C4 June 2014 (R) Q2

EdexcelOld spec9 marksIntegration Numerical Methods

Figure 1: curve y = (2 - x)e^(2x) crossing the y-axis at 2 and the x-axis at 2, with region R shaded between the curve and the axes — Figure 1

Figure 1 shows a sketch of part of the curve with equation \[y = (2 - x)\mathrm{e}^{2x}, \qquad x \in \mathbb{R}\]

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

The table below shows corresponding values of \(x\) and \(y\) for \(y = (2 - x)\mathrm{e}^{2x}\)

\(x\)	0	0.5	1	1.5	2
\(y\)	2	4.077	7.389	10.043	0

(a) Use the trapezium rule with all the values of \(y\) in the table, to obtain an approximation for the area of \(R\), giving your answer to 2 decimal places. (3)

(b) Explain how the trapezium rule can be used to give a more accurate approximation for the area of \(R\). (1)

(c) Use calculus, showing each step in your working, to obtain an exact value for the area of \(R\). Give your answer in its simplest form. (5)

Mark scheme (a)

Scheme	Marks
\(\text{Area} \approx \dfrac{1}{2} \times 0.5;\ \times \left[2 + 2\left(4.077 + 7.389 + 10.043\right) + 0\right]\)	B1; M1
\(= \dfrac{1}{4} \times 45.018 = 11.2545 = 11.25\) (2 dp) 11.25	A1 cao
	(3)

Notes

B1: Outside brackets \(\dfrac{1}{2} \times 0.5\) or \(\dfrac{0.5}{2}\) or 0.25 or \(\dfrac{1}{4}\).

M1: For structure of trapezium rule \(\left[\ \ldots\ldots\ldots\ldots\ \right]\). Condone missing 0.

Note: No errors are allowed [eg. an omission of a \(y\)-ordinate or an extra \(y\)-ordinate or a repeated \(y\) ordinate].

A1: 11.25 cao

Note: Working must be seen to demonstrate the use of the trapezium rule. The actual area is 12.39953751…

Note: Award B1M1A1 for \(\dfrac{0.5}{2}(2 + 0) + \dfrac{1}{2}\left(4.077 + 7.389 + 10.043\right) = 11.25\)

Bracketing mistake: Unless the final answer implies that the calculation has been done correctly.
Award B1M0A0 for \(\dfrac{1}{2} \times 0.5 + 2 + 2\left(4.077 + 7.389 + 10.043\right) + 0\) (nb: answer of 45.268).

Alternative method for part (a): Adding individual trapezia

\(\text{Area} \approx 0.5 \times \left[\dfrac{2 + 4.077}{2} + \dfrac{4.077 + 7.389}{2} + \dfrac{7.389 + 10.043}{2} + \dfrac{10.043 + 0}{2}\right] = 11.2545 = 11.25\) (2 dp) cao

B1: 0.5 and a divisor of 2 on all terms inside brackets.

M1: First and last ordinates once and the middle ordinates twice inside brackets ignoring the 2.

A1: 11.25 cao

Mark scheme (b)

Scheme	Marks
Any one of Increase the number of strips Use more trapezia Make \(h\) smaller Increase the number of \(x\) and/or \(y\) values used Shorter /smaller intervals for \(x\) More values of \(y\). More intervals of \(x\) Increase \(n\)	B1
	(1)

Notes

B0: Give B0 for

smaller values of \(x\) and/or \(y\).
use more decimal places

Mark scheme (c)

Scheme	Marks
\(\left\{\displaystyle\int (2 - x)\mathrm{e}^{2x}\,\mathrm{d}x\right\},\ \left\{\begin{aligned} u &= 2 - x &\Rightarrow\ \dfrac{\mathrm{d}u}{\mathrm{d}x} &= -1\\ \dfrac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^{2x} &\Rightarrow\ v &= \dfrac{1}{2}\mathrm{e}^{2x}\end{aligned}\right\}\)
\(= \dfrac{1}{2}(2 - x)\mathrm{e}^{2x} - \displaystyle\int -\dfrac{1}{2}\mathrm{e}^{2x}\{\mathrm{d}x\}\) Either \((2 - x)\mathrm{e}^{2x} \to \pm\lambda(2 - x)\mathrm{e}^{2x} \pm \displaystyle\int \mu\mathrm{e}^{2x}\{\mathrm{d}x\}\) or \(\pm x\mathrm{e}^{2x} \to \pm\lambda x\mathrm{e}^{2x} \pm \displaystyle\int \mu\mathrm{e}^{2x}\{\mathrm{d}x\}\) \((2 - x)\mathrm{e}^{2x} \to \dfrac{1}{2}(2 - x)\mathrm{e}^{2x} - \displaystyle\int -\dfrac{1}{2}\mathrm{e}^{2x}\{\mathrm{d}x\}\)	M1 A1
\(= \dfrac{1}{2}(2 - x)\mathrm{e}^{2x} + \dfrac{1}{4}\mathrm{e}^{2x}\) \(\dfrac{1}{2}(2 - x)\mathrm{e}^{2x} + \dfrac{1}{4}\mathrm{e}^{2x}\)	A1 oe
\(\text{Area} = \left\{\left[\dfrac{1}{2}(2 - x)\mathrm{e}^{2x} + \dfrac{1}{4}\mathrm{e}^{2x}\right]_0^2\right\}\) \(= \left(0 + \dfrac{1}{4}\mathrm{e}^4\right) - \left(\dfrac{1}{2}(2)\mathrm{e}^0 + \dfrac{1}{4}\mathrm{e}^0\right)\) Applies limits of 2 and 0 to all terms* and subtracts the correct way round.*	dM1
\(= \dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\) \(\dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\) or \(\dfrac{\mathrm{e}^4 - 5}{4}\) cao	A1 oe
	(5)
	(9 marks)

Notes

M1: Either \((2 - x)\mathrm{e}^{2x} \to \pm\lambda(2 - x)\mathrm{e}^{2x} \pm \displaystyle\int \mu\mathrm{e}^{2x}\{\mathrm{d}x\}\) or \(\pm x\mathrm{e}^{2x} \to \pm\lambda x\mathrm{e}^{2x} \pm \displaystyle\int \mu\mathrm{e}^{2x}\{\mathrm{d}x\}\)

A1: \((2 - x)\mathrm{e}^{2x} \to \dfrac{1}{2}(2 - x)\mathrm{e}^{2x} - \displaystyle\int -\dfrac{1}{2}\mathrm{e}^{2x}\{\mathrm{d}x\}\) either un-simplified or simplified.

A1: Correct expression, i.e. \(\dfrac{1}{2}(2 - x)\mathrm{e}^{2x} + \dfrac{1}{4}\mathrm{e}^{2x}\) or \(\dfrac{5}{4}\mathrm{e}^{2x} - x\mathrm{e}^{2x}\) (or equivalent)

dM1: which is dependent on the 1^st M1 mark being awarded.
Complete method of applying limits of 2 and 0 to all terms and subtracting the correct way round.

Note: Evidence of a proper consideration of the limit of 0 is needed for M1. So, just subtracting zero is M0.

A1: \(\dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\) or \(\dfrac{\mathrm{e}^4 - 5}{4}\). Do not allow \(\dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\mathrm{e}^0\) unless simplified to give \(\dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\)

Note: 12.39953751... without seeing \(\dfrac{1}{4}\mathrm{e}^4 - \dfrac{5}{4}\) is A0.

Note: 12.39953751... from NO working is M0A0A0M0A0.