2024 H2 Maths Paper 1 Q10 — Area and Volume of Revolution

The region, livethe area

Sweep across the region to

x

6.00

The curve y = 0.5 The region R One washer

Part (a)

3 marks

Using the substitution $u=\sqrt{4x+1}$ , show that $\displaystyle\int\frac{\sqrt{4x+1}}{x-2}\,dx=\int\frac{2u^{2}}{u^{2}-9}\,du$ .

Approach

A substitution needs three separate pieces before the integrand is touched: the numerator in terms of $u$ , the denominator in terms of $u$ , and $dx$ in terms of $du$ . Assemble all three first and the rewrite is mechanical; go hunting for them mid-line and something gets dropped.

Squaring the substitution and making $x$ the subject:

u=\sqrt{4x+1}\quad\Longrightarrow\quad u^{2}=4x+1\quad\Longrightarrow\quad x=\frac{u^{2}-1}{4}

so the denominator becomes

x-2=\frac{u^{2}-1}{4}-2=\frac{u^{2}-1-8}{4}=\frac{u^{2}-9}{4}.

Differentiating $u^{2}=4x+1$ with respect to $x$ gives the third piece:

2u\frac{du}{dx}=4\quad\Longrightarrow\quad \frac{du}{dx}=\frac{2}{u}\quad\Longrightarrow\quad dx=\frac{u}{2}\,du

Putting the three together, with the numerator simply $u$ :

\begin{aligned}\int\frac{\sqrt{4x+1}}{x-2}\,dx&=\int\frac{u}{\dfrac{u^{2}-9}{4}}\cdot\frac{u}{2}\,du\\[4pt]&=\int\frac{4u}{u^{2}-9}\cdot\frac{u}{2}\,du\\[2pt]&=\int\frac{2u^{2}}{u^{2}-9}\,du\ \ (\text{shown})\end{aligned}

To check your answer:

At $x=6$ the substitution gives $u=\sqrt{25}=5$ , and then $\frac{u^{2}-9}{4}=\frac{16}{4}=4$ , which is indeed $x-2$ . The rewritten denominator is right.

Part (b)(i)

3 marks

Sketch the curve $y=\frac{\sqrt{4x+1}}{x-2}$ for $x\geq0$ , giving intercepts and asymptotes, and shade the region $R$ .

Setting $x=0$ gives the $y$ -intercept:

y=\frac{\sqrt{1}}{0-2}=-\frac{1}{2}\quad\Longrightarrow\quad\left(0,\ -\frac{1}{2}\right)

There is no $x$ -intercept: $\sqrt{4x+1}=0$ needs $x=-\frac{1}{4}$ , which is outside $x\geq0$ .

The denominator vanishes at $x=2$ while the numerator there is $\sqrt{9}=3\neq0$ , so that is a vertical asymptote. For large $x$ ,

y=\frac{\sqrt{4x+1}}{x-2}\approx\frac{2\sqrt{x}}{x}=\frac{2}{\sqrt{x}}\longrightarrow0,

so $y=0$ is a horizontal asymptote, approached from above.

$\therefore$ the curve meets the axes only at $\left(0,-\frac{1}{2}\right)$ , with asymptotes $x=2$ and $y=0$ .

The sketch the question asks for: both branches for

x\geq0

, the vertical asymptote at

x=2

, the curve settling onto

y=0

from above, the intercept at

\left(0,-\tfrac{1}{2}\right)

, and

R

shaded between the curve and

y=0.5

Note

$R$ sits between the curve above and $y=0.5$ below. Worth confirming that the curve really is the upper boundary across the whole strip: at $x=6$ it is $\frac{5}{4}$ and at $x=12$ it is $\frac{7}{10}$ , both comfortably above $0.5$ , and the curve is decreasing in between.

Part (b)(ii)

5 marks

Find the exact area of $R$ , in the form $a+b\ln c$ .

Thought process

The region is the strip between two curves, so the area is the integral of the difference. Part (a) has already done the hard half of the work — but only for the curve, so take the two pieces separately: integrate the curve with the substitution, then subtract the rectangle underneath.

The region runs from $x=6$ to $x=12$ , between the curve and $y=0.5$ :

\text{Area}=\int_{6}^{12}\left(\frac{\sqrt{4x+1}}{x-2}-0.5\right)dx

Splitting the integrand from part (a) into a proper fraction and using the standard result from MF27:

\frac{2u^{2}}{u^{2}-9}=2+\frac{18}{u^{2}-9},\qquad\int\frac{1}{u^{2}-9}\,du=\frac{1}{6}\ln\left|\frac{u-3}{u+3}\right|+C

\int\frac{2u^{2}}{u^{2}-9}\,du=2u+3\ln\left|\frac{u-3}{u+3}\right|+C

The limits change with the variable: $x=6$ gives $u=\sqrt{25}=5$ , and $x=12$ gives $u=\sqrt{49}=7$ .

\begin{aligned}\int_{6}^{12}\frac{\sqrt{4x+1}}{x-2}\,dx&=\left[2u+3\ln\left(\frac{u-3}{u+3}\right)\right]_{5}^{7}\\[2pt]&=\left(14+3\ln\frac{4}{10}\right)-\left(10+3\ln\frac{2}{8}\right)\\[2pt]&=4+3\ln\!\left(\frac{2/5}{1/4}\right)\\[2pt]&=4+3\ln\frac{8}{5}\end{aligned}

The rectangle beneath the region has area $\int_{6}^{12}0.5\,dx=3$ , so

\text{Area}=\left(4+3\ln\frac{8}{5}\right)-3=1+3\ln\frac{8}{5}

To check your answer:

$1+3\ln\frac{8}{5}=1+3(0.4700)=2.4100\ldots$ — and the strip is roughly 6 wide by an average height of about 0.4, which is the same size. Sweep the region in the model and the running total lands on the same number.

$\therefore$ the exact area of $R$ is $1+3\ln\dfrac{8}{5}$ , so $a=1$ , $b=3$ and $c=\dfrac{8}{5}$ .

Where the marks go

Change the limits when you change the variable. Leaving $[\,\cdot\,]_{6}^{12}$ on an integral now written in $u$ is the single most common way this question is lost — the limits are $5$ and $7$ . And the answer wanted is exact: combine the two logarithms into one before writing it down, or the form $a+b\ln c$ is not met.

Part (b)(iii)

3 marks

Find the volume of the solid generated when $R$ is rotated through $2\pi$ radians about the $x$ -axis, to 2 decimal places.

Approach

Rotating a region that does not touch the axis leaves a hole down the middle. Each slice is a washer: a disc of radius the curve, with a disc of radius $0.5$ punched out of it. Its area is the difference of the two squares, and the volume is that summed along the strip.

V=\pi\int_{6}^{12}\left[\left(\frac{\sqrt{4x+1}}{x-2}\right)^{2}-(0.5)^{2}\right]dx=\pi\int_{6}^{12}\left[\frac{4x+1}{(x-2)^{2}}-0.25\right]dx

The first term is improper as it stands, so force the numerator to show the denominator's factor — write $4x+1=4(x-2)+9$ :

\frac{4x+1}{(x-2)^{2}}=\frac{4(x-2)+9}{(x-2)^{2}}=\frac{4}{x-2}+\frac{9}{(x-2)^{2}}

Both pieces now integrate directly:

\int\left[\frac{4}{x-2}+\frac{9}{(x-2)^{2}}-0.25\right]dx=4\ln|x-2|-\frac{9}{x-2}-0.25x+C

Evaluating between the limits:

\begin{aligned}\text{at }x=12:&\quad 4\ln10-\frac{9}{10}-3=4\ln10-3.9\\\text{at }x=6:&\quad 4\ln4-\frac{9}{4}-1.5=4\ln4-3.75\end{aligned}

\begin{aligned}V&=\pi\left[\left(4\ln10-3.9\right)-\left(4\ln4-3.75\right)\right]\\[2pt]&=\pi\left(4\ln\frac{10}{4}-0.15\right)\\[2pt]&=\pi\left(4\ln\frac{5}{2}-0.15\right)\end{aligned}

To check your answer:

$4\ln2.5=3.6652\ldots$ , so the bracket is $3.5152\ldots$ and $V=11.0432\ldots$ A rough sanity check: the solid is about 6 long with an outer radius averaging near 0.9 and a 0.5 hole, giving roughly $\pi(0.9^{2}-0.5^{2})(6)\approx10.6$ — the right size. The hole itself is a plain cylinder all the way through, $\pi(0.5)^{2}(12-6)=1.5\pi$ .

$\therefore\ V=\pi\left(4\ln\dfrac{5}{2}-0.15\right)=11.04$ (2 d.p.).

The error that costs the whole part

$\pi\displaystyle\int_{6}^{12}\left(y-0.5\right)^{2}dx$ is not the volume. That expression rotates the gap itself, which is a different solid altogether. The washer subtracts the squares — $\pi\int\left(y^{2}-0.5^{2}\right)dx$ — and the two differ by $\pi\int y\,dx$ worth of material.

What this question was really testing

Assembling a substitution completely. New numerator, new denominator, new $dx$ — and, in a definite integral, new limits. Four things, and the fourth is the one that goes missing.
Recognising a standard form. $\int\frac{du}{u^{2}-9}$ is in MF27. Splitting $\frac{2u^{2}}{u^{2}-9}$ into $2+\frac{18}{u^{2}-9}$ is what makes it reachable.
Reading "exact" and reading the form. $a+b\ln c$ is an instruction: one logarithm, so combine them.
Knowing which solid you are making. A region that stands off the axis sweeps a washer, not a disc — and the difference of squares is not the square of the difference.

One substitution, one region, and the solid it sweeps out

Part (a)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

What this question was really testing

This is how we mark, and how we teach.