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Question 570

The work done WWW by a magnetic force on a micro-particle is determined by its displacement sss (in mm). For 0≤s≤20 \le s \le 20≤s≤2, the work required is given by the integral:

W=∫023s+4(16−s2)32 ds W = \int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds W=∫02​(16−s2)23​3s+4​ds
a.

Use the substitution s=4sin⁡θs = 4 \sin \thetas=4sinθ to show that

∫023s+4(16−s2)32 ds=∫0p(34sec⁡θtan⁡θ+14sec⁡2θ) dθ \int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds = \int_{0}^{p} \left( \frac{3}{4} \sec \theta \tan \theta + \frac{1}{4} \sec^2 \theta \right) \, d\theta ∫02​(16−s2)23​3s+4​ds=∫0p​(43​secθtanθ+41​sec2θ)dθ

where ppp is a constant to be found.

[5]
b.

Hence find the exact value of

∫023s+4(16−s2)32 ds \int_{0}^{2} \frac{3s+4}{(16-s^2)^{\frac{3}{2}}} \, ds ∫02​(16−s2)23​3s+4​ds
[3]
Markscheme

Integration Questions

  1. A Level
  2. /Maths
  3. /Integration

864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.

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