2020 H2 Mathematics Paper 1 Question 7

Integration Techniques

Answers

(i)

2x+14cos⁡4x+C{2x + \frac{1}{4} \cos 4 x + C}

(ii)

π24+π8{\frac{\pi^2}{4} + \frac{\pi}{8}}

(iii)

9π4{\frac{9\pi}{4}}

Full solutions

(i)

∫f(x)  dx∫2−sin⁡4x  dx=2x+14cos⁡4x+C  ■\begin{align*} & \int f(x) \; \mathrm{d}x \\ & \int 2 - \sin 4 x \; \mathrm{d}x \\ & = 2x + \frac{1}{4} \cos 4 x + C \; \blacksquare \end{align*}

(ii)

∫012πxf(x)  dx=∫012π2x−xsin⁡4x  dx=[x2]012π−[x(−cos⁡4x4)]012π+∫012π−cos⁡4x4  dx=π24+π2cos⁡2π4−[sin⁡4x16]012π=π24+π8−0+0=π24+π8  ■\begin{align*} & \int_0^{\frac{1}{2} \pi} x f(x) \; \mathrm{d}x \\ & = \int_0^{\frac{1}{2} \pi} 2x - x \sin 4x \; \mathrm{d}x \\ & = \left[ x^2 \right]_0^{\frac{1}{2} \pi} - \left[ x \left( \frac{-\cos 4x}{4} \right) \right]_0^{\frac{1}{2} \pi} + \int_0^{\frac{1}{2} \pi} - \frac{\cos 4x}{4} \; \mathrm{d}x \\ & = \frac{\pi^2}{4} + \frac{\pi}{2} \frac{\cos 2 \pi}{4} - \left[ \frac{\sin 4x}{16} \right]_0^{\frac{1}{2} \pi} \\ & = \frac{\pi^2}{4} + \frac{\pi}{8} - 0 + 0 \\ & = \frac{\pi^2}{4} + \frac{\pi}{8} \; \blacksquare \end{align*}

(iii)

∫012π(f(x))2  dx=∫012π(2−sin⁡4x)2  dx=∫012π4−4sin⁡4x+sin⁡24x  dx=∫012π4−4sin⁡4x+1−cos⁡8x2  dx=∫012π92−4sin⁡4x−cos⁡8x2  dx=[92x+cos⁡4x−sin⁡8x16]012π=9π4+1−0−0−1+0=9π4  ■\begin{align*} & \int_0^{\frac{1}{2} \pi} \Big( f(x) \Big)^2 \; \mathrm{d}x \\ & = \int_0^{\frac{1}{2} \pi} \left( 2 - \sin 4x \right)^2 \; \mathrm{d}x \\ & = \int_0^{\frac{1}{2} \pi} 4 - 4 \sin 4x + \sin^2 4x \; \mathrm{d}x \\ & = \int_0^{\frac{1}{2} \pi} 4 - 4 \sin 4x + \frac{1-\cos 8x}{2} \; \mathrm{d}x \\ & = \int_0^{\frac{1}{2} \pi} \frac{9}{2} - 4 \sin 4x - \frac{\cos 8x}{2} \; \mathrm{d}x \\ & = \left[ \frac{9}{2}x + \cos 4x - \frac{\sin 8x}{16} \right]_0^{\frac{1}{2} \pi} \\ & = \frac{9\pi}{4} + 1 - 0 - 0 - 1 + 0 \\ & = \frac{9\pi}{4} \; \blacksquare \end{align*}