since f > g on (0,1), the area is just
∫[0,1] (x-x^3)-(x^2-x) dx = 5/12
find the area of the region bounded by the curves f(x)=x-x^3 ; g(x)=x^2-x ; over [0,1]
5 answers
ty!
i got 5/6?
i used the equation 2x-x^5, is that wrong?
it sure is
(x-x^3)-(x^2-x) = -x^3-x^2+2x
You can't combine x^2 and x^3 to make x^5!! They're added, not multiplied.
(x-x^3)-(x^2-x) = -x^3-x^2+2x
You can't combine x^2 and x^3 to make x^5!! They're added, not multiplied.