A-LevelMathematicsIntegration9709/12 · Oct/Nov 2010 · Q1113 Marks

The diagram shows part of the curve y = 1/(3x + 1)⁴. The curve cuts the y-axis at A and the line x = 5 at B. [Figure 11.1]

(i)Show that the equation of the line AB is y = -1/10x + 1.[4]
(ii)Find the volume obtained when the shaded region is rotated through 360° about the x-axis.[9]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2010). The mark or mark type is shown in brackets.

[B1] A = (0, 1)
[B1] B = (5, 2)
[M1] y-1 = (-1/10)(x- 0) ft their A,B
[A1] y = -1/10x + 1 AG
[M1] Curve: (π) ∫(3x + 1)^(-1/2) dx Attempt ∫y²dx (π not vital)
[A1A1] 2π/3 [(3x+1)^(-1/2)] from 0 to 5 (πnot vital). 2nd A mark is for ÷ 3.
[DM1] 2π/3 [4-1] Application of limits to their integral (in either integral). Limits 0 to 5 only.
[A1] [2π]
[M1] Line: (π) ∫(1/10x - 1/2x + 1)² dx Attempt ∫y²dx (not vital)
[A2,1] Also directly 10/3 (-1/10x + 1)³ or -10/3 (-1/10x + 1)³ from 0 to 5 (πnot vital) - applying limits to their integral
[DM1] Subtraction of their volumes
[A1] 11π/12 co

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