When we integrate a function with respect to dx, we are breaking the area under the function into many pieces of width dx and finding the area of each piece and then take the limit of the sum as dx becomes 0. In this case dx is constant so this is pretty intuitive.
When we use u-substitution, we replace dx with du. But clearly du is not constant as it changes with x. To me, this doesn't make sense because how do we integrate with respect to something that is not constant? Would appreciate if someone can give a visualisation of what is going on (graphically) when we integrate by substitution. For instance if we graph out x, g(x) and f(x) parametrically, how can we show that ∫f(u)du and ∫f(g(x)*g'(x)dx are the same thing?
I also found this related thread: Is there a way to graphically visually integration by substitution?, but I don't really understand the answer.
Of course the indefinite integrals $\int f(x)\>dx$ and $\int f\bigl(g(x)\bigr)\>g'(x)\>dx$ are not the same. As an example, let $f(x):=x$ and $g(x):=x^2$. Then $\int f(x)\>dx={1\over2}x^2+C$, and $\int f\bigl(g(x)\bigr)\>g'(x)\>dx={1\over2}x^4+C$.
The correct way to write this substitution formula for indefinite integrals is $$\int f\bigl(g(t)\bigr)\>g'(t)\>dt=\int f(x)\>dx\biggr|_{x:=g(t)}\ .\tag{1}$$ Formula $(1)$ is an immediate consequence of the chain rule. I don't think there is a graphical visualization.
It is another thing with the substitution formula for definite integrals: $$\int_a^b f\bigl(g(t)\bigr)\>g'(t)\>dt=\int_{g(a)}^{g(b)}f(x)\>dx\ .$$ Here we can invoke Riemann sums and draw narrow rectangles. In this way we arrive at the following chain of approximations: $$\eqalign{\int_{g(a)}^{g(b)}f(x)\>dx&\approx\sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\cr &\approx\sum_{k=1}^N f\bigl(g(\tau_k)\bigr)\>g'(\tau_k)\>(t_k-t_{k-1})\cr &\approx\int_a^b f\bigl(g(t)\bigr)\>g'(t)\>dt\ .\cr}$$ Here we have assumed that the partition points are $x_k=g(t_k)$, and the sampling points are $\xi_k=g(\tau_k)$. Furthermore we have used the MVT formula $x_k-x_{k-1}\approx g'(\tau_k)(t_k-t_{k-1})$ in order to handle the width of corresponding rectangles on the $x$-, resp., the $t$-axis.