Question #aa131

2 Answers
May 14, 2017

#-(1-x)^4/4+C#

Explanation:

Substitute #u=1-x# and #(du)/dx[1-x]=-1#

This simplifies to:
#-int[u^3]du#
Power Rule:
#=-(u^4/4)#
Undo subtitution:
#=-(1-x)^4/4+C#

May 14, 2017

#-1/4*(1-x)^4 + C#

Explanation:

Two ways to solve this :
1) You can develop #(1-x)^3 # and integrate term by term but that isn't really clean...

2) Remember the formula : #(f(g))' = g'*f'(g)# ?
In your case :
#f : u -> u^4# and #g : x -> 1-x#
#g'(x) = -1 #

then #f(g(x)) = (1-x)^4#
#(f(g(x)))' = - 4*(1-x)^3 #
And because you don't want the -4 factor, you divide each side by this and you get :

#int (1-x)^3 dx = [-1/4*(1-x)^4] #