I've been trying to prove or have an intuitive understanding of the change of variables, and I tried it for the function $f(x)=x^2$ using $u(x)=x^2$, the transformed function then becomes $g(u)=u$.
Now I understand that a rectangle with a width dx in the first function transforms to a rectangle with a width of $2x\,dx$ in the second function (with the same height) and if I want to evaluate the area of the second, we must add $2x$ to the integral of $f(x)$. But if I want to evaluate the area of the first function using the second function, what does $du$ transform to in the first function?
There are two approaches.
We may express $f$ as a composition, $f(x)=g(u)$ where $u$ is a function of $x$.
Let $g(y):=y$ and $u(x):=x^2$
$$\begin{align} \int_a^b f(x)\,\mathrm d x &= \int_a^b g(u(x))\,\mathrm d x\\&= \int_{u(a)}^{u(b)} ~g(u)\,\left(\dfrac{\mathrm d u}{\mathrm d x}\right)^{-1}\,\mathrm d u \\[1ex] &=\int_{a^2}^{b^2} u\cdot\dfrac 1{2\surd u}\,\mathrm d u\\ &= \tfrac 12\int_{a^2}^{b^2} \sqrt{ u\,}\,\mathrm d u \end{align}$$
Alternatively: We express $x$ as a function of $v,$ so $x= h(v) =\surd v$
$$\begin{align} \int_a^b f(x) \,\mathrm d x &= \int_{h^{-1}(a)}^{h^{-1}(b)} f(h(v))\,\dfrac{\mathrm d h(v)}{\mathrm d v}\,\mathrm d v\\ &=\int_{h^{-1}(a)}^{h^{-1}(b)} f(h(v))\,h'(v)\,\mathrm d v \\ &= \int_{a^2}^{b^2} f(\surd v)\,\dfrac 1{2\surd v}\,\mathrm d v\\ &= \tfrac 12\int_{a^2}^{b^2}\surd v\,\mathrm d v\end{align}$$