∫₀^a x² dx = a³/3

By the power rule, $\int x^2\,dx = \frac{x^3}{3} + C$, so $\int_0^a x^2\,dx = \frac{a^3}{3}$. For $a = 3$, the area is $9$. This is one of the most-asked definite integrals on board papers.

Slide the upper limit $a$ and the lab shades the region under $y = x^2$ from $0$ to $a$ and computes $\frac{a^3}{3}$. Try $a = 2$: area $\approx 2.67$. The lab confirms that doubling $a$ multiplies area by $8$ — the cubic scaling.

Because the height of the curve also grows like $a^2$ at the right edge, multiplying by width $a$ gives $a\cdot a^2 = a^3$ (up to the constant $1/3$). For linear $y = x$, area grows as $a^2/2$. The lab makes the scaling difference between linear and quadratic intuitive.

Moment of inertia: $I = \int r^2\,dm$ involves integrating $r^2$. Kinetic energy in rotational motion, deflection of beams, and the volume of a paraboloid via the disk method all use $\int x^2\,dx$. It is a quietly important integral.