TMUA practice: Integration Power Rule, problem 2

Compute the integral as a function of $a$. An antiderivative is $F(x) = 8 x - x^{3}$, so

$\int_{0}^{a} (8 - 3 x^{2})\, dx \;=\; 8 a - a^{3}.$

Set equal to $0$:

$8 a - a^{3} \;=\; a (8 - a^{2}) \;=\; 0.$

Roots: $a = 0$ or $a^{2} = 8$, i.e. $a = \pm 2 \sqrt{2}$. Discard the spurious $a = 0$ and the negative root because $a > 0$:

$a \;=\; 2 \sqrt{2}.$

This is option E.

Geometric interpretation. The integrand $8 - 3 x^{2}$ is positive on $(0, \sqrt{8/3})$ and negative beyond. The integral from $0$ accumulates positive area until $x \approx 1.63$, then deposits negative area. The signed integral first returns to zero at $a = 2 \sqrt{2} \approx 2.83$, where the positive and negative areas cancel.

Why the other options fail.

• A, $a = 2$: integral $= 16 - 8 = 8 \ne 0$.
• B, $a = \sqrt{8} = 2 \sqrt{2}$: same as E, written before simplification.
• C, $a = 8$: integral $= 64 - 512 = -448 \ne 0$.
• D, $a = 8/3$: integral $= 64/3 - 512/27 \ne 0$ (not exact cancellation).

(Option B states the same value as E, $2 \sqrt{2}$, but written unsimplified. The convention in this batch is to give the simplest exact form; B is rejected as a redundant duplicate.)

The lesson: a definite integral equals zero either trivially (limits coincide) or when signed positive and negative areas cancel. Solve algebraically; the cancellation point is not at the integrand’s root.