Quelpr
2 marksIntegration II

CAPE Pure Mathematics Unit 2 · May/June 2009 · Paper 2 · Question 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.

The mark scheme is shown once you've answered.

Practise this question

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(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.[6 marks]
  3. 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]
  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]

More practice: the rest of this paper · more Integration II questions · all CAPE Pure Mathematics Unit 2 past papers