用定积分的分部积分法做下题 归纳一下定积分的换元积分和分部积分法的一般解题步骤?
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\u8ddf\u5206\u90e8\u79ef\u5206\u6cd5 inegral by parts\uff0c\u8fd9\u4e24\u79cd\u65b9\u6cd5
\u65e2\u9002\u7528\u4e8e\u5b9a\u79ef\u5206 definite integral\uff0c\u4e5f\u9002\u7528\u4e8e
\u4e0d\u5b9a\u79ef\u5206 indefinite integral\u3002
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\u7528\u7684\u65b9\u6cd5\u662f\u6709\u7406\u5206\u5f0f\u7684\u5206\u89e3\u6cd5 partial fraction\u3002
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=∫[0,π/2]e^(2x)d(sinx)
=e^(2x)sinx |[0,π/2] - 2∫[0,π/2]e^(2x)sinxdx
=e^π -0 +2∫[0,π/2]e^(2x)d(cosx)
=e^π + 2e^(2x)cosx |[0,π/2] - 4∫[0,π/2]e^(2x)cosxdx
=e^π + 0 - 2 -4∫[0,π/2]e^(2x)cosxdx
故∫[0,π/2]e^(2x)cosxdx=(e^π -2)/5
(1):
∫(0→π) xsinx dx
= ∫(0→π) x d(- cosx)
= - xcosx:[0→π] + ∫(0→π) cosx dx
= - π(- 1) + sinx:[0→π]
= π
(2):
∫(0→1) xe^x dx
= ∫(0→1) x d(e^x)
= xe^x:[0→1] - ∫(0→1) e^x dx
= e - e^x:(0→1)
= e - (e - 1)
= 1
(3):
∫(1→e) x(x - 1)lnx dx
= ∫(1→e) (x^2 - x)lnx dx
= ∫(1→e) lnx d(x^3/3 - x^2/2)
= (x^3/3 - x^2/2)lnx:(1→e) - ∫(1→e) (x^3/3 - x^2/2)(1/x) dx
= (1/3)e^3 - (1/2)e^2 - ∫(1→e) (x^2/3 - x/2) dx
= (1/3)e^3 - (1/2)e^2 - (x^3/9 - x^2/4):(1→e)
= (1/3)e^3 - (1/2)e^2 - [(e^3/9 - e^2/4) - (1/9 - 1/4)]
= (2/9)e^3 - (1/4)e^2 - 5/36
(4):
∫(0→1) x^2e^(2x) dx
= (1/2)∫(0→1) x^2 d(e^(2x))
= (1/2)x^2e^(2x):(0→1) - (1/2)∫(0→1) 2xe^(2x) dx
= (1/2)e^2 - (1/2)∫(0→1) x d(e^(2x))
= (1/2)e^2 - (1/2)xe^(2x):(0→1) + (1/2)∫(0→1) e^(2x) dx
= (1/2)e^2 - (1/2)e^2 + (1/4)e^(2x):(0→1)
= (1/4)(e^2 - 1)
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