Using the part of the Maclaurin series of cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... up to only the second term, i.e., the x^2 term, an approximate value of the integral√x cos(x^2)dx: [asnwer: 19/33]
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Maclaurin Series 1
2014-05-14 1:53 pm
回答 (1)
2014-05-14 8:22 pm
✔ 最佳答案
cos x = 1 - x^2/2! cos(x^2) = 1 - x^4/2!
sqrt x cos(x^2) = x^(1/2)[ 1 - x^4/2!] = x^(1/2) - x^(9/2)/2!
∫ sqrt x cos(x^2) dx = ∫ x^(1/2) - x^(9/2)/2! dx
= 2x^(3/2)/3 - x^(11/2)/11
when x = 0 the integral = 0
when x = 1, the integral = 2/3 - 1/11 = (22 - 3)/33 = 19/33.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140514000051KK00028