A-LevelMathematicsIntegration9709/12 · Feb/Mar 2016 · Q1012 Marks

The diagram shows part of the curve y = 1/16 (3x – 1)², which touches the x-axis at the point P. The point Q (3, 4) lies on the curve and the tangent to the curve at Q crosses the x-axis at R. [Figure 10]

(i)State the x-coordinate of P.[1]
(ii)Showing all necessary working, find by calculation the x-coordinate of R,[5]
(iii)the area of the shaded region PQR.[6]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2016). The mark or mark type is shown in brackets.

[B1] x=1/3
[B1] dy/dx = 2/16 (3x-1) [3] (Power and chain rule applied)
[B1] dy/dx = 3/8 (3x-1)
[M1] When x = 3, dy/dx = 3/8 (3(3)-1) = 3/8 (8) = 3 (soi)
[M1] Equation of tangent QR is y - 4 = 3(x - 3)
[A1] When y = 0, 0 - 4 = 3(x - 3) => -4 = 3x - 9 => 3x = 5 => x = 5/3 (x-coordinate of R)
[B1] Area under curve = ∫[from 1/3 to 3] 1/16 (3x-1)² dx. Integration: 1/16 * (3x-1)³/3 * 1/3. (Correct integral form with 1/3 for chain rule. M1 for integration method.)
[B1] Correct integral: 1/144 (3x-1)³
[B1] Apply limits: 1/144 (3(3)-1)³ - 1/144 (3(1/3)-1)³ = 1/144 (8)³ - 1/144 (0)³ = 512/144 = 32/9.
[B1] Area of triangle QPR: Base PR = 3 - 5/3 = 4/3. Height = 4. Area = 1/2 * (3 - 5/3) * 4 = 1/2 * (4/3) * 4 = 8/3.
[M1] Shaded area = Area under curve - Area of triangle PQR = 32/9 - 8/3.
[A1] = 32/9 - 24/9 = 8/9 (or 0.889)

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