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)](https://www.tinkutara.com/question/Q130286.png)
$$\Omega\:=\:\int_{\mathrm{2}} ^{\mathrm{6}} \underset{{k}=\mathrm{1}} {\overset{\mathrm{9}} {\prod}}\left({x}−{k}\right){dx} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}−\mathrm{5} \\ $$$$\Omega\:=\:\int_{−\mathrm{3}} ^{+\mathrm{1}} \left({u}+\mathrm{4}\right)…\left({u}+\mathrm{1}\right){u}\left({u}−\mathrm{1}\right)…\left({u}−\mathrm{4}\right){du} \\ $$$$\Omega\:=\:\int_{−\mathrm{3}} ^{+\mathrm{1}} {u}\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} −\mathrm{4}\right)\left({u}^{\mathrm{2}} −\mathrm{9}\right)\left({u}^{\mathrm{2}} −\mathrm{16}\right){du} \\ $$$$\Omega\:=\:\int_{−\mathrm{3}} ^{+\mathrm{1}} {u}\left({u}^{\mathrm{4}} −\mathrm{5}{u}^{\mathrm{2}} +\mathrm{4}\right)\left({u}^{\mathrm{4}} −\mathrm{25}{u}^{\mathrm{2}} +\mathrm{144}\right){du} \\ $$$$\Omega\:=\:\int_{−\mathrm{3}} ^{+\mathrm{1}} {u}\left({u}^{\mathrm{8}} −\mathrm{30}{u}^{\mathrm{6}} +\mathrm{273}{u}^{\mathrm{4}} −\mathrm{820}{u}^{\mathrm{2}} +\mathrm{576}\right){du} \\ $$$$\Omega\:=\:\left[\frac{{u}^{\mathrm{10}} }{\mathrm{10}}−\mathrm{30}\frac{{u}^{\mathrm{8}} }{\mathrm{8}}+\mathrm{273}\frac{{u}^{\mathrm{6}} }{\mathrm{6}}−\mathrm{820}\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{576}\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{3}} ^{+\mathrm{1}} \\ $$$$\Omega\:=\:\left[\frac{{u}^{\mathrm{10}} }{\mathrm{10}}−\mathrm{15}\frac{{u}^{\mathrm{8}} }{\mathrm{4}}+\mathrm{91}\frac{{u}^{\mathrm{6}} }{\mathrm{2}}−\mathrm{205}{u}^{\mathrm{4}} +\mathrm{288}{u}^{\mathrm{2}} \right]_{−\mathrm{3}} ^{+\mathrm{1}} \\ $$$$\Omega\:=\:\left(\frac{\mathrm{1}}{\mathrm{10}}−\frac{\mathrm{15}}{\mathrm{4}}+\frac{\mathrm{91}}{\mathrm{2}}−\mathrm{205}+\mathrm{288}\right) \\ $$$$−\left(\frac{\mathrm{3}^{\mathrm{10}} }{\mathrm{10}}−\mathrm{15}\frac{\mathrm{3}^{\mathrm{8}} }{\mathrm{4}}+\mathrm{91}\frac{\mathrm{3}^{\mathrm{6}} }{\mathrm{2}}−\mathrm{205}.\mathrm{3}^{\mathrm{4}} +\mathrm{288}.\mathrm{9}\right) \\ $$$$\Omega\:=\:\frac{\mathrm{2497}}{\mathrm{20}}−\frac{\mathrm{9153}}{\mathrm{20}}\:=\:−\frac{\mathrm{6656}}{\mathrm{20}}\:=\:−\frac{\mathrm{1664}}{\mathrm{5}} \\ $$
Commented by Lordose last updated on 24/Jan/21
$$\mathrm{Same}\:\mathrm{approacb} \\ $$