Calculus Help: Finding the Area?

Question

Find the area bounded by graphs of the indicated equations over the given interval

y= x^2 - 20 ; y = 0 ; -3 < or = x which is < or = 0

can anyone help me with this? i know that x^2 is a parabola. but i'm stuck

Answer 1

y = x² - 20 is symmetrical about y axis.
Cuts x axis at x = ± √20 = ± 2√5

A = ∫ x ² - 20 dx between - 3 and 0
A = [ x ³ / 3 - 20 x ]
A = (0 - (- 9 + 60 ) )
A = - 51 units²

The - ve sign denotes that area lies below x axis.

Answer 2

integrate the function with the limits from -3 to 0 for x
(do you see that since it is a parabola, the limits could be 0 to 3 as well)
∫(x² - 20) dx = (1/3)x³ - 20x
evaluate from -3 to 0
0 - [(-9) + 60] = - 51....since area cannot be negative
area = 51 square units

Answer 3

For - 3 <= x <= 0, y < 0. So the required area A is:

A = - integral(-3-->0) (x^2 - 20) dx
= - [(x^3)/3 - 20x](-3-->0)
= - (0 - (((-3)^3)/3 - 20(-3)))
= 51 square units.