What is the area of the region completely bounded by the curve y=-x^2+x+6?

Firstly, factorise the equation:
y = -1 (x² - x - 6)
y = -1 (x - 3)(x + 2)
From this, we can tell the x-axis intercepts are -2 and 3.
(To do this, simply equate any of the expressions involving x to 0 and rearrange to give x).

Now, integrate between the limits -2 and 3:
R = area of region bounded
R = ³∫₋₂ -x² + x + 6 .dx
R = ³[-1/3x³ + 1/2x² + 6x]₋₂
R = (-1/3(3)³ + 1/2(3)² + 6(3)) - (-1/3(-2)³ + 1/2(-2)² +6(-2))
R = (-9 + 9/2 + 18) - (8/3 + 2 - 12)
R = (27/2) - (-22/3)
R = 27/2 + 22/3 = 125/6 units²