題目
Problem
6. (a) ∫ 0 1 x ( sin x + sin − 1 x ) d x = ( 9 ) ‾ . \displaystyle\int_0^1 x(\sin x + \sin^{-1} x)\,\mathrm{d}x = \underline{\quad (9) \quad} \,. ∫ 0 1 x ( sin x + sin − 1 x ) d x = ( 9 ) .
(b) Let D D D y = ( 10 x − x 2 − 21 ) 1 4 y = (10x-x^2-21)^{\frac{1}{4}} y = ( 10 x − x 2 − 21 ) 4 1 x x x D D D x x x
解答
(a)
解法一
思路
展開
將積分拆開成兩部分：
∫ 0 1 x sin x d x + ∫ 0 1 x sin − 1 x d x \int_0^1 x\sin x\,\mathrm{d}x + \int_0^1 x\sin^{-1}x\,\mathrm{d}x ∫ 0 1 x sin x d x + ∫ 0 1 x sin − 1 x d x
第一部分 x sin x x\sin x x sin x 分部積分法 (Integration by parts) 求解。
第二部分 x sin − 1 x x\sin^{-1}x x sin − 1 x x = sin θ x = \sin\theta x = sin θ
答題過程
展開
原積分拆開：
I = ∫ 0 1 x sin x d x ⏟ I 1 + ∫ 0 1 x sin − 1 x d x ⏟ I 2 I = \underbrace{\int_0^1 x\sin x\,\mathrm{d}x}_{I_1} + \underbrace{\int_0^1 x\sin^{-1}x\,\mathrm{d}x}_{I_2} I = I 1 ∫ 0 1 x sin x d x + I 2 ∫ 0 1 x sin − 1 x d x 計算 I 1 I_1 I 1
使用分部積分，令 u = x , d v = sin x d x ⟹ d u = d x , v = − cos x u = x, \mathrm{d}v = \sin x\,\mathrm{d}x \implies \mathrm{d}u = \mathrm{d}x, v = -\cos x u = x , d v = sin x d x ⟹ d u = d x , v = − cos x
I 1 = [ − x cos x ] 0 1 + ∫ 0 1 cos x d x = − cos ( 1 ) + 0 + [ sin x ] 0 1 = sin 1 − cos 1 I_1 = \left[ -x\cos x \right]_0^1 + \int_0^1 \cos x\,\mathrm{d}x = -\cos(1) + 0 + \left[ \sin x \right]_0^1 = \sin 1 - \cos 1 I 1 = [ − x cos x ] 0 1 + ∫ 0 1 cos x d x = − cos ( 1 ) + 0 + [ sin x ] 0 1 = sin 1 − cos 1 計算 I 2 I_2 I 2
令 x = sin θ ⟹ d x = cos θ d θ x = \sin\theta \implies \mathrm{d}x = \cos\theta\,\mathrm{d}\theta x = sin θ ⟹ d x = cos θ d θ 0 → π 2 0 \to \frac{\pi}{2} 0 → 2 π
I 2 = ∫ 0 π / 2 θ sin θ cos θ d θ = 1 2 ∫ 0 π / 2 θ sin ( 2 θ ) d θ I_2 = \int_0^{\pi/2} \theta \sin\theta \cos\theta\,\mathrm{d}\theta = \frac{1}{2} \int_0^{\pi/2} \theta \sin(2\theta)\,\mathrm{d}\theta I 2 = ∫ 0 π /2 θ sin θ cos θ d θ = 2 1 ∫ 0 π /2 θ sin ( 2 θ ) d θ 使用分部積分，令 u = θ , d v = sin ( 2 θ ) d θ ⟹ d u = d θ , v = − 1 2 cos ( 2 θ ) u = \theta, \mathrm{d}v = \sin(2\theta)\,\mathrm{d}\theta \implies \mathrm{d}u = \mathrm{d}\theta, v = -\frac{1}{2}\cos(2\theta) u = θ , d v = sin ( 2 θ ) d θ ⟹ d u = d θ , v = − 2 1 cos ( 2 θ )
I 2 = 1 2 ( [ − θ 2 cos ( 2 θ ) ] 0 π / 2 + 1 2 ∫ 0 π / 2 cos ( 2 θ ) d θ ) = 1 2 ( − π 4 cos ( π ) + 0 + 1 4 [ sin ( 2 θ ) ] 0 π / 2 ) = 1 2 ( π 4 + 0 ) = π 8 \begin{align*}
I_2 =&\, \frac{1}{2} \left( \left[ -\frac{\theta}{2}\cos(2\theta) \right]_0^{\pi/2} + \frac{1}{2} \int_0^{\pi/2} \cos(2\theta)\,\mathrm{d}\theta \right) \\[4mm]
=&\, \frac{1}{2} \left( -\frac{\pi}{4}\cos(\pi) + 0 + \frac{1}{4}\left[ \sin(2\theta) \right]_0^{\pi/2} \right) \\[4mm]
=&\, \frac{1}{2} \left( \frac{\pi}{4} + 0 \right) = \frac{\pi}{8}
\end{align*} I 2 = = = 2 1 ( [ − 2 θ cos ( 2 θ ) ] 0 π /2 + 2 1 ∫ 0 π /2 cos ( 2 θ ) d θ ) 2 1 ( − 4 π cos ( π ) + 0 + 4 1 [ sin ( 2 θ ) ] 0 π /2 ) 2 1 ( 4 π + 0 ) = 8 π 合併：
I = I 1 + I 2 = sin 1 − cos 1 + π 8 I = I_1 + I_2 = \sin 1 - \cos 1 + \frac{\pi}{8} I = I 1 + I 2 = sin 1 − cos 1 + 8 π 故 (9) 處應填入 sin 1 − cos 1 + π 8 \displaystyle\boxed{\sin 1 - \cos 1 + \frac{\pi}{8}} sin 1 − cos 1 + 8 π
(b)
解法一
思路
展開
利用圓盤法（Disk method）公式求繞 x x x V = ∫ a b π y 2 d x V = \int_{a}^{b} \pi y^2\,\mathrm{d}x V = ∫ a b π y 2 d x
首先解出 y = 0 y=0 y = 0 10 x − x 2 − 21 = 0 10x-x^2-21 = 0 10 x − x 2 − 21 = 0
代入公式得： V = π ∫ 3 7 10 x − x 2 − 21 d x V = \pi \int_3^7 \sqrt{10x-x^2-21}\,\mathrm{d}x V = π ∫ 3 7 10 x − x 2 − 21 d x
配方後利用幾何意義（半圓面積）求出積分值，避免繁瑣的三角代換。
答題過程
展開
解 y = ( 10 x − x 2 − 21 ) 1 / 4 = 0 ⟹ x 2 − 10 x + 21 = ( x − 3 ) ( x − 7 ) = 0 y = (10x-x^2-21)^{1/4} = 0 \implies x^2 - 10x + 21 = (x-3)(x-7) = 0 y = ( 10 x − x 2 − 21 ) 1/4 = 0 ⟹ x 2 − 10 x + 21 = ( x − 3 ) ( x − 7 ) = 0 x = 3 x = 3 x = 3 x = 7 x = 7 x = 7
利用圓盤法：
V = π ∫ 3 7 y 2 d x = π ∫ 3 7 10 x − x 2 − 21 d x V = \pi \int_3^7 y^2\,\mathrm{d}x = \pi \int_3^7 \sqrt{10x-x^2-21}\,\mathrm{d}x V = π ∫ 3 7 y 2 d x = π ∫ 3 7 10 x − x 2 − 21 d x 將根號內配方：
10 x − x 2 − 21 = 4 − ( x − 5 ) 2 10x-x^2-21 = 4 - (x-5)^2 10 x − x 2 − 21 = 4 − ( x − 5 ) 2 故：
V = π ∫ 3 7 4 − ( x − 5 ) 2 d x V = \pi \int_3^7 \sqrt{4 - (x-5)^2}\,\mathrm{d}x V = π ∫ 3 7 4 − ( x − 5 ) 2 d x 被積函數 h ( x ) = 4 − ( x − 5 ) 2 h(x) = \sqrt{4 - (x-5)^2} h ( x ) = 4 − ( x − 5 ) 2 ( 5 , 0 ) (5, 0) ( 5 , 0 ) R = 2 R=2 R = 2 [ 3 , 7 ] [3, 7] [ 3 , 7 ]
∫ 3 7 4 − ( x − 5 ) 2 d x = 1 2 π R 2 = 1 2 π ( 2 2 ) = 2 π \int_3^7 \sqrt{4 - (x-5)^2}\,\mathrm{d}x = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi (2^2) = 2\pi ∫ 3 7 4 − ( x − 5 ) 2 d x = 2 1 π R 2 = 2 1 π ( 2 2 ) = 2 π 因此，旋轉體體積為：
V = π ⋅ ( 2 π ) = 2 π 2 V = \pi \cdot (2\pi) = \boxed{2\pi^2} V = π ⋅ ( 2 π ) = 2 π 2 