※ 引述《gogoabc (神阿 请多给我一点体重)》之铭言： : The area of the region bounded by the curve y=|x^2- 1| and the line y=3 : ^^^^^^^^ : 绝对值x平方减1 : 答案是8 谢谢 : ∞ dx : ∫ -------- : 4 x(1-√x) : ∞ : 我算出来 2 ln √x + 2ln (1-√x) / : 4 : 怎麽看都是发散 答案是-2ln2 dx 先计算 ∫-------------- (x)(1 - √x) 令 y = √x , 则 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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