Integration: Question 10

Worked solution

Part (a): Intersection points of the two curves

Set the two expressions for $y$ equal to each other:

$x^2 - 2x = 4x - x^2$

Rearrange so one side is zero:

$x^2 - 2x - 4x + x^2 = 0 \implies 2x^2 - 6x = 0$

Factorise:

$2x(x-3) = 0 \implies x = 0 \ \text{ or } \ x = 3$

Recompute independently by substituting back: at $x=0$, both curves give $y=0$ ✓. At $x=3$, the first curve gives $y=9-6=3$ and the second gives $y=12-9=3$ ✓. Both intersection points check out.

Part (b): Area enclosed between the two curves

First confirm which curve is on top between $x=0$ and $x=3$: testing $x=1.5$, $y=x^2-2x=2.25-3=-0.75$, while $y=4x-x^2=6-2.25=3.75$, so $y=4x-x^2$ lies above $y=x^2-2x$ on this interval. The enclosed area is therefore

$\text{Area} = \int_0^3 \Big[(4x-x^2) - (x^2-2x)\Big]\,dx = \int_0^3 (6x - 2x^2)\,dx$

Integrate:

$\int (6x-2x^2)\,dx = 3x^2 - \frac{2x^3}{3}$

Evaluate at the limits:

At $x=3$: $\ 3(3)^2 - \dfrac{2(3)^3}{3} = 3(9) - \dfrac{2(27)}{3} = 27 - 18 = 9$

At $x=0$: $\ 0$

$\text{Area} = 9 - 0 = 9$

Recompute independently two ways:

1. Differentiate the antiderivative back: $\dfrac{d}{dx}\left(3x^2 - \dfrac{2x^3}{3}\right) = 6x - 2x^2$, which matches the integrand exactly, confirming the antiderivative is correct.
2. Use the “parabola hump” shortcut: $6x-2x^2 = -2x(x-3) = -2(x-0)(x-3)$, a scaled version of $-(x-p)(x-q)$ with $p=0$, $q=3$. Since the standard shortcut gives $\dfrac{(q-p)^3}{6}$ for coefficient $-1$, scaling by the factor $2$ here gives Area $= 2\times\dfrac{(3-0)^3}{6} = 2\times\dfrac{27}{6} = 2\times4.5 = 9$.

Both checks agree with the direct calculation, confirming the area is $9$.

Final answers

• (a) The curves intersect at $x=0$ and $x=3$
• (b) Area enclosed $= 9$ square units