2020 H2 Maths Paper 1 Q7 — Worked Solution

Part (i)

1 mark

\int f(x)\, dx = 2x + \frac{1}{4}\cos 4x + C

Explanation: integrating term by term, $\int 2\, dx = 2x$ , and $\int\sin 4x\, dx = - \frac{1}{4}\cos 4x$ by the reverse chain rule (dividing by the derivative of the angle $4x$ ). Hence $\int\left( 2 - \sin 4x \right)\, dx = 2x - \left( - \frac{1}{4}\cos 4x \right) + C = 2x + \frac{1}{4}\cos 4x + C$ .

To check your answer: differentiating

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

gives

2 - \frac{1}{4}\left( 4\sin 4x \right) = 2 - \sin 4x = f(x)

, matching the original integrand. ✓

$\therefore$ $\int f(x)\, dx = 2x + \frac{1}{4}\cos 4x + C$ .

Part (ii)

4 marks

Approach

$xf(x) = 2x - x\sin 4x$ splits into an easy polynomial integral and a product $x\sin 4x$ , which needs integration by parts (a polynomial times a sine cannot be integrated directly).

\int_{0}^{\frac{1}{2}\pi}xf(x)\, dx = \int_{0}^{\frac{1}{2}\pi}2x\, dx - \int_{0}^{\frac{1}{2}\pi}x\sin 4x\, dx

The first integral:

\int_{0}^{\frac{1}{2}\pi}2x\, dx = \left\lbrack x^{2} \right\rbrack_{0}^{\frac{1}{2}\pi} = \left( \frac{\pi}{2} \right)^{2} = \frac{\pi^{2}}{4}

For the second, we use integration by parts with $u = x$ (so $du = dx$ ) and $dv = \sin 4x\, dx$ (so $v = - \frac{1}{4}\cos 4x$ ):

\int x\sin 4x\, dx = - \frac{x}{4}\cos 4x + \int\frac{1}{4}\cos 4x\, dx

= - \frac{x}{4}\cos 4x + \frac{1}{16}\sin 4x

Evaluating from $0$ to $\frac{\pi}{2}$ ,

\left\lbrack - \frac{x}{4}\cos 4x + \frac{1}{16}\sin 4x \right\rbrack_{0}^{\frac{\pi}{2}} = \left( - \frac{\pi/2}{4}\cos 2\pi + \frac{1}{16}\sin 2\pi \right) - (0 + 0)

= - \frac{\pi}{8}(1) + 0 - 0

= - \frac{\pi}{8}

Combining the two integrals,

\int_{0}^{\frac{1}{2}\pi}xf(x)\, dx = \frac{\pi^{2}}{4} - \left( - \frac{\pi}{8} \right) = \frac{\pi^{2}}{4} + \frac{\pi}{8}

To check your answer: numerical integration of

x\left( 2 - \sin 4x \right)

from

0

to

\frac{\pi}{2}

gives

\text{≈}2.860

, and

\frac{\pi^{2}}{4} + \frac{\pi}{8} \approx 2.467 + 0.393 \approx 2.860

. ✓

$\therefore$ $\int_{0}^{\frac{1}{2}\pi}xf(x)\, dx = \frac{\pi^{2}}{4} + \frac{\pi}{8}$ .

Part (iii)

4 marks

Approach

$\left( f(x) \right)^{2}$ contains $\sin^{2}(4x)$ , which is not directly integrable; the double angle formula $\sin^{2}(4x) = \frac{1 - \cos 8x}{2}$ turns it into a sum of integrable terms.

\left( f(x) \right)^{2} = \left( 2 - \sin 4x \right)^{2} = 4 - 4\sin 4x + \sin^{2}(4x)

Using $\sin^{2}(4x) = \frac{1 - \cos 8x}{2}$ ,

\left( f(x) \right)^{2}

= 4 - 4\sin 4x + \frac{1}{2} - \frac{1}{2}\cos 8x

= \frac{9}{2} - 4\sin 4x - \frac{1}{2}\cos 8x

Integrating term by term,

\int_{0}^{\frac{1}{2}\pi}\left( f(x) \right)^{2}dx = \left\lbrack \frac{9}{2}x + \cos 4x - \frac{1}{16}\sin 8x \right\rbrack_{0}^{\frac{\pi}{2}}

At $x = \frac{\pi}{2}$ :

\frac{9}{2}\left( \frac{\pi}{2} \right) + \cos 2\pi - \frac{1}{16}\sin 4\pi = \frac{9\pi}{4} + 1 - 0 = \frac{9\pi}{4} + 1

At $x = 0$ :

0 + \cos 0 - \frac{1}{16}\sin 0 = 0 + 1 - 0 = 1

Subtracting,

\int_{0}^{\frac{1}{2}\pi}\left( f(x) \right)^{2}dx = \left( \frac{9\pi}{4} + 1 \right) - 1 = \frac{9\pi}{4}

To check your answer: numerical integration of

\left( 2 - \sin 4x \right)^{2}

from

0

to

\frac{\pi}{2}

gives

\text{≈}7.069

, and

\frac{9\pi}{4} \approx 7.069

. ✓

$\therefore$ $\int_{0}^{\frac{1}{2}\pi}\left( f(x) \right)^{2}dx = \frac{9\pi}{4}$ .

2020 H2 Maths Paper 1, Question 7

Part (i)

Part (ii)

Part (iii)

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