The first thing we need to do for this problem is to find the zeros of x3-2x2-5x+6. Since this doesn't factor in any obvious way, we guess at the possible zeros, using the Rational Zero Theorem. If we test x=1 by synthetic division, we find that 1 is a zero and that x3-2x2-5x+6=(x-1)(x2-x-6)=(x-1)(x+2)(x-3). So, the zeros are 1,-2, and 3. After testing points between -2 and 1, and between 1 and 3 we find that the polynomial is positive in the interval (-2,1) and negative in the interval (1,3).
Since the graph is both above and below the x-axis, we need to use absolute value to calculate the area.
A=∫3-2|x3-2x2-5x+6| dx=∫1-2(x3-2x2-5x+6) dx +∫31(-x^3+2x2+5x-6) dx.
This comes out to be 253/12.