Further Mathematics Paper 2, May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength

Question 11

(a) Ify = (2x + 3)7 + x-l ,find the value of dywhen x
2x-l                         dx
Using the substitution, u = x + 2, evaluate f12 X-14dx.
(x+2)

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Observation

The report stated that while the part (a) was satisfactorily handled by majority of the candidates who attempted this question, the part (b) was poorly handled. They reportedly displayed poor knowledge of integration by substitution. Majority of them could not obtain the new limits of integration. They were expected to respond as follows:
I art ().                    = (2 + 3)7 + x-1 dy= 7(2 + 3)6(2) + (Zx-1)(1)- (x-1)(Z)
n p       a , smce y         x             Zx-1' dx                 X      (2X-1)2
= 14(2x +3)6 + 1 2'Hence dY/ ]= 14'/9.
(Zx-1)                       dx x =-
In part (b), candidates were expected to use the substitution u = x + 2 to solve the
integration J12 X-14 dx. They were to obtain the new limits by substituting the
(x+Z)4
old limits (1 and 2) into the expression u = x+ 2. These limits were (3 and 4).
Furthermore, since u = x + 2, then x = u - 2 and dx = du

2(X-1)                  _ f4U-3        _ [1                             1J4 - 5                                                  8
Thus 1 ~x - 3 ~u - - -2 + 3" 3 - -- or 0.002 94.
(x+Z)                        U                       2u          U               1728