C2 June 2007 Q5

EdexcelOld spec9 marksIntegration

5. The curve \(C\) has equation\[y = x\sqrt{(x^3 + 1)}, \qquad 0 \leqslant x \leqslant 2.\]

(a) Complete the table below, giving the values of \(y\) to 3 decimal places at \(x = 1\) and \(x = 1.5\).

\(x\)	0	0.5	1	1.5	2
\(y\)	0	0.530			6

(2)

(b) Use the trapezium rule, with all the \(y\) values from your table, to find an approximation for the value of \(\displaystyle\int_0^2 x\sqrt{(x^3 + 1)}\,\mathrm{d}x\), giving your answer to 3 significant figures. (4)

Figure 2: the curve C from O to (2, 6) and the straight line l joining O to (2, 6), with the region R between them hatched — Figure 2

Figure 2 shows the curve \(C\) with equation \(y = x\sqrt{(x^3 + 1)},\ 0 \leqslant x \leqslant 2\), and the straight line segment \(l\), which joins the origin and the point \((2, 6)\). The finite region \(R\) is bounded by \(C\) and \(l\).

(c) Use your answer to part (b) to find an approximation for the area of \(R\), giving your answer to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks
1.414 (allow also exact answer \(\sqrt{2}\)), 3.137 Allow awrt	B1, B1
	(2)

Notes

If answers are given to only 2 d.p. (1.41 and 3.14), this is B0 B0, but full marks can be given in part (b) if 4.04 is achieved.

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2}(0.5)\ \ldots\)	B1
\(\ldots\ \{0 + 6 + 2(0.530 + 1.414 + 3.137)\}\)	M1 A1ft
\(= 4.04\) (Must be 3 s.f.)	A1
	(4)

Notes

Bracketing mistake: i.e. \(\dfrac{1}{2}(0.5)(0 + 6) + 2(0.530 + 1.414 + 3.137)\)
scores B1 M1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given).

Alternative (finding and adding separate areas):
\(\dfrac{1}{2}\times\dfrac{1}{2}\) (Triangle/trapezium formulae, and height of triangle/trapezium) [B1]
Fully correct method for total area, with values from table. [M1, A1ft]
4.04 [A1]

Mark scheme (c)

Scheme	Marks
Area of triangle \(= \dfrac{1}{2}(2\times 6)\) (Could also be found by integration, or even by the trapezium rule on \(y = 3x\))	B1
Area required = Area of triangle − Answer to (b) (Subtract either way round)	M1
\(6 - 4.04 = 1.96\) Allow awrt (ft from (b), dependent on the B1, and on answer to (b) less than 6)	A1ft
	(3)
	9

Notes

B1: Can be given for 6 with no working, but should not be given for 6 obtained from wrong working.

A1ft: This is a dependent follow-through: the B1 for 6 must have been scored, and the answer to (b) must be less than 6.