Find the area between a parabola y+-x^2+x+30 and the x-axis.?

This is a simple integral calculus problem. Just integrate y= x^2 + x + 30, which I get x^3/3 +x^2/2 +30x +c, if I remember correctly.

Find the points of intersection of the parabola and x-axis:
-x^2 + x + 30 = 0, so x^2 - x - 30 = 0; thus, (x+5)(x-6) = 0, so -5 and six will suffice as lower and upper limits of integration.
The antiderivative of (-x^2 + x + 30), -(1/3)x^3 +(1/2)x^2 + 30x, evaluated via the fundamental theorem of calculus between -5 and 6, i.e. the area under the parabola and above the x-axis, is 1331/6.

I'll fix up some of the errors in your question.

Integral -x^2 + x + 30 dx = AREA
(-1/3)(x^3) + (1/2)(x^2) + 30x = AREA

y = -x^2 + x + 30

Find the points it intersect at the x-axis

-x^2 + x + 30 = 0
x^2 - x - 30 = 0
(x-6)(x+5) = 0

x = -5, 6

So we're integrating from -5 to 6.

I'll leave that for you to calculate. Just plug in.

integrate -x^2+x+30 between its x intercepts

0=-x^2+x+30 =-(x^2-x-30) = -(x-6)(x+5)
x interecepts = -5 ,6

integrating -x^2+x+30 you get (-1/3)x^3 + (1/2)x^2+30x
the defensite integral becomes

(-1/3)x^3 + (1/2)x^2+30x }x=-5 to 6

-[(-1/3)(-5)^3 + (1/2)(-5)^2+30(-5)] +[(-1/3)(6)^3 + (1/2)(6)^2+30(6)]
= evaluate yourself

there's a typo in your equation