Maths

Finding out A first:

-2x = x2 - x - 20

x2 + x - 20 = 0

(x + 5)(x - 4) = 0

x = 4 (Since A is on the right of the y-axis)

y = -8

Also B can be found by solving x2 - x - 20 = 0

(x - 5)(x + 4) = 0

x = 5 (Since B is on the right of the y-axis)

This, the triangle's base and height are 5 and 8 resp.

Area = 20

You may find out the area of ∆AOB by the following ways:

Method 1: Find the coordinate of A first

At point A,
y = -2x
y = x^2 - x - 20

-2x = x^2 - x - 20
x^2 + x - 20 = 0
(x - 4)(x + 5) = 0
x = 4 or -5 (rejected, as x > 0 in the figure)

So, y = -2(4) = -8

A = (4, -8)

At point B,
y = x^2 - x - 20

y = 0
So, x^2 - x - 20 = 0
(x - 5)(x + 4) = 0
x = -4(rejected, as x>0) or 5

B = (5,0)

OB = 5,
height of triangle = vertical distance between A and the x-axis
= 0-(-8) = 8

Area = 1/2 * 5 * 8 = 20 units^2


Method 2

Area = 1/2 * |(0 0) (4 -8) (5 0) (0 0)|
= 1/2 * | [(4)(0) + (5)(-8) + (0)(0)] - [(0)(-8) + (4)(0) + (5)(0)] |
= 1/2 * | (0 - 40 + 0) - (0 + 0 + 0) |
= 1/2 * |-40|
= 1/2 * 40
= 20 units^2



If the required area is not a triangle, but is the shape surrounded by OB, OA and the arc AB, then you may find out its area in this way:

Area = ∫[4 0] (y = -2x) dx + ∫[5 4] (y = x^2 - x - 20) dx
= [-2 (x^2 / 2)] [4 0] + [x^3 / 3 - x^2 / 2 - 20x] [5 4]
= [-x^2] [4 0] + [x^3 / 3 - x^2 / 2 - 20x] [5 4]
= (-16 - 0) + [(5^3 / 3 - 5^2 / 2 - 20*5) - (4^3 / 3 - 4^2 / 2 - 20*4)]
= -16 + (-25/6)

However, you would notice the area is negative, which is impossible. So, you will take its absolute value.

That is, area = |-16| + |-25/6|
= 16 + 25/6
= 121/6 units^2