Edexcel IAL Mathematics P4.6.5 parametric areas

Practise finding areas from parametric equations by forming y dx/dt integrals and evaluating exact results without sketching.

[Maximum number: 11]

Figure 2

Figure 2 shows a sketch of the curve defined by the parametric equations

x=t^{2}+2 t \quad y=\frac{2}{t(3-t)} \quad a \leqslant t \leqslant b

where a and b are constants.
The ends of the curve lie on the line with equation y=1

(a)

Show that the area of region R is given by

M-k \int_{a}^{b} \frac{t+1}{t(3-t)} \mathrm{d} t

where M and k are constants to be found.

[ 5 ]

Area under curve $=\int_{t=11}^{t=" 2 "} y \mathrm{~d} x=\int_{1}^{2} y \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$

M1
A1

$t=1 \Rightarrow x=3, \quad t=2 \Rightarrow x=8$

B1

So area of $R=1 \times(8-3)-\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$

M1

$=5-4 \int_{1}^{2} \frac{t+1}{t(3-t)} \mathrm{d} t$

A1

(5)

(b)

Use algebraic integration to find the exact area of R, giving your answer in simplest form.

[ 6 ]

$\int\frac{t+1}{t(3-t)}\,dt =\int\left(\frac1{3t}+\frac4{3(3-t)}\right)\,dt =\frac13\ln t-\frac43\ln(3-t)$
M1
$\begin{aligned} \text{Area} &=5-4\left[\frac13\ln t-\frac43\ln(3-t)\right]_1^2\\ &=5-4\left(\frac13\ln2+\frac43\ln2\right) \end{aligned}$
dM1
$=5-\frac{20}{3}\ln2$
A1

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P4.6.5 - Use integration to find the area question 1

Question (a)

Question (b)