TMUA practice: Integration Power Rule, problem 4

Write the integral as a function of $c$. Antiderivative: $F(x) = x^{3} - 3 x^{2} + 4 x$, so

$\int_{0}^{c} (3 x^{2} - 6 x + 4)\, dx \;=\; c^{3} - 3 c^{2} + 4 c.$

Set equal to $4$:

$c^{3} - 3 c^{2} + 4 c - 4 \;=\; 0.$

Try $c = 2$: $8 - 12 + 8 - 4 = 0$ ✓. So $(c - 2)$ is a factor. Polynomial-divide:

$c^{3} - 3 c^{2} + 4 c - 4 \;=\; (c - 2)(c^{2} - c + 2).$

The quadratic factor $c^{2} - c + 2$ has discriminant $1 - 8 = -7 < 0$, so no real roots. Therefore $c = 2$ is the only real solution. Since the problem states $c > 0$, $c = 2$ satisfies the constraint.

This is option D.

Why the other options fail.

• A, $c = 1$: integral $= 1 - 3 + 4 = 2 \ne 4$.
• B, $c = \sqrt{2}$: integral $= 2 \sqrt{2} - 6 + 4 \sqrt{2} = 6 \sqrt{2} - 6 \approx 2.49 \ne 4$.
• C, $c = 3/2$: integral $= 27/8 - 27/4 + 6 = 27/8 - 54/8 + 48/8 = 21/8 \ne 4$.
• E, $c = 3$: integral $= 27 - 27 + 12 = 12 \ne 4$.

The lesson: back-solving for an upper limit usually produces a polynomial equation. Try small integer candidates (here $c = 2$ works) and verify by substitution. Factoring out the found root reveals whether other real solutions exist.