Find adjoint operator defined on element

Question

Let $E = L^2 (0, 1)$. Given $u ∈ E$, set $Tu(x)=\int_0^x u(t)dt$.

Find $T^*$.

Solution says only $(u, T^* v) = \int_t^1 v(x)dx$.

I dont understand. adjoint operator is defined by $(Tu,v)=(u,T^*v)$. How do we get the formulae above?

Edit

Wew have $(Tu,v)=(u,T^*v)=\int_0^1 (\int_0^x u(t)dt)v(x)dx=\int_0^x u(t)(\int_0^1 v(x)dx)dt$ by interchanging the integrals but It looks like there is a variable change to do but I can't see where.

Thank you for any help/hints.

Answer 1

$$ \langle Tu,v\rangle = \int_0^1 \int_0^x u(y)dy\; \overline{v(x)}dx \\ = \int_0^1\int_0^1\chi_{[0,\infty)}(x-y)u(y)dy\,\overline{v(x)}dx \\ = \int_0^1\int_0^1\chi_{[0,\infty)}(x-y)\overline{v(x)}dx\, u(y)dy \\ = \int_0^1 u(y) \overline{\int_y^1 v(x)dx} dy = \langle u,T^*v\rangle $$ Therefore $$ T^*v = \int_y^1 v(x)dx. $$