The question
Draw the rough sketch of the curve y = 20 cos 2 x y = 20\cos 2x y = 20 cos 2 x ( where π 6 ≤ x ≤ π 3 ) \left(\text{where } \dfrac{\pi}{6} \le x \le \dfrac{\pi}{3}\right) ( where 6 π ≤ x ≤ 3 π )
Using integration, find the area of the region bounded by the curve y = 20 cos 2 x y = 20\cos 2x y = 20 cos 2 x x = π 6 x = \dfrac{\pi}{6} x = 6 π x = π 3 x = \dfrac{\pi}{3} x = 3 π x x x
Exam answer Show all explanations
At x = π 6 x = \frac{\pi}{6} x = 6 π y = 20 cos π 3 = 10 y = 20\cos\frac{\pi}{3} = 10 y = 20 cos 3 π = 10
at x = π 4 x = \frac{\pi}{4} x = 4 π y = 20 cos π 2 = 0 y = 20\cos\frac{\pi}{2} = 0 y = 20 cos 2 π = 0
at x = π 3 x = \frac{\pi}{3} x = 3 π y = 20 cos 2 π 3 = − 10 y = 20\cos\frac{2\pi}{3} = -10 y = 20 cos 3 2 π = − 10
Why?
To sketch the curve, find y y y x x x x x x
cos π 3 = 1 2 , cos π 2 = 0 cos 2 π 3 = − 1 2 \begin{aligned} \cos\frac{\pi}{3} &= \frac12,\quad \cos\frac{\pi}{2} = 0 \\ \cos\frac{2\pi}{3} &= -\frac12 \end{aligned} cos 3 π cos 3 2 π = 2 1 , cos 2 π = 0 = − 2 1 x = π 6 : 2 x = 2 π 6 = π 3 y = 20 × 1 2 = 10 x = π 4 : 2 x = 2 π 4 = π 2 y = 20 × 0 = 0 x = π 3 : 2 x = 2 π 3 y = 20 × ( − 1 2 ) = − 10 \begin{aligned} x = \tfrac{\pi}{6}&: 2x = \tfrac{2\pi}{6} = \tfrac{\pi}{3} \\ y &= 20 \times \tfrac12 = 10 \\[4pt] x = \tfrac{\pi}{4}&: 2x = \tfrac{2\pi}{4} = \tfrac{\pi}{2} \\ y &= 20 \times 0 = 0 \\[4pt] x = \tfrac{\pi}{3}&: 2x = \tfrac{2\pi}{3} \\ y &= 20 \times \left(-\tfrac12\right) = -10 \end{aligned} x = 6 π y x = 4 π y x = 3 π y : 2 x = 6 2 π = 3 π = 20 × 2 1 = 10 : 2 x = 4 2 π = 2 π = 20 × 0 = 0 : 2 x = 3 2 π = 20 × ( − 2 1 ) = − 10 As x x x π 6 \frac{\pi}{6} 6 π π 3 \frac{\pi}{3} 3 π 2 x 2x 2 x π 3 \frac{\pi}{3} 3 π 2 π 3 \frac{2\pi}{3} 3 2 π − 10 -10 − 10
10
−10
π/6
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Y
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π/4
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y = 20 cos 2x
The curve is above the x x x π 6 ≤ x < π 4 \frac{\pi}{6} \le x < \frac{\pi}{4} 6 π ≤ x < 4 π π 4 < x ≤ π 3 \frac{\pi}{4} < x \le \frac{\pi}{3} 4 π < x ≤ 3 π
Why?
cos 2 x > 0 \cos 2x > 0 cos 2 x > 0 2 x < π 2 2x < \frac{\pi}{2} 2 x < 2 π x < π 4 x < \frac{\pi}{4} x < 4 π π 2 < 2 x ≤ 2 π 3 \frac{\pi}{2} < 2x \le \frac{2\pi}{3} 2 π < 2 x ≤ 3 2 π cos 2 x < 0 \cos 2x < 0 cos 2 x < 0
So the region has two pieces, one above and one below the x x x x = π 4 x = \frac{\pi}{4} x = 4 π
Required area = 20 ∫ π 6 π 4 cos 2 x d x + ∣ 20 ∫ π 4 π 3 cos 2 x d x ∣ \begin{aligned} &\text{Required area} \\ &= 20\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\cos 2x\,dx \\ &\quad + \left\lvert 20\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cos 2x\,dx\right\rvert \end{aligned} Required area = 20 ∫ 6 π 4 π cos 2 x d x + 20 ∫ 4 π 3 π cos 2 x d x Why?
An area is never negative. Where the curve is below the axis, the integral of y y y x = π 4 x = \frac{\pi}{4} x = 4 π
Watch out: one integral from π 6 \frac{\pi}{6} 6 π π 3 \frac{\pi}{3} 3 π
= 20 [ sin 2 x 2 ] π 6 π 4 + ∣ 20 [ sin 2 x 2 ] π 4 π 3 ∣ \begin{aligned} &= 20\left[\frac{\sin 2x}{2}\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ &\quad + \left\lvert 20\left[\frac{\sin 2x}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\right\rvert \end{aligned} = 20 [ 2 sin 2 x ] 6 π 4 π + 20 [ 2 sin 2 x ] 4 π 3 π Why?
Standard integral:
∫ cos a x d x = sin a x a + C \int\cos ax\,dx = \frac{\sin ax}{a} + C ∫ cos a x d x = a sin a x + C Here a = 2 a = 2 a = 2 sin 2 x 2 \frac{\sin 2x}{2} 2 s i n 2 x 2 cos 2 x 2 = cos 2 x \frac{2\cos 2x}{2} = \cos 2x 2 2 c o s 2 x = cos 2 x
= 10 [ sin π 2 − sin π 3 ] + ∣ 10 [ sin 2 π 3 − sin π 2 ] ∣ \begin{aligned} &= 10\left[\sin\frac{\pi}{2} - \sin\frac{\pi}{3}\right] \\ &\quad + \left\lvert 10\left[\sin\frac{2\pi}{3} - \sin\frac{\pi}{2}\right]\right\rvert \end{aligned} = 10 [ sin 2 π − sin 3 π ] + 10 [ sin 3 2 π − sin 2 π ] Why?
20 × 1 2 = 10 20 \times \frac12 = 10 20 × 2 1 = 10 sin 2 x \sin 2x sin 2 x
2 × π 4 = π 2 , 2 × π 6 = π 3 2 \times \frac{\pi}{4} = \frac{\pi}{2},\quad 2 \times \frac{\pi}{6} = \frac{\pi}{3} 2 × 4 π = 2 π , 2 × 6 π = 3 π and 2 × π 3 = 2 π 3 2 \times \frac{\pi}{3} = \frac{2\pi}{3} 2 × 3 π = 3 2 π
= 10 ( 1 − 3 2 ) + ∣ 10 ( 3 2 − 1 ) ∣ = 10 ( 1 − 3 2 ) + 10 ( 1 − 3 2 ) \begin{aligned} &= 10\left(1 - \frac{\sqrt3}{2}\right) \\ &\quad + \left\lvert 10\left(\frac{\sqrt3}{2} - 1\right)\right\rvert \\ &= 10\left(1 - \frac{\sqrt3}{2}\right) + 10\left(1 - \frac{\sqrt3}{2}\right) \end{aligned} = 10 ( 1 − 2 3 ) + 10 ( 2 3 − 1 ) = 10 ( 1 − 2 3 ) + 10 ( 1 − 2 3 ) Why?
sin π 2 = 1 \sin\frac{\pi}{2} = 1 sin 2 π = 1 sin π 3 = 3 2 \sin\frac{\pi}{3} = \frac{\sqrt3}{2} sin 3 π = 2 3 sin 2 π 3 = sin ( π − π 3 ) = 3 2 \sin\frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \frac{\sqrt3}{2} sin 3 2 π = sin ( π − 3 π ) = 2 3
3 2 ≈ 0.866 \frac{\sqrt3}{2} \approx 0.866 2 3 ≈ 0.866 3 2 − 1 \frac{\sqrt3}{2} - 1 2 3 − 1 1 − 3 2 1 - \frac{\sqrt3}{2} 1 − 2 3
∣ 10 ( 3 2 − 1 ) ∣ = 10 ( 1 − 3 2 ) \begin{aligned} &\left\lvert 10\left(\frac{\sqrt3}{2} - 1\right)\right\rvert \\ &= 10\left(1 - \frac{\sqrt3}{2}\right) \end{aligned} 10 ( 2 3 − 1 ) = 10 ( 1 − 2 3 ) The two pieces have equal areas. Adding them, 2 × 10 = 20 2 \times 10 = 20 2 × 10 = 20
20 × 3 2 = 10 3 20 \times \frac{\sqrt3}{2} = 10\sqrt3 20 × 2 3 = 10 3 10 ( 1 − 3 2 ) + 10 ( 1 − 3 2 ) = 20 ( 1 − 3 2 ) = 20 − 20 × 3 2 = 20 − 10 3 \begin{aligned} &10\left(1 - \tfrac{\sqrt3}{2}\right) + 10\left(1 - \tfrac{\sqrt3}{2}\right) \\ &= 20\left(1 - \tfrac{\sqrt3}{2}\right) \\ &= 20 - 20 \times \tfrac{\sqrt3}{2} \\ &= 20 - 10\sqrt3 \end{aligned} 10 ( 1 − 2 3 ) + 10 ( 1 − 2 3 ) = 20 ( 1 − 2 3 ) = 20 − 20 × 2 3 = 20 − 10 3 ∴ \therefore ∴ = 20 ( 1 − 3 2 ) = 20\left(1 - \dfrac{\sqrt3}{2}\right) = 20 ( 1 − 2 3 ) = ( 20 − 10 3 ) = (20 - 10\sqrt3) = ( 20 − 10 3 )
Where this comes from Chapter 8, Application of Integrals : Area between y = 20 cos 2x and the x-axis.
Question type: A curve that dips below the axis
How to spot it: click to see Type 5. A curve that dips below the axis: split where it crosses
Spot it: part of the curve is below the x x x sin x \sin x sin x 2 π 2\pi 2 π
Here: 20 cos 2 x 20\cos 2x 20 cos 2 x x = π 4 x = \frac\pi4 x = 4 π π 6 \frac\pi6 6 π π 3 \frac\pi3 3 π π 4 \frac\pi4 4 π π 3 \frac\pi3 3 π
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 7, Integrals .