f(1/8) = (1/8)^(1/3) = 1/2
g(1/8) = 0; therefore, f(x) is higher.
A[second] = Integral (0 to 1, [f(x) - g(x)] dx )
Part 3: Find the area between 1 and 15.
Test x = 8.
f(8) = 2
g(8) = 0
So f(x) is higher than g(x), and the third area is
A[third] = Integral (1 to 15, [f(x) - g(x)] dx )
That means the area is
A = A[first] + A[second] + A[third]
A = Integral (-1 to 0, [g(x) - f(x)] dx ) + Integral (0 to 1, [f(x) - g(x)] dx ) + Integral (1 to 15, [f(x) - g(x)] dx )
A = Integral (-1 to 0, [0 - (x^(1/3) - x)] dx ) + Integral (0 to 1, [x^(1/3) - x - 0] dx ) + Integral (1 to 15, [x^(1/3) - x - 0] dx )
And we get some simplifications.
A = Integral (-1 to 0, [-x^(1/3) + x)] dx ) + Integral (0 to 1, [x^(1/3) - x] dx ) + Integral (1 to 15, [x^(1/3) - x] dx )
The integral of x^(1/3) is solved using the reverse power rule; it is equal to (3/4)x^(4/3). Similarly, the integral of x is (1/2)x^2.
A = [ (-3/4)x^(4/3) + (1/2)x^2 ]{evaluated from -1 to 0} + [(3/4)x^(4/3) - (1/2)x^2] {evaluated from 0 to 1} + [(3/4)x^(4/3) - (1/2)x^2] {evaluated from 1 to 15}
A = [ (-3/4)(0)^(4/3) + (1/2)0^2 ] - [(-3/4)(-1)^(4/3) + (1/2)(-1)^2] + [(3/4)(1)^(4/3) - (1/2)1^2] - [(3/4 (0)^(4/3) - (1/2)0^2] + [(3/4)15^(4/3) - (1/2)15^2] - [(3/4)1^(4/3) - (1/2)1^2]
A = [0 + 0] - [(-3/4) + (1/2)] + [(3/4) - (1/2)] - [0 - 0] + [(3/4)(15)^(4/3) - (1/2)(225)] - [(3/4) - (1/2)]
A = [0] - [-1/4] + [1/4] - [0] + [(3/4)(15)^(4/3)] - [225/2] - [1/4]
A = [1/4] + [1/4] + [(3/4)(15)^(4/3)] - [225/2] - [1/4]
A = [1/4] + [(3/4)(15)^(4/3)] - [225/2]
A = [451/4] + [(3/4)(15)^(4/3)]
A = [451/4] + [(3 (15)^(4/3)]/4
A = [ 451 + 3(15)^(4/3) ]/4