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