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6 marksIntegration II

CAPE Pure Mathematics Unit 2 · May/June 2009 · Paper 2 · Question 2(b)

Using FIVE vertical strips, apply the trapezium rule to show that \int_0^1 \sqrt{1 - x^2}\, \mathrm{d}x \approx 0.759.

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Other parts of this question

  1. 2(a)Sketch the region whose area is defined by the integral \int_0^1 \sqrt{1 - x^2}\, \mathrm{d}x.[3 marks]
  2. 2(c)(i)Use integration by parts to show that, if I = \int \sqrt{1 - x^2}\, \mathrm{d}x, then I = x\sqrt{1 - x^2} - I + \int \frac{1}{\sqrt{1 - x^2}}\, \mathrm{d}x.[9 marks]
  3. 2(c)(ii)Deduce that I = \frac{x\sqrt{1 - x^2} + \sin^{-1} x}{2} + c, where c is an arbitrary constant of integration.[2 marks]
  4. 2(c)(iii)Hence, find \int_0^1 \sqrt{1 - x^2}\, \mathrm{d}x.[3 marks]
  5. 2(c)(iv)Use the results in Parts (b) and (c)(iii) above to find an approximation to \pi.[2 marks]

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