Area enclosed by both curves
Find the Point of intersections of both graphs
y=x^2,y=x+6
x^2=x+6
x^2 -x -6 =0
(x + 2) , (x - 3)
x=-2 , x=3
3
∫ (x+6) - (x^2) dx Area under y=x+6 is larger than area under y=x^2
-2
3
∫ x+6 - x^2 dx
-2
3
= (x^2)/2 +6x - (x^3)/3 ]
-2
= (3^2)/2 +6(3) - (3^3)/3 - [ ((-2)^2)/2 +6(-2) - ((-2)^3)/3 ]
= 27/2 - ( - 22/3 )
= 125/6 u^2
Is that right? maybe i did something wrong :/