Let $C[0,1]$ be the collection of continuous functions on the unit closed interval $[0,1]$. It is a vector space with respect to the usual addition and scaler multiplication. Let $\Phi:C[0,1]\to\mathbb R$ be defined by
$$\Phi(f)=\int_0^1f(x)\,dx.$$
(whether Riemann or Lebesgue, not important) Then, it is immediate that $\Phi$ is a linear functional.
Question) What is the dual of $\Phi$? More precisely, what is the dual element $\phi\in C[0,1]$ corresponding to the linear functional $\Phi$? Is it $f$? Why?
By dual element, I mean the following
(1) If $V$ is a vector space over a field $F$, there corresponds the dual space $V^\ast$ of all linear functionals of $V$ into $F$.
(2) If V is a finite dimensional vector space with basis $\{v_1,\cdots,v_n\}$, the correspondence between $V$ and $V^\ast$ can be easily described by the dual bases as $$f_i(v_j)=\delta_{ij}$$ where $\{f_1,\cdots,f_n\}$ is the corresponding basis for $V^\ast$ and $\delta_{ij}$ is the Kronecker delta.
(3) The correspondence in the finite dimensional case is actually an one-to-one correspondence (bijection). So $f_i\in V^\ast$ is the dual element of $v_i\in V$ and $v_i\in V$ is the dual element of $f_i\in V^\ast$.
(4) I don't know if there is such a bijection if $V$ is infinite-dimensional and how to set the basis for such a function space.
This is not quite what you wanted but in a very meaningful sense does answer the question of "what is the dual?".
For $C(X)$ where $X$ is a compact Hausdorff space, such as $[0,1]$, we have a lovely description of the positive elements of $(C(X))^\ast$ up to isomorphism in the Riesz-Markov-Kakutani representation theorem (actually this talks about LCH spaces and $C_c(X)$ but if $X$ is compact Hausdorff $C(X)=C_c(X)$ and it is also LCH). The theorem relates positive linear functionals on $C(X)$ with Radon measures on $X$.
Under this, the dual of integration is simply the Lebesgue-1 measure on $[0,1]$. If you like, the dual is the 'uniform distribution' on $[0,1]$.