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1.8 Integration

1.8 Integration

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Question 221

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

1.8 Integration Questions

  1. A Level
  2. /Maths
  3. /1.8 Integration

438 exam-style questions on OCR A Level Maths 1.8 Integration, covering 1.8.1 Fundamental theorem of calculus (A-level only), 1.8.2 Integrating x^n, 1.8.3 Integrating standard functions (A-level only), 1.8.4 Evaluating definite integrals, 1.8.5 Area between a curve and the x-axis, 1.8.6 Area between two curves, 1.8.7 Integration as the limit of a sum (A-level only), 1.8.8 Integration by substitution (A-level only), 1.8.9 Integration by parts (A-level only), 1.8.10 Use of partial fractions in integration (A-level only), 1.8.11 Differential equations with separable variables (A-level only), 1.8.12 Interpreting the solution of a differential equation (A-level only), and 1.8 Integration. Each one has a worked solution and a mark scheme showing where the marks go.

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