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A-LevelFurther MathematicsIntegration9231/12 · May/June 2013 · Q1114 Marks

Answer only one of the following two alternatives. EITHER. The curve C has equation y = 2 sec x, for 0 ≤ x ≤ ¼π. Show that the arc length s of C is given by s = ∫₀^(¼π) (2 sec² x - 1) dx.

  1. (a)The curve C has equation y = 2 sec x, for 0 ≤ x ≤ ¼π. Show that the arc length s of C is given by s = ∫₀^(¼π) (2 sec² x - 1) dx.[4]
  2. (b)Find the exact value of s.[2]
  3. (c)(i)The surface area generated when C is rotated through 2π radians about the x-axis is denoted by S. Show that S = 4π ∫₀^(¼π) (2 sec³ x - sec x) dx,[3]
  4. (c)(ii)d/dx(sec x tan x) = 2 sec³ x - sec x.[3]
  5. (c)(iii)Hence find the exact value of S.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.

  1. M1 Differentiation. y = 2 sec x ⇒ y' = 2 sec x tan x
  2. M1 Use of sec² x = 1 + tan² x. 1+(y')² = 1+4 sec² x(sec² x−1)
  3. A1 = 4 sec⁴ x - 4 sec² x + 1
  4. A1 Substitute in arc length formula. s = ∫₀^(π/4) (2 sec² x - 1) dx
  5. M1 Integrate. [2 tan x - x]₀^(π/4)
  6. A1 Substitute limits. [2 - π/4]
  7. M1A1 Use surface area formula. S = 2π ∫₀^(π/4) y√(1+(y')²) dx = 2π ∫₀^(π/4) 2 sec x (2 sec² x - 1) dx
  8. A1 Obtain correct form. = 4π ∫₀^(π/4) (2 sec³ x - sec x) dx (AG)
  9. M1A1 Differentiates d/dx(sec x tan x) = sec x tan² x + sec³ x
  10. A1 and obtains printed result. = sec x (sec² x - 1) + sec³ x = 2 sec³ x - sec x (AG)
  11. M1 Uses result to deduce surface area. S = 4π [sec x tan x]₀^(π/4)
  12. A1 = 4π√2

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About this question

Topic
Integration, A-Level Further Mathematics (syllabus 9231). Structured question, 14 marks.
Source
A-Level Further Mathematics, May/June 2013 Paper 1 Variant 2, question 11.

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