Find the area between a parabola y=-4 x^2 + 0 x +36 and the x-axis.?

Answer 1

Assume what is meant is:-
y = - 4x² + 10x + 36
Cuts x axis when y = 0
- 4x² + 10x + 36 = 0
4x² - 10x - 36 = 0
2 x ² - 5 x - 18 = 0
( 2 x - 9 ) ( x + 2 ) = 0
x = 4.5 , x = - 2 (limits of integration)
A = ∫ - 4 x ² + 10x + 36 dx
A = - 4 x ³ / 3 + 5 x ² + 36 x
A = [ -121.5 + 101.25 + 162 ] - [ - 16 - 20 - 72 ]
A = 141.75 - 108
A = 33.75 units ²

Answer 2

y = -4x^2 + 36
When y= 0, -4x^2 + 36 = 0, so x = ± 3. These are the zeros.

To find the area under y between x = -3 and x = 3, find the
integral of y :

= Integral (from -3 to 3) of (-4x^2 + 36) dx
= -4x^3/3 + 36x (from -3 to 3)
= [-4*3^3/3 + 36*3] - [-4(-3)^3/3 + 36(-3)
= [-36 + 108] - [36 - 108]
= -72 + 216
= 144
EDIT: OK, the other answerer's evaluation of the equation
does seem more logical than mine.