Let $E = L^p(0,1)$ with $1 ≤ p < ∞$. Given $u ∈ E$, set
$$Tu(x):=\int_0^x u(t)dt$$
Find the adjoint of $T$.
I know how to this in the case $p=2$ as shown here. But in general $L^p$ is not an Hilbert space and the definition of adjoint is different: $$\langle Tx,y^*\rangle =\langle x,T^*y^*\rangle $$ with $T:X\to Y$, $y^*\in Y^*$ and $T^*:Y^*\to X^*$.
In this case I do not necessarily have an integral representation of the duality and therefore I do not know how to compute the adjoint.
As it turns out, you do still have integral representation of the duality. If $1\leq p<\infty$, the dual of $L^p$ is isometrically isomorphic to $L^{p'}$ where $p'$ is the dual exponent, i.e. $\frac1p+\frac1{p'}=1$. (The dual exponent of $1$ is $\infty$.) The duality is given by $$\langle f,g\rangle=\int_0^1f(x)g(x)dx$$ for $f\in L^{p'}(0,1),g\in L^p(0,1)$. One can show that $T\in\mathcal{L}(L^p,L^q)$ for all $1\leq p,q\leq\infty$, so assuming $p,q<\infty$ we have for all $f\in L^{q'},g\in L^p$, $$\langle f,Tg\rangle_{L^q}=\int_0^1\int_0^xf(x)g(y)dydx=\int_0^1\int_y^1f(x)g(y)dxdy=\langle T^*f,g\rangle_{L^p}$$ if we define $T^*f:=\int_y^1f(x)dx$. This defines the adjoint operator $T^*\,:\,(L^q)^*\rightarrow(L^p)^*$.