Finding the adjoint of an operator

This is from my homework, I'm totally lost as to how to proceed. Consider the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $(Tf)(x) = \int^x_0 f(s) \ ds$ What is the adjoint of $T$?

This operator doesn't seem to be an orthogonal projection, nor is it self-adjoint. How does one find the adjoint of an operator in general? Thanks in advance!

Using the fact that

$$ \langle Tf , g \rangle=\langle f , T^{*}g \rangle, $$

we have

$$ \langle Tf, g\rangle = \int_{0}^{1} (Tf)(t)g(t)\,dt =\int_{0}^{1} \int_{0}^{t} f(\tau)\,d \tau\, g(t)\, dt = \int_{0}^{1} f(\tau)\, \left(\int_{\tau}^{1} g(t) \,dt\right)\, d \tau $$ $$ = \langle f, T^{*}g\rangle $$

From the last integral, we can see that the adjoint is given by

$$ (T^{*}f) (x) = \int_{x}^{1} f(s)\, ds $$

We can find adjoint for kernel operators, that is, operators given by $$T(f)(x)=\int_{[0,1]}K(x,y)f(y)dy,$$ with $K$ satisfying good conditions. We should have $$\int_{[0,1]}T^*(f)(x)\overline{g(x)}dx=\int_{[0,1]}f(x)\overline{T(g)(x)}dx.$$ Since $$\int_{[0,1]^2}f(x)\overline{K(x,y)g(y)}dxdy=\int_{[0,1]}\left(\int_{[0,1]}\widetilde K(y,x)f(x)dx\right)\overline{g(y)}dy,$$ where $\widetilde K(x,y)=\overline{K(y,x)}$. Since it's true for any $g$, we have $$T^*(f)(x)=\int_{[0,1]}\widetilde K(x,y)f(y)dy.$$