Re: [積分] 積面積與瑕積分
※ 引述《gogoabc (神阿 請多給我一點體重)》之銘言：
: The area of the region bounded by the curve y=|x^2- 1| and the line y=3
: ^^^^^^^^
: 絕對值x平方減1
: 答案是8 謝謝
: ∞ dx
: ∫ --------
: 4 x(1-√x)
: ∞
: 我算出來 2 ln √x + 2ln (1-√x) /
: 4
: 怎麼看都是發散 答案是-2ln2
dx
先計算 ∫--------------
(x)(1 - √x)
令 y = √x , 則 x = y^2 => dx = 2y dy
dx
∫--------------
(x)(1 - √x)
2y
= ∫-------------- dy
(y^2)(1 - y)
2
= ∫---------- dy
(y)(1 - y)
1
= (2)(∫---------- dy)
(y)(1 - y)
1 1
= (2)(∫--- + ------- dy)
y 1 - y
= (2)(ln|y| - ln|1-y|) + c
| y |
= (2)(ln|-------|) + c
| 1 - y |
| √x |
= (2)(ln|---------|) + c
| 1 - √x |
∞ 1
∫ -------------- dx
4 (x)(1 - √x)
R 1
= lim ∫ -------------- dx
R→∞ 4 (x)(1 - √x)
| √x | |R
= lim (2)(ln|---------|) |
R→∞ | 1 - √x | |4
| √R | | √4 |
= lim (2)((ln|---------|) - ln|---------|)
R→∞ | 1 - √R | | 1 - √4 |
| 1 | | 2 |
= lim (2)(ln|---------------| - ln|----|)
R→∞ | (1/(√R)) - 1 | | -1 |
= (2)(ln1 - ln2) = (2)(0 - ln2) = (-2)(ln2)
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09/19 21:42, , 1F
09/19 21:42, 1F
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