Question Number 214917 by Spillover last updated on 23/Dec/24
Answered by cemoosky last updated on 23/Dec/24
![∫_0 ^1 ∫_0 ^1 ∙∙∙∫_0 ^1 (x_1 ^3 +x_2 ^3 +∙∙∙+x_n ^3 )dx_1 dx_2 ∙∙∙dx_n for n = 2018 n∫_0 ^1 ∫_0 ^1 ∙∙∙∫_0 ^1 (x_n ^3 )dx_1 dx_2 ∙∙∙dx_n = n∫_0 ^1 (x_n ^3 )dx_n = n[(x^4 /4)]_0 ^1 = n[(((1)^4 )/4) − (((0)^4 )/4)] = n[(1/4)] = 2018[(1/4)] = ((2018)/4) = 504.5](Q214922.png)
$$\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \centerdot\centerdot\centerdot\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}_{\mathrm{1}} ^{\mathrm{3}} +{x}_{\mathrm{2}} ^{\mathrm{3}} +\centerdot\centerdot\centerdot+{x}_{{n}} ^{\mathrm{3}} \right){dx}_{\mathrm{1}} {dx}_{\mathrm{2}} \centerdot\centerdot\centerdot{dx}_{{n}} \\ $$$$\:{for}\:{n}\:=\:\mathrm{2018} \\ $$$$\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \centerdot\centerdot\centerdot\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}_{{n}} ^{\mathrm{3}} \right){dx}_{\mathrm{1}} {dx}_{\mathrm{2}} \centerdot\centerdot\centerdot{dx}_{{n}} \:=\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}_{{n}} ^{\mathrm{3}} \right){dx}_{{n}} \: \\ $$$$\:=\:{n}\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:{n}\left[\frac{\left(\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{4}}\:−\:\frac{\left(\mathrm{0}\right)^{\mathrm{4}} }{\mathrm{4}}\right]\:=\:{n}\left[\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$$$\:=\:\mathrm{2018}\left[\frac{\mathrm{1}}{\mathrm{4}}\right]\:=\:\frac{\mathrm{2018}}{\mathrm{4}}\:=\:\mathrm{504}.\mathrm{5} \\ $$
Commented by Spillover last updated on 24/Dec/24
$${great} \\ $$
Answered by Spillover last updated on 24/Dec/24
