题目
Problem
(a) Use the substitution u = x 2 − 3 u=x^2-3 u = x 2 − 3
∫ 7 12 4 x 3 x 2 − 3 d x = ∫ a b 2 ( u + 3 ) u d u , \int_{\sqrt7}^{\sqrt{12}}\frac{4x^3}{\sqrt{x^2-3}}\,\mathrm{d}x=\int_a^b\frac{2(u+3)}{\sqrt u}\,\mathrm{d}u, ∫ 7 12 x 2 − 3 4 x 3 d x = ∫ a b u 2 ( u + 3 ) d u , where a a a b b b
(4)
(b) Hence, find the exact value of ∫ 7 12 4 x 3 x 2 − 3 d x \displaystyle\int_{\sqrt7}^{\sqrt{12}}\frac{4x^3}{\sqrt{x^2-3}}\,\mathrm{d}x ∫ 7 12 x 2 − 3 4 x 3 d x
(4)
题目中文翻译
(a) 使用换元 u = x 2 − 3 u=x^2-3 u = x 2 − 3
∫ 7 12 4 x 3 x 2 − 3 d x = ∫ a b 2 ( u + 3 ) u d u , \int_{\sqrt7}^{\sqrt{12}}\frac{4x^3}{\sqrt{x^2-3}}\,\mathrm{d}x=\int_a^b\frac{2(u+3)}{\sqrt u}\,\mathrm{d}u, ∫ 7 12 x 2 − 3 4 x 3 d x = ∫ a b u 2 ( u + 3 ) d u , 其中 a a a b b b
(4)
(b) 由此求 ∫ 7 12 4 x 3 x 2 − 3 d x \displaystyle\int_{\sqrt7}^{\sqrt{12}}\frac{4x^3}{\sqrt{x^2-3}}\,\mathrm{d}x ∫ 7 12 x 2 − 3 4 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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指定换元后保留一个 x 2 = u + 3 x^2=u+3 x 2 = u + 3 2 x d x 2x\,\mathrm{d}x 2 x d x
答题过程
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Let u = x 2 − 3 u=x^2-3 u = x 2 − 3
d u = 2 x d x , x 2 = u + 3. \mathrm{d}u=2x\,\mathrm{d}x,\qquad x^2=u+3. d u = 2 x d x , x 2 = u + 3. The limits are
x = 7 ⟹ u = 4 , x = 12 ⟹ u = 9. x=\sqrt7\ \Longrightarrow\ u=4,\qquad
x=\sqrt{12}\ \Longrightarrow\ u=9. x = 7 ⟹ u = 4 , x = 12 ⟹ u = 9. Also,
4 x 3 d x = 2 x 2 ( 2 x d x ) = 2 ( u + 3 ) d u . 4x^3\,\mathrm{d}x=2x^2(2x\,\mathrm{d}x)=2(u+3)\,\mathrm{d}u. 4 x 3 d x = 2 x 2 ( 2 x d x ) = 2 ( u + 3 ) d u . Therefore
∫ 7 12 4 x 3 x 2 − 3 d x = ∫ 4 9 2 ( u + 3 ) u d u , \begin{align*}
&\,\int_{\sqrt7}^{\sqrt{12}}\frac{4x^3}{\sqrt{x^2-3}}\,\mathrm{d}x\\[4mm]
=&\,\int_4^9\frac{2(u+3)}{\sqrt u}\,\mathrm{d}u,
\end{align*} = ∫ 7 12 x 2 − 3 4 x 3 d x ∫ 4 9 u 2 ( u + 3 ) d u , so a = 4 , b = 9 \boxed{a=4,\ b=9} a = 4 , b = 9
(b)
解法一
思路
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沿用(a)的积分，把分式拆成两个幂函数再积分。
答题过程
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From (a),
I = ∫ 4 9 ( 2 u 1 / 2 + 6 u − 1 / 2 ) d u = [ 4 3 u 3 / 2 + 12 u 1 / 2 ] 4 9 = 4 3 ( 27 − 8 ) + 12 ( 3 − 2 ) = 76 3 + 12 = 112 3 . \begin{align*}
I=&\,\int_4^9\big(2u^{1/2}+6u^{-1/2}\big)\,\mathrm{d}u\\
=&\,\bigg[\frac43u^{3/2}+12u^{1/2}\bigg]_4^9\\
=&\,\frac43(27-8)+12(3-2)\\
=&\,\frac{76}{3}+12\\
=&\,\boxed{\frac{112}{3}}.
\end{align*} I = = = = = ∫ 4 9 ( 2 u 1/2 + 6 u − 1/2 ) d u [ 3 4 u 3/2 + 12 u 1/2 ] 4 9 3 4 ( 27 − 8 ) + 12 ( 3 − 2 ) 3 76 + 12 3 112 . 