Integration: Question 3

Worked solution

Part (a): Intersection points of C and L

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

$8x - x^2 = 2x$

Rearrange so one side is zero:

$8x - x^2 - 2x = 0 \implies 6x - x^2 = 0$

Factorise:

$x(6-x) = 0 \implies x = 0 \ \text{ or } \ x = 6$

Recompute independently by substituting back: at $x=0$, both $C$ and $L$ give $y=0$ ✓. At $x=6$, $C$ gives $y = 8(6)-6^2 = 48-36=12$ and $L$ gives $y=2(6)=12$ ✓. Both intersection points check out.

Part (b): Area enclosed between C and L

First confirm which curve is on top between $x=0$ and $x=6$: testing $x=3$, $C$ gives $y=8(3)-3^2=24-9=15$ while $L$ gives $y=2(3)=6$, so $C$ lies above $L$ on this interval. The enclosed area is therefore

$\text{Area} = \int_0^6 \Big[(8x-x^2) - 2x\Big]\,dx = \int_0^6 (6x - x^2)\,dx$

Integrate:

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

Evaluate at the limits:

At $x=6$: $3(6)^2 - \dfrac{6^3}{3} = 3(36) - \dfrac{216}{3} = 108 - 72 = 36$

At $x=0$: $0$

$\text{Area} = 36 - 0 = 36$

Recompute independently two ways:

1. Differentiate the antiderivative back: $\dfrac{d}{dx}\left(3x^2 - \dfrac{x^3}{3}\right) = 6x - x^2$, which matches the integrand, confirming the antiderivative is correct.
2. Use the “parabola hump” shortcut: for $-(x-p)(x-q)$ integrated between the roots $p$ and $q$, the enclosed area equals $\dfrac{(q-p)^3}{6}$. Here $6x-x^2 = -(x-0)(x-6)$ with $p=0$, $q=6$, so Area $= \dfrac{6^3}{6} = \dfrac{216}{6} = 36$.

Both checks agree with the direct calculation, confirming the area is $36$.

Final answers

• (a) $C$ and $L$ intersect at $x=0$ and $x=6$
• (b) Area enclosed $= 36$ square units