**Find the total area of the region under the curve.** The curve meets the $x$-axis where $$4x - x^2 = x(4 - x) = 0 \quad \Rightarrow \quad x = 0, \; x = 4$$ The parabola opens downwards, so it is above the axis on $0 < x < 4$ and no splitting is needed: $$A_{\text{total}} = \int_0^4\left(4x - x^2\right)dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3}$$ **Each half must therefore have area $\dfrac{16}{3}$.** **Find where the line meets the curve.** Setting them equal: $$4x - x^2 = mx$$ $$x\left(4 - m - x\right) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 4 - m$$ For the line to cut the region, we need $0 < 4 - m < 4$, so **$0 < m < 4$** — a constraint to check against at the end. **Find the area between the curve and the line.** On $0 < x < 4 - m$ the curve lies above the line (both pass through the origin, and the parabola rises faster initially since its gradient there is 4 > m). So: $$A_{\text{upper}} = \int_0^{4-m}\left[\left(4x - x^2\right) - mx\right]dx = \int_0^{4-m}\left[(4 - m)x - x^2\right]dx$$ **Let $p = 4 - m$** to keep the algebra clean. Then: $$A_{\text{upper}} = \int_0^{p}\left(px - x^2\right)dx = \left[\frac{px^2}{2} - \frac{x^3}{3}\right]_0^{p}$$ $$= \frac{p^3}{2} - \frac{p^3}{3} = p^3\left(\frac{3 - 2}{6}\right) = \frac{p^3}{6}$$ **Set this equal to half the total area:** $$\frac{p^3}{6} = \frac{16}{3}$$ $$p^3 = 32$$ $$p = \sqrt[3]{32} = \sqrt[3]{32}$$ Simplifying: $32 = 2^5$, so $p = 2^{5/3} = 2 \cdot 2^{2/3} = 2\sqrt[3]{4}$. **Recover $m$:** $$m = 4 - p = 4 - 2\sqrt[3]{4}$$ **Check the constraint.** $\sqrt[3]{4} \approx 1.5874$, so $p \approx 3.175$ and $$m \approx 4 - 3.175 = 0.825$$ This satisfies $0 < m < 4$ ✓ and $m > 0$ as required ✓ **Verify numerically.** With $m = 0.8250$, the intersection is at $x = 3.175$, and $$A_{\text{upper}} = \frac{(3.175)^3}{6} = \frac{32.00}{6} = 5.333... = \frac{16}{3} \;\checkmark$$ which is exactly half of $\frac{32}{3} = 10.667$ ✓ $$m = 4 - 2\sqrt[3]{4} \approx 0.825$$ **The elegant step.** Substituting $p = 4 - m$ collapsed a messy expansion in $m$ into the single clean result $A = \frac{p^3}{6}$. Recognising that the integrand $(px - x^2)$ has the *same* form as the original with 4 replaced by $p$ — and that its enclosed area is therefore $\frac{p^3}{6}$ by the same calculation — avoids expanding a cubic in $m$ entirely.

Integration — practice questions

Anti-differentiation

Integrating powers and polynomials

Integrating exponential, trigonometric and rational functions

The reverse chain rule and trigonometric formulae

Definite Integrals & Area

Definite integrals

Area under a curve

Area between two curves

Numerical integration — the trapezium rule

Differential Equations & Rates

Differential equations of the form y′ = f(x)

Separable differential equations

Growth, decay, inflation and Newton's law of cooling

Kinematics and rates of change