It's visually easy to see how the integral of a constant is $x=y$ as the area is growing linearly as you sum the dx boxes.
Similarly the integral for $y=x$ is $\frac{x^2}{2}$ which is the triangle formed by dissecting the ${x}$ x ${y}$ area.
But I'm not able to visualize the integer of $x^2$ as $\frac{x^3}{3}$.
I imagine a square extruded in the 3D axis but I can't see there the $\frac{1}{3}$ is coming from.
I tried to draw it but no help. Is there a visual trick to this?
It would be helpful to show this in a video I'm making about cross vs. dot product.
For any $r > 0$, consider the cube $[0,r] \times [0,r]\times[0,r]$.
This cube has $3$ faces opposite to the the origin: $$\{r\}\times[0,r]\times[0,r],\quad [0,r]\times \{r\}\times[0,r]\quad\text{ and }\quad [0,r]\times[0,r]\times\{r\}$$ Using origin as apex and these 3 faces as bases, we can decompose the cube into $3$ pyramids. These pyramid clearly have same volume. This mean the volume of each pyramid is $\frac13 r^3$.
Let's look at the first pyramid, if we intersect it with a plane $x = t$ for $t \in [0,r]$, we get $\{t\} \times [0,t] \times [0,t]$, a square of side $t$. This means we can rewrite $\frac13 r^3$, the volume of first pyramid, as an integral. The end result is
$$\int_0^r x^2 dx = \frac13 r^3$$