Using the Fundamental Theorem of Calculus find the area of the region bounded by the x-axis and the graph of f(x)=−x^2+3x+10.

Answer 1

A quick sketch will show a downwards opening parabola with
x - intercepts of -2 and 6

So you simply need
∫ (−x^2+3x+10) dx from -2 to 6
= [ -(1/3)x^3 + (3/2)x^2 + 10x] from -2 to 6
= ...

you do the arithmetic

Answer 2

f(x) = -(x+2)(x-5)
so the area is
∫[-2,5] (-x^2+3x+10) dx = 343/6