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Question Number 93299 by Rio Michael last updated on 12/May/20
find a reduction formulae for I_n = ∫_(−1) ^0 x^n (1 + x)^2 dx
findareductionformulaeforIn=∫−10xn(1+x)2dx
Commented by mathmax by abdo last updated on 12/May/20
I_n =∫_(−1) ^0 x^n (1+x)^2 dx ⇒I_n =_(x=−t) ∫_0 ^1 (−1)^n t^n (1−t)^2 dt =(−1)^n ∫_0 ^1 t^n (t−1)^2 dt by parts we get ∫_0 ^1 t^n (t−1)^2 dt =[(1/(n+1))t^(n+1) (t−1)^2 ]_0 ^1 −(2/(n+1))∫_0 ^1 t^(n+1) (t−1)dt =−(2/(n+1)) { [(1/(n+2))t^(n+2) (t−1)]_0 ^1 −(1/(n+2))∫_0 ^1 t^(n+2) dt} =−(2/(n+1))(−(1/(n+2)))×(1/(n+3)) =(2/((n+1)(n+2)(n+3))) ⇒ I_n =((2(−1)^n )/((n+1)(n+2)(n+3)))
In=∫−10xn(1+x)2dx⇒In=x=−t∫01(−1)ntn(1−t)2dt=(−1)n∫01tn(t−1)2dtbypartsweget∫01tn(t−1)2dt=[1n+1tn+1(t−1)2]01−2n+1∫01tn+1(t−1)dt=−2n+1{[1n+2tn+2(t−1)]01−1n+2∫01tn+2dt}=−2n+1(−1n+2)×1n+3=2(n+1)(n+2)(n+3)⇒In=2(−1)n(n+1)(n+2)(n+3)
Commented by Rio Michael last updated on 12/May/20
thank you sir
thankyousir

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