N2020 P1 Q7

N2020 P1 Q7

Junior College 2
9 marks
Free

Do not use a calculator in answering this question.

It is given that \(f(x)=2-\sin 4x\).

  1. Find \(\int f(x)\,\mathrm dx\).[1]
  2. Find the exact value, in terms of \(\pi\), of \(\int_0^{\frac12\pi}xf(x)\,\mathrm dx\).[4]
  3. Find the exact value, in terms of \(\pi\), of \(\int_0^{\frac12\pi}(f(x))^2\,\mathrm dx\).[4]

Default solution

(i) \(\int(2-\sin4x)\,\mathrm dx=2x+\frac14\cos4x+C\).

(ii) By parts, \(\int_0^{\pi/2}x\sin4x\,\mathrm dx=\left[-\frac{x\cos4x}{4}+\frac{\sin4x}{16}\right]_0^{\pi/2}=-\frac\pi8\).

Thus \(\int_0^{\pi/2}x(2-\sin4x)\,\mathrm dx=\frac{\pi^2}{4}+\frac\pi8\).

(iii) \((2-\sin4x)^2=4-4\sin4x+\frac12(1-\cos8x)\). Over \([0,\pi/2]\), the sine and cosine terms integrate to zero, giving \(\frac{9\pi}{4}\).

Answer:(i) \(2x+\frac14\cos4x+C\); (ii) \(\frac{\pi^2}{4}+\frac\pi8\); (iii) \(\frac{9\pi}{4}\)

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