C4 June 2005 Q5

EdexcelOld spec10 marksIntegration

Figure 1: the curve y = xe^(2x) for x from 0 to 1, with the region R between the curve, the x-axis and x = 1 shaded — Figure 1

Figure 1 shows the graph of the curve with equation

\[y = x\mathrm{e}^{2x}, \qquad x \geqslant 0.\]

The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.

(a) Use integration to find the exact value for the area of \(R\). (5)

(b) Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and 0.8.

\(x\)	0	0.2	0.4	0.6	0.8	1
\(y = x\mathrm{e}^{2x}\)	0	0.29836		1.99207		7.38906

(1)

(c) Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int x\mathrm{e}^{2x}\,\mathrm{d}x = \frac{1}{2}x\mathrm{e}^{2x} - \frac{1}{2}\int \mathrm{e}^{2x}\,\mathrm{d}x\) Attempting parts in the right direction	M1 A1
\(= \dfrac{1}{2}x\mathrm{e}^{2x} - \dfrac{1}{4}\mathrm{e}^{2x}\)	A1
\(\left[\dfrac{1}{2}x\mathrm{e}^{2x} - \dfrac{1}{4}\mathrm{e}^{2x}\right]_0^1 = \dfrac{1}{4} + \dfrac{1}{4}\mathrm{e}^2\)	M1 A1
	(5)

Scheme	Marks
\(x = 0.4 \Rightarrow y \approx 0.89022\) \(x = 0.8 \Rightarrow y \approx 3.96243\) Both are required to 5 d.p.	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(I \approx \dfrac{1}{2} \times 0.2 \times [\ \ldots\ ]\)	B1
\(\approx \ldots \times \left[0 + 7.38906 + 2(0.29836 + .89022 + 1.99207 + 3.96243)\right]\) ft their answers to (b)	M1 A1ft
\(\approx 0.1 \times 21.67522\)
\(\approx 2.168\) cao	A1
	(4)
	(10 marks)

Notes

Note \(\dfrac{1}{4} + \dfrac{1}{4}\mathrm{e}^2 \approx 2.097\ \ldots\)