First, set the two functions equal to each other to determine the endpoints of the enclosed region:
(1/2)x^2 = x^2 -3x
x^2 - 6x =0
x(x-6) =0
x=0, x=6
Now, integrate the higher function (plug in a number such as 1 into each function to determine which is higher), and subtract the integral of the lower function. Since the integral of each function is the area under the curve, the difference between the two integrals yields the area enclosed between the curves.
Integral of (1/2)x^2: (x^3)/6, evaluated at 6 and 0. This definite integral yields a value of 36.
Integral of (x2-3x): (x^3)/3 - (3/2)(x^2), evaluated at 6 and 0. This definite integral yield a value of 18.
Area enclosed between curves = 36-18 = 36
