Find the area bounded by the curves: y = x^ 2 and y = x + 12.?

Answer 1

Intersect when:-
x² = x + 12
x² - x - 12 = 0
(x - 4)(x + 3) = 0
x = 4 , x = - 3
A = ∫ (x + 12) - x ² dx between lims of - 3 and 4
A = x²/2 + 12x - x³ / 3
A = (8 + 48 - 64/3) - (9/2 - 36 + 9)
A = 56 - 64/3 - 9/2 + 27
A = 83 - 128/6 - 27/6
A = 83 - 155/6
A = 498/6 - 155/6
A = 343 / 6 units ²

Answer 2

Find the domain of the function by setting the two functions equal to eachother:

x^2 = x + 12
x^2 - x -12 = 0
(x - 4) * (x + 3) = 0
x = 4
x = -3

Subtract the top curve from the bottom curve and integrate between x=4 and x=-3:

INT[x^2 - x -12] on the interval of -3 to 4

Answer 3

x^2 = x + 12
x^2 - x - 12 = 0
(x-4)(x+3) = 0
x = 4, -3

Integral (-3 to 4) x^2 dx =
(1/3)(x^3) from -3 to 4
64/3 - (1/3)(-27) =
64/3 + 27/3 =
91/3

Integral (-3 to 4) x + 12 dx =
(1/2)(x^2) + 12x from -3 to 4
(8 + 48) - (9/2 - 36) =
7/2 + 84

(7/2 + 84) - (91/3) =
343 / 6

Answer 4

http://archives.math.utk.edu/visual.calculus/5/area2curves.3/

this is realy good to teach how to find the area between 2 curves.

Answer 5

y=x+12 isnt a curve?