题目
Problem
(a) Use the substitution x = 2 sin u x=2\sin u x = 2 sin u
∫ 0 1 3 x 3 4 − x 2 d x = ∫ a b 24 sin 3 u d u , \int_0^1\frac{3x^3}{\sqrt{4-x^2}}\,\mathrm{d}x=\int_a^b24\sin^3u\,\mathrm{d}u, ∫ 0 1 4 − x 2 3 x 3 d x = ∫ a b 24 sin 3 u d u , where a a a b b b
(4)
(b) Hence, find the exact value of ∫ 0 1 3 x 3 4 − x 2 d x \displaystyle\int_0^1\frac{3x^3}{\sqrt{4-x^2}}\,\mathrm{d}x ∫ 0 1 4 − x 2 3 x 3 d x
(4)
题目中文翻译
(a) 使用代换 x = 2 sin u x=2\sin u x = 2 sin u
∫ 0 1 3 x 3 4 − x 2 d x = ∫ a b 24 sin 3 u d u , \int_0^1\frac{3x^3}{\sqrt{4-x^2}}\,\mathrm{d}x=\int_a^b24\sin^3u\,\mathrm{d}u, ∫ 0 1 4 − x 2 3 x 3 d x = ∫ a b 24 sin 3 u d u , 其中 a a a b b b
(4)
(b) 由此求 ∫ 0 1 3 x 3 4 − x 2 d x \displaystyle\int_0^1\frac{3x^3}{\sqrt{4-x^2}}\,\mathrm{d}x ∫ 0 1 4 − x 2 3 x 3 d x
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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按指定代换求微分，并把两个积分限一并换成 u u u
答题过程
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Let x = 2 sin u x=2\sin u x = 2 sin u 0 ≤ u ≤ π / 6 0\le u\le\pi/6 0 ≤ u ≤ π /6 d x = 2 cos u d u \mathrm{d}x=2\cos u\,\mathrm{d}u d x = 2 cos u d u 4 − x 2 = 2 cos u \sqrt{4-x^2}=2\cos u 4 − x 2 = 2 cos u 0 0 0 π / 6 \pi/6 π /6
3 ( 2 sin u ) 3 2 cos u ( 2 cos u ) = 24 sin 3 u . \frac{3(2\sin u)^3}{2\cos u}(2\cos u)=24\sin^3u. 2 cos u 3 ( 2 sin u ) 3 ( 2 cos u ) = 24 sin 3 u . Therefore
∫ 0 1 3 x 3 4 − x 2 d x = ∫ 0 π / 6 24 sin 3 u d u . \int_0^1\frac{3x^3}{\sqrt{4-x^2}}\,\mathrm{d}x
=\int_0^{\pi/6}24\sin^3u\,\mathrm{d}u. ∫ 0 1 4 − x 2 3 x 3 d x = ∫ 0 π /6 24 sin 3 u d u . Thus a = 0 a=0 a = 0 b = π / 6 b=\pi/6 b = π /6
(b)
解法一
思路
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承接 (a)，把奇次正弦写成 sin u ( 1 − cos 2 u ) \sin u(1-\cos^2u) sin u ( 1 − cos 2 u )
答题过程
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By part (a), the integral equals
∫ 0 π / 6 24 sin u ( 1 − cos 2 u ) d u . \int_0^{\pi/6}24\sin u(1-\cos^2u)\,\mathrm{d}u. ∫ 0 π /6 24 sin u ( 1 − cos 2 u ) d u . An antiderivative is − 24 cos u + 8 cos 3 u -24\cos u+8\cos^3u − 24 cos u + 8 cos 3 u
[ − 24 cos u + 8 cos 3 u ] 0 π / 6 = − 24 3 2 + 8 ( 3 2 ) 3 − ( − 24 + 8 ) = − 12 3 + 3 3 + 16 = 16 − 9 3 . \begin{align*}
&\,\bigg[-24\cos u+8\cos^3u\bigg]_0^{\pi/6}\\[4mm]
=&\,-24\frac{\sqrt3}{2}+8\bigg(\frac{\sqrt3}{2}\bigg)^3-(-24+8)\\[4mm]
=&\,-12\sqrt3+3\sqrt3+16\\[4mm]
=&\,16-9\sqrt3.
\end{align*} = = = [ − 24 cos u + 8 cos 3 u ] 0 π /6 − 24 2 3 + 8 ( 2 3 ) 3 − ( − 24 + 8 ) − 12 3 + 3 3 + 16 16 − 9 3 . 