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Question Number 130285 by Lordose last updated on 23/Jan/21

Answered by Olaf last updated on 24/Jan/21

Ω = ∫_2 ^6 Π_(k=1) ^9 (x−k)dx  Let u = x−5  Ω = ∫_(−3) ^(+1) (u+4)...(u+1)u(u−1)...(u−4)du  Ω = ∫_(−3) ^(+1) u(u^2 −1)(u^2 −4)(u^2 −9)(u^2 −16)du  Ω = ∫_(−3) ^(+1) u(u^4 −5u^2 +4)(u^4 −25u^2 +144)du  Ω = ∫_(−3) ^(+1) u(u^8 −30u^6 +273u^4 −820u^2 +576)du  Ω = [(u^(10) /(10))−30(u^8 /8)+273(u^6 /6)−820(u^4 /4)+576(u^2 /2)]_(−3) ^(+1)   Ω = [(u^(10) /(10))−15(u^8 /4)+91(u^6 /2)−205u^4 +288u^2 ]_(−3) ^(+1)   Ω = ((1/(10))−((15)/4)+((91)/2)−205+288)  −((3^(10) /(10))−15(3^8 /4)+91(3^6 /2)−205.3^4 +288.9)  Ω = ((2497)/(20))−((9153)/(20)) = −((6656)/(20)) = −((1664)/5)

Ω=269k=1(xk)dxLetu=x5Ω=3+1(u+4)...(u+1)u(u1)...(u4)duΩ=3+1u(u21)(u24)(u29)(u216)duΩ=3+1u(u45u2+4)(u425u2+144)duΩ=3+1u(u830u6+273u4820u2+576)duΩ=[u101030u88+273u66820u44+576u22]3+1Ω=[u101015u84+91u62205u4+288u2]3+1Ω=(110154+912205+288)(3101015384+91362205.34+288.9)Ω=249720915320=665620=16645

Commented by Lordose last updated on 24/Jan/21

Same approacb

Sameapproacb

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