Question

计算下列定积分：

如图

Answer 1

(3) 令 √(5-4x) = u, 则 x = (5-u^2)/4, dx = -udu/2
I = ∫<下3，上1>(5-u^2)(-du/)/8 = (1/8) ∫<下1，上3>(5-u^2)-du
= (1/8)[5u-u^3/3]<下1，上3> = 1/6
(6) 令 x = secu, 则 dx = secutanudu
I = ∫<下2π/3，上π>secutanudu/[secu(-tanu)]
= - ∫<下2π/3，上π>du = -π/3
(7) I = ∫<下0，上2>(x-1+1)dx/(x^2-2x+2)
= (1/2)∫<下0，上2>d(x^2-2x+2)/(x^2-2x+2)
+ ∫<下0，上2>d(x-1)/[1+(x-1)^2]
= (1/2)[ln(x^2-2x+2)]<下0，上2> + [arctan(x-1)]<下0，上2>
= 0 + π/4 - (-π/4) = π/2

Answer 2

掌握定积分的基本公式就ok了

Answer 3

我会做

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