Integration: Question 8

Worked solution

Part (a): Where the two lines meet

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

$x+5 = 2x+1$

Rearrange:

$5-1 = 2x-x \implies x = 4$

Recompute independently by substituting back: at $x=4$, $y=x+5=9$ and $y=2x+1=9$ ✓, both lines give the same point, confirming $x=4$.

Part (b): Volume of the solid of revolution

For $0\le x\le 4$, check which line is on top: at $x=0$, $y=x+5=5$ is above $y=2x+1=1$; at $x=2$, $y=7$ is above $y=5$. So $y=x+5$ is the outer boundary (radius $R$) and $y=2x+1$ is the inner boundary (radius $r$). Since $R$ and $r$ are both positive throughout $[0,4]$ (minimum value $r=1$ at $x=0$), the region $R$ never touches the $x$-axis, so the solid is a “washer”, the volume of revolution of the outer curve minus the volume of revolution of the inner curve:

$V = \pi\int_0^4 \Big[(x+5)^2 - (2x+1)^2\Big]\,dx$

Using the $(ax+b)^n$ rule term by term, with $n=2$:

$\int (x+5)^2\,dx = \frac{(x+5)^3}{3} \qquad \int (2x+1)^2\,dx = \frac{(2x+1)^3}{2\times3} = \frac{(2x+1)^3}{6}$

So

$V = \pi\left[\frac{(x+5)^3}{3} - \frac{(2x+1)^3}{6}\right]_0^4$

At $x=4$: $x+5=9$, $2x+1=9$, so $\dfrac{9^3}{3} - \dfrac{9^3}{6} = \dfrac{729}{3} - \dfrac{729}{6} = 243 - 121.5 = 121.5$

At $x=0$: $x+5=5$, $2x+1=1$, so $\dfrac{5^3}{3} - \dfrac{1^3}{6} = \dfrac{125}{3} - \dfrac{1}{6} = \dfrac{250}{6} - \dfrac{1}{6} = \dfrac{249}{6} = 41.5$

$V = \pi\left(121.5 - 41.5\right) = 80\pi$

Recompute independently by expanding the brackets first

$(x+5)^2 - (2x+1)^2 = \left(x^2+10x+25\right) - \left(4x^2+4x+1\right) = -3x^2+6x+24$

$\int_0^4 \left(-3x^2+6x+24\right)dx = \Big[-x^3+3x^2+24x\Big]_0^4$

At $x=4$: $-64+48+96 = 80$. At $x=0$: $0$.

$V = \pi(80-0) = 80\pi$

This matches the result from the $(ax+b)^n$ method exactly, confirming the volume is correct. Both methods also agree that the differentiated check holds: $\dfrac{d}{dx}\left(-x^3+3x^2+24x\right) = -3x^2+6x+24$, matching the integrand above.

Final answers

• (a) The lines meet at $x=4$
• (b) $V = 80\pi$ cubic units