Calculus Problem~

[匿名]

2008-02-24 12:15 pm

1. Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

integral from negative infinite to 20; re^(r/5)

2. Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

integral from 0 to infinite; (xarctanx)/(1+x^2)^2

回答 (2)

用戶8745

2008-02-24 9:58 pm

✔ 最佳答案

1.∫ (a to 20) rer/5 dr
= (1/5)∫ (a to 20) rder/5
= (1/5)(20e20/5 - aea/5) -(1/5)∫ (a to 20) er/5 dr
= (1/5)(20e4 - aea/5) - (1/25)(e4 - ea/5)
= 99e4/25 - (a/5 - 1/25)ea/5

So
∫ (-∞ to 20) rer/5 dr
= lim(a→-∞) (99e4/25 - (a/5 - 1/25)ea/5)
= 99e4/25 .

2. Note that xdx/(1+x2)2 = (1/2)d(1+x2)/(1+x2)2 = (-1/2)d(1+x2)-1. So
∫ (0 to a) xarctanx dx/(1+x2)2
= (-1/2)∫ (0 to a) arctanx d(1+x2)-1
= (-1/2)(arctan a / (1+a2) - 0) + (1/2)∫ (0 to a) dx / (1+x2)2

Let x = tan u, dx = sec2u du,
x=a, u=arctan a.
x=0, u=0.

So,
∫ (0 to a) dx / (1+x2)2
= ∫ (0 to arctan a) sec2u du / sec2u
= ∫ (0 to arctan a) du
= arctan a.

Therefore,

∫ (0 to a) xarctanx dx/(1+x2)2
= (-1/2)arctan a / (1+a2) + (1/2)arctan a.

∫ (0 to ∞) xarctanx dx/(1+x2)2
= lim(a→∞) ((-1/2)arctan a / (1+a2) + (1/2)arctan a)
= π/4.

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myisland8132

2008-02-28 2:28 am

好

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