How do you find the area between #f(x)=(x-1)^3# and #g(x)=x-1#?

First, find the bounds given by #f(x) = g(x)#

#(x-1)^3 = x-1#

#(x-1)^3 - (x-1) = 0#

#(x-1)((x-1)^2-1) = 0#

#(x-1)(x^2-2x) = 0#

#x(x-1)(x-2) = 0#

#x = 0,1,2#

This means the graphs actually close off 2 separate areas, since there are three intersection points. To see what I mean by this, here is a graph of #color(blue)(y=x-1# and #color(red)(y = (x-1)^3#:

So, to find the total area, we need to find the area of both sections and then add them together.

From #x = 0# to #x = 1#, we can see that #(x-1)^3 > x-1#, so in order to find the POSITIVE area between the two curves, we will subtract #x-1# (smaller) from #(x-1)^3# (bigger).

#int_0^1[(x-1)^3 - (x-1)]dx#

#int_0^1[x^3-3x^2+3x-1-x+1]dx#

#int_0^1(x^3-3x^2+2x)dx#

#[x^4/4-x^3+x^2]_0^1 = (1/4-1+1)-(0/4-0+0) = 1/4#

From #x=1# to #x=2#, we can see that #x-1 > (x-1)^3#, so in order to find the POSITIVE area between the two curves, we will subtract #(x-1)^3# (smaller) from #x-1# (bigger).

#int_1^2[(x-1)-(x-1)^3]dx#

#int_1^2[x-1-x^3+3x^2-3x+1]dx#

#int_1^2(-x^3+3x^2-2x)dx#

#[-x^4/4+x^3-x^2]_1^2 = (-16/4+8-4) - (-1/4+1-1) = 1/4#

So the area of the first section is #1/4# and the area of the second section is #1/4#. Therefore, the total area between #f(x)=(x-1)^3# and #g(x) = x-1# is #1/2#

Final Answer