Part (a) 5 marks
f ( x ) = e x cos x f(x) = e^{x}\cos x Using the product rule:
f ′ ( x ) = e x cos x − e x sin x = e x ( cos x − sin x ) f'(x) = e^{x}\cos x - e^{x}\sin x = e^{x}\left( \cos x - \sin x \right) At a stationary point, f ′ ( x ) = 0 f'(x) = 0 e x ≠ 0 e^{x} \neq 0
cos x − sin x = 0 \cos x - \sin x = 0 ⇒ tan x = 1 \Rightarrow \tan x = 1 ⇒ x = π 4 \Rightarrow x = \frac{\pi}{4} (within the given domain 0 ≤ x ≤ π 2 0 \leq x \leq \frac{\pi}{2}
Substituting into f ( x ) f(x)
f ( π 4 ) = e π / 4 cos π 4 = 2 2 e π / 4 f\left( \frac{\pi}{4} \right) = e^{\pi/4}\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}e^{\pi/4} To determine the nature, we differentiate again:
f ′ ′ ( x ) = d d x [ e x ( cos x − sin x ) ] f''(x) = \frac{d}{dx}\left\lbrack e^{x}\left( \cos x - \sin x \right) \right\rbrack = e x ( cos x − sin x ) + e x ( − sin x − cos x ) = e^{x}\left( \cos x - \sin x \right) + e^{x}\left( - \sin x - \cos x \right) = − 2 e x sin x = - 2e^{x}\sin x At x = π 4 x = \frac{\pi}{4}
f ′ ′ ( π 4 ) = − 2 e π / 4 sin π 4 = − 2 e π / 4 < 0 f''\left( \frac{\pi}{4} \right) = - 2e^{\pi/4}\sin\frac{\pi}{4} = - \sqrt{2}\, e^{\pi/4} < 0 Since f ′ ′ ( π 4 ) < 0 f''\left( \frac{\pi}{4} \right) < 0
To check your answer: f ( π 4 ) = 2 2 e π / 4 ≈ 1.551 f\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}e^{\pi/4} \approx 1.551 f ( 0.7 ) ≈ 1.548 f(0.7) \approx 1.548 f ( 0.85 ) ≈ 1.545 f(0.85) \approx 1.545
∴ \therefore ( π 4 , 2 2 e π / 4 ) \left( \frac{\pi}{4},\ \frac{\sqrt{2}}{2}e^{\pi/4} \right)
Part (b) 4 marks
Approach
integrating by parts once returns an integral of the same “e 2 x × e^{2x} \times J J I I J J I I
Let I = ∫ e 2 x cos 2 x d x I = \int e^{2x}\cos 2x\, dx u = e 2 x u = e^{2x} d v = cos 2 x d x dv = \cos 2x\, dx d u = 2 e 2 x d x du = 2e^{2x}\, dx v = 1 2 sin 2 x v = \frac{1}{2}\sin 2x
I = 1 2 e 2 x sin 2 x − ∫ e 2 x sin 2 x d x I = \frac{1}{2}e^{2x}\sin 2x - \int e^{2x}\sin 2x\, dx Let J = ∫ e 2 x sin 2 x d x J = \int e^{2x}\sin 2x\, dx u = e 2 x u = e^{2x} d v = sin 2 x d x dv = \sin 2x\, dx d u = 2 e 2 x d x du = 2e^{2x}\, dx v = − 1 2 cos 2 x v = - \frac{1}{2}\cos 2x
J = − 1 2 e 2 x cos 2 x + ∫ e 2 x cos 2 x d x = − 1 2 e 2 x cos 2 x + I J = - \frac{1}{2}e^{2x}\cos 2x + \int e^{2x}\cos 2x\, dx = - \frac{1}{2}e^{2x}\cos 2x + I Substituting back into the expression for I I
I = 1 2 e 2 x sin 2 x − J I = \frac{1}{2}e^{2x}\sin 2x - J = 1 2 e 2 x sin 2 x − ( − 1 2 e 2 x cos 2 x + I ) = \frac{1}{2}e^{2x}\sin 2x - \left( - \frac{1}{2}e^{2x}\cos 2x + I \right) = 1 2 e 2 x sin 2 x + 1 2 e 2 x cos 2 x − I = \frac{1}{2}e^{2x}\sin 2x + \frac{1}{2}e^{2x}\cos 2x - I 2 I = 1 2 e 2 x ( sin 2 x + cos 2 x ) 2I = \frac{1}{2}e^{2x}\left( \sin 2x + \cos 2x \right) ⇒ I = 1 4 e 2 x ( sin 2 x + cos 2 x ) + c \Rightarrow I = \frac{1}{4}e^{2x}\left( \sin 2x + \cos 2x \right) + c
To check your answer: differentiating 1 4 e 2 x ( sin 2 x + cos 2 x ) \frac{1}{4}e^{2x}\left( \sin 2x + \cos 2x \right) 1 4 [ 2 e 2 x ( sin 2 x + cos 2 x ) + e 2 x ( 2 cos 2 x − 2 sin 2 x ) ] = e 2 x cos 2 x \frac{1}{4}\left\lbrack 2e^{2x}\left( \sin 2x + \cos 2x \right) + e^{2x}\left( 2\cos 2x - 2\sin 2x \right) \right\rbrack = e^{2x}\cos 2x
∴ \therefore ∫ e 2 x cos 2 x d x = 1 4 e 2 x ( sin 2 x + cos 2 x ) + c \int e^{2x}\cos 2x\, dx = \frac{1}{4}e^{2x}\left( \sin 2x + \cos 2x \right) + c
Part (c) 4 marks
Approach
rotating y = f ( x ) y = f(x) x x V = π ∫ y 2 d x V = \pi\int y^{2}\, dx y 2 = e 2 x cos 2 x y^{2} = e^{2x}\cos^{2}x e 2 x cos 2 x e^{2x}\cos 2x e 2 x cos 2 x e^{2x}\cos^{2}x cos 2 x = 1 + cos 2 x 2 \cos^{2}x = \frac{1 + \cos 2x}{2}
V = π ∫ 0 π / 2 [ f ( x ) ] 2 d x = π ∫ 0 π / 2 e 2 x cos 2 x d x V = \pi\int_{0}^{\pi/2}\left\lbrack f(x) \right\rbrack^{2}dx = \pi\int_{0}^{\pi/2}e^{2x}\cos^{2}x\, dx Using cos 2 x = 1 + cos 2 x 2 \cos^{2}x = \frac{1 + \cos 2x}{2}
V = π 2 ∫ 0 π / 2 e 2 x d x + π 2 ∫ 0 π / 2 e 2 x cos 2 x d x V = \frac{\pi}{2}\int_{0}^{\pi/2}e^{2x}\, dx + \frac{\pi}{2}\int_{0}^{\pi/2}e^{2x}\cos 2x\, dx For the first integral:
∫ 0 π / 2 e 2 x d x = [ 1 2 e 2 x ] 0 π / 2 = 1 2 ( e π − 1 ) \int_{0}^{\pi/2}e^{2x}\, dx = \left\lbrack \frac{1}{2}e^{2x} \right\rbrack_{0}^{\pi/2} = \frac{1}{2}\left( e^{\pi} - 1 \right) For the second integral, using the result of part (b):
∫ 0 π / 2 e 2 x cos 2 x d x = [ 1 4 e 2 x ( sin 2 x + cos 2 x ) ] 0 π / 2 \int_{0}^{\pi/2}e^{2x}\cos 2x\, dx = \left\lbrack \frac{1}{4}e^{2x}\left( \sin 2x + \cos 2x \right) \right\rbrack_{0}^{\pi/2} At x = π 2 x = \frac{\pi}{2} sin π = 0 \sin\pi = 0 cos π = − 1 \cos\pi = - 1 1 4 e π ( 0 − 1 ) = − 1 4 e π \frac{1}{4}e^{\pi}(0 - 1) = - \frac{1}{4}e^{\pi}
At x = 0 x = 0 sin 0 = 0 \sin 0 = 0 cos 0 = 1 \cos 0 = 1 1 4 ( 1 ) ( 0 + 1 ) = 1 4 \frac{1}{4}(1)(0 + 1) = \frac{1}{4}
∫ 0 π / 2 e 2 x cos 2 x d x = − 1 4 e π − 1 4 = − 1 4 ( e π + 1 ) \int_{0}^{\pi/2}e^{2x}\cos 2x\, dx = - \frac{1}{4}e^{\pi} - \frac{1}{4} = - \frac{1}{4}\left( e^{\pi} + 1 \right) Substituting back:
V = π 2 × 1 2 ( e π − 1 ) + π 2 × [ − 1 4 ( e π + 1 ) ] V = \frac{\pi}{2} \times \frac{1}{2}\left( e^{\pi} - 1 \right) + \frac{\pi}{2} \times \left\lbrack - \frac{1}{4}\left( e^{\pi} + 1 \right) \right\rbrack = π 4 ( e π − 1 ) − π 8 ( e π + 1 ) = \frac{\pi}{4}\left( e^{\pi} - 1 \right) - \frac{\pi}{8}\left( e^{\pi} + 1 \right) Combining over a common denominator of 8 8
V = 2 π ( e π − 1 ) − π ( e π + 1 ) 8 V = \frac{2\pi\left( e^{\pi} - 1 \right) - \pi\left( e^{\pi} + 1 \right)}{8} = π [ 2 e π − 2 − e π − 1 ] 8 = \frac{\pi\left\lbrack 2e^{\pi} - 2 - e^{\pi} - 1 \right\rbrack}{8} = π ( e π − 3 ) 8 = \frac{\pi\left( e^{\pi} - 3 \right)}{8}
To check your answer: π ( e π − 3 ) 8 ≈ 7.91 \frac{\pi\left( e^{\pi} - 3 \right)}{8} \approx 7.91 π ∫ 0 π / 2 e 2 x cos 2 x d x \pi\int_{0}^{\pi/2}e^{2x}\cos^{2}x\, dx
∴ \therefore V = π ( e π − 3 ) 8 units 3 V = \frac{\pi\left( e^{\pi} - 3 \right)}{8}\ \text{units}^{3}
← Question 8 Vectors Question 10 → Differential equations
The Learning Cave