FP3 June 2015 Q4

EdexcelOld spec7 marksHyperbolic Functions Integration

4. The curve \(C\) has equation \[y = \frac{1}{\sqrt{x^2 + 2x - 3}}, \qquad x > 1\]

(a) Find \(\displaystyle\int y\,\mathrm{d}x\) (3)

The region \(R\) is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = 2\) and \(x = 3\). The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.

(b) Find the volume of the solid generated. Give your answer in the form \(p\pi\ln q\), where \(p\) and \(q\) are rational numbers to be found. (4)

Mark scheme (a)

Scheme	Marks
\(x^2 + 2x - 3 = (x + 1)^2 - 4\)	M1
\(\displaystyle\int\frac{1}{\sqrt{(x + 1)^2 - 4}}\,\mathrm{d}x = \text{arcosh}\,\frac{(x + 1)}{2}\ (+c)\) or \(\ln\left\{(x + 1) + \sqrt{(x + 1)^2 - 4}\right\}\)	M1 A1
	(3)

Notes

M1: \(x^2 + 2x - 3 = (x \pm 1)^2 \pm \alpha \pm 3,\ \alpha \neq 0\)

M1: Use of arcosh (allow arccosh, \(\cosh^{-1}\)) Or uses \(\ln\left\{x + \sqrt{x^2 - a^2}\right\}\)

A1: \(\text{arcosh}\,\dfrac{(x + 1)}{2}\) (+ c not required) Or \(\ln\left\{(x + 1) + \sqrt{(x + 1)^2 - 4}\right\}\)

(Corrected from the printed mark scheme: the scheme column prints the alternative as \(\ln\left\{(x + 1)^2 + \sqrt{(x + 1)^2 - 4}\right\}\); the notes column and the correct integral have \((x + 1)\), not \((x + 1)^2\), before the square root.)

Mark scheme (b)

Scheme	Marks
\(\displaystyle S = \pi\int y^2\,\mathrm{d}x = \pi\int\left(\frac{1}{\sqrt{x^2 + 2x - 3}}\right)^2\mathrm{d}x\)	M1
\(\displaystyle = \int\frac{1}{(x + 1)^2 - 4}\,\mathrm{d}x = \left[\frac{1}{4}\ln\left(\frac{x - 1}{x + 3}\right)\right]\)	M1A1
\(= \dfrac{\pi}{4}\left(\ln\dfrac{1}{3} - \ln\dfrac{1}{5}\right) = \dfrac{\pi}{4}\ln\dfrac{5}{3}\)	A1
	(4)
	(7 marks)

Notes

M1: Use of \(\int\pi y^2\,\mathrm{d}x\)

M1: Use of \(\ln\left(\dfrac{x \pm p}{x \pm q}\right)\)

A1: \(\displaystyle\int\frac{1}{(x + 1)^2 - 4}\,\mathrm{d}x = \frac{1}{4}\ln\left(\frac{x - 1}{x + 3}\right)\)

A1: \(\dfrac{\pi}{4}\ln\dfrac{5}{3}\)

Special case: Uses \(S = k\int y^2\,\mathrm{d}x\) scores a maximum M0M1A1A0

NB: May use partial fractions in (b) for middle M1A1

Scheme	Marks
\(\dfrac{1}{x^2 + 2x - 3} \equiv \dfrac{1}{(x + 3)(x - 1)} \equiv \dfrac{1}{4}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 3}\right)\)
\(\displaystyle\int\frac{1}{(x + 3)(x - 1)}\,\mathrm{d}x = \left[\frac{1}{4}\ln\left(\frac{x - 1}{x + 3}\right)\right]\)	M1A1

M1: Use of \(\ln\left(\dfrac{x \pm p}{x \pm q}\right)\)

A1: \(\dfrac{1}{4}\ln\left(\dfrac{x - 1}{x + 3}\right)\)

Alternative for (b) by substitution

Scheme	Marks
\(\displaystyle S = \pi\int y^2\,\mathrm{d}x = \pi\int\left(\frac{1}{\sqrt{x^2 + 2x - 3}}\right)^2\mathrm{d}x\)	M1
\(\displaystyle u = x + 1 \Rightarrow \int\frac{1}{(x + 1)^2 - 4}\,\mathrm{d}x = \int\frac{1}{u^2 - 4}\,\mathrm{d}u\)
\(\displaystyle\int\frac{1}{u^2 - 4}\,\mathrm{d}u = \left[\frac{1}{4}\ln\frac{u - 2}{u + 2}\right]\)	M1A1
\(\pi\left[\dfrac{1}{4}\ln\dfrac{u - 2}{u + 2}\right]_3^4 = \dfrac{\pi}{4}\left(\ln\dfrac{1}{3} - \ln\dfrac{1}{5}\right) = \dfrac{\pi}{4}\ln\dfrac{5}{3}\)	A1

M1: Use of \(\int\pi y^2\,\mathrm{d}x\)

M1: Use of \(\ln\left(\dfrac{u \pm p}{u \pm q}\right)\)

A1: \(\dfrac{1}{4}\ln\dfrac{u - 2}{u + 2}\)

A1: \(\dfrac{\pi}{4}\ln\dfrac{5}{3}\)