hey totally stuck on this problem!
Find the area of the region that is enclosed by the horizontal line y=3 and the parabola y=x^2-1?
help would be mucho appreciated!
(don't know how to put in squared..new to this!)
First, look at the graph:
Identify the endpoints of the region (I'll call them a and b) and which curve is "above" and which is "below". (I'll call them T(x) for top and B(x) for bottom.)
Then we set up the integral:
$\displaystyle \int_a^b \left[T(x) - B(x)\right] dx = \int_{-2}^2 \left[3 - (x^2-1)\right] dx = \int_{-2}^2 (4 - x^2) \, dx$
Do you know how to solve this?