Find the area of the region enclosed by the curves y=x+9, y=22-x^2, x=-1, and x=2, accurate to two decimal places.
First thing's first: draw the graphs. This will let you know which graph is above which. The area of the region enclosed will be the area under the higher graph minus the area under the lower graph. We take the integral between the limits given, that is x=-1 to x=2.
Plotting the graphs, you will realize that the 22 - x^2 graph is the higher one. So the area between the enclosed region will be:
int{22 - x^2 - (x + 9)}dx evaluated between x=-1 and x=2
= int{-x^2 - x + 13}dx
= -(x^3)/3 - (x^2)/2 + 13x + C
so the area is:
A = [-(2^3)/3 - (2^2)/2 + 13(2)] - [-(-1^3)/3 - (-1^2)/2 + 13(-1)] ...Note, we dont have to write the C's in here because they'll cancel anyway.
=> A = -8/3 - 4/2 + 26 - (1/3 - 1/2 - 13)
= -8/3 - 2 + 26 - 1/3 - 1/2 - 13
= 15/2 = 7.50 squared units
The method is right, but double check my computation, 'cause i'm half asleep right now