

Conveniently we always integrate over the second argument, and $t$ is the first's only value, and the only value of any integral's lower limit. Define $x(u):=X(t,\,u),\,y(u):=\int_t^u x(v)dv$ so $$I=\int_t^T y'(s)y(s)ds=\frac{y^2(T)-y^2(t)}{2}.$$By definition $y(t)=0$, while $y(T)=\int_t^T X(t,\,u) du$, completing the proof.