C2 January 2012 Q6

EdexcelOld spec11 marksIntegration

Figure 1: decreasing curve y = 16/x^2 - x/2 + 1 meeting the x-axis at x = 4, with region R shaded between x = 1, the x-axis and the curve — Figure 1

Figure 1 shows the graph of the curve with equation\[y = \frac{16}{x^2} - \frac{x}{2} + 1, \qquad x \gt 0\]The finite region \(R\), bounded by the lines \(x = 1\), the \(x\)-axis and the curve, is shown shaded in Figure 1. The curve crosses the \(x\)-axis at the point \((4, 0)\).

(a) Complete the table with the values of \(y\) corresponding to \(x = 2\) and 2.5

\(x\)	1	1.5	2	2.5	3	3.5	4
\(y\)	16.5	7.361			1.278	0.556	0

(2)

(b) Use the trapezium rule with all the values in the completed table to find an approximate value for the area of \(R\), giving your answer to 2 decimal places. (4)

Scheme	Marks
	B1, B1
	(2)

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2} \times 0.5,\ \left\{(16.5 + 0) + 2\left(7.361 + 4 + 2.31 + 1.278 + 0.556\right)\right\}\)	B1, M1A1ft
\(= 11.88\) (or answers listed below in note)	A1
	(4)

Notes

B1: Need 0.25 or ½ of 0.5

M1: requires first bracket to contain first \(y\) value plus last \(y\) value (0 may be omitted or be at end) and second bracket to include no additional \(y\) values from those in the scheme. They may however omit one value as a slip.

N.B. Special Case - Bracketing mistake \(\dfrac{1}{2} \times 0.5(16.5 + 0) + 2\left(7.361 + 4 + 2.31 + 1.278 + 0.556\right)\) scores B1 M1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks )

A1ft: This should be correct but ft their 4 and 2.31

A1: Accept 11.8775 or 11.878 or 11.88 only

Alternative Method for (b)

Separate trapezia may be used : B1 for 0.25, M1 for \(\tfrac{1}{2}h(a + b)\) used 5 or 6 times ( and A1ft all correct for their “4” and “2.31” ) final A1 for 11.88 etc. as before

In part (b) Need to use trapezium rule – answer only (with no working) is 0/4 -any doubts send to review

Mark scheme (c)

Scheme	Marks
\(\displaystyle\int_1^4 \frac{16}{x^2} - \frac{x}{2} + 1\,\mathrm{d}x = \left[-\frac{16}{x} - \frac{x^2}{4} + x\right]_1^4\)	M1 A1 A1
\(= \left[-4 - 4 + 4\right] - \left[-16 - \tfrac{1}{4} + 1\right]\)	dM1
\(= 11\tfrac{1}{4}\) or equivalent	A1
	(5)
	11

Notes

M1 Attempt to integrate ie power increased by 1 or 1 becomes \(x\),

A1 two correct terms, next A1 all three correct unsimplified (ignore +c) (Allow \(-16x^{-1} - 0.25x^2 + 1x\) or equivalent)

dM1 (This cannot be earned if previous M mark has not been awarded) Uses limits 4 and 1 in their integrated expression and subtracts (either way round)

A1 11.25 or 11 ¼ or 45/4 or equivalent (penalise negative final answer here)

In part (c) need to see integration