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111 學年度台綜大微積分 B 第 5 題

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111學年度 · 111台綜大微積分B · 第 5 題

題目

Problem

5. Compute the integral 0216x2+7x+2dx\int_0^2 \frac{1}{6x^2 + 7x + 2}\,\mathrm{d}x. (10%)

解答

思路

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  1. 此為有理函數積分。先將分母因式分解: 6x2+7x+2=(2x+1)(3x+2)6x^2 + 7x + 2 = (2x+1)(3x+2)
  2. 使用部分分式法(Partial Fractions)拆分: 1(2x+1)(3x+2)=A2x+1+B3x+2\frac{1}{(2x+1)(3x+2)} = \frac{A}{2x+1} + \frac{B}{3x+2}
  3. 利用通分或留數法(Heaviside Cover-up Method)求得常數 A=2A = 2B=3B = -3
  4. 對各項進行積分,並代入上下限。

答題過程

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將分母進行因式分解:

6x2+7x+2=(2x+1)(3x+2)6x^2 + 7x + 2 = (2x+1)(3x+2)

設部分分式:

1(2x+1)(3x+2)=A2x+1+B3x+2\frac{1}{(2x+1)(3x+2)} = \frac{A}{2x+1} + \frac{B}{3x+2}

同乘分母可得恆等式:

1=A(3x+2)+B(2x+1)1 = A(3x+2) + B(2x+1)
  • x=12    1=A(32+2)=12A    A=2x = -\frac{1}{2} \implies 1 = A\left(-\frac{3}{2} + 2\right) = \frac{1}{2}A \implies A = 2
  • x=23    1=B(43+1)=13B    B=3x = -\frac{2}{3} \implies 1 = B\left(-\frac{4}{3} + 1\right) = -\frac{1}{3}B \implies B = -3

因此:

16x2+7x+2=22x+133x+2\frac{1}{6x^2+7x+2} = \frac{2}{2x+1} - \frac{3}{3x+2}

計算定積分:

0216x2+7x+2dx=02(22x+133x+2)dx\int_0^2 \frac{1}{6x^2+7x+2}\,\mathrm{d}x = \int_0^2 \left( \frac{2}{2x+1} - \frac{3}{3x+2} \right) \mathrm{d}x =[ln2x+1ln3x+2]02= \Big[ \ln|2x+1| - \ln|3x+2| \Big]_0^2 =(ln5ln8)(ln1ln2)= (\ln 5 - \ln 8) - (\ln 1 - \ln 2) =ln5ln8+ln2=ln53ln2+ln2= \ln 5 - \ln 8 + \ln 2 = \ln 5 - 3\ln 2 + \ln 2 =ln52ln2= \ln 5 - 2\ln 2

結論: 積分值為 ln52ln2\displaystyle \ln 5 - 2\ln 2