A curve has equation y = 10 + 8x + x^2 - x^3. X >= 0

Question

a)Find the coordinates of the turning point. and show whether it is maximum or minimum.

b) Hence, Find the area of the region bounded by the curve, the line y = 11x and the y axis?

I manage to complete a) which I got :

dy/dx = 8 + 2x - 3x^2

dy/dx = 0

(-3x^2 + 2x + 8) x-1 = 0
= 3x^2 - 2x -8 = 0
(3x + 4) ( x - 2) = 0
x = -4/3 or x = 2

since x >= 0 I chose 2
x = 2
y = 10 + 8(2) + (2)^2 - (2)^3
y = 22

so turning point is (2, 22)

//finding max or min :

f(-4/3) = 10 + 8(-4/3) + (-4/3)^2 - (-4/3)^3
= -50/27, f(-4/3) < 0, max

f(2) = 10 + 8(2) + (2)^2 + (2)^3
=22, f(2) > 0, min

The part I am stuck is part b) I don't quite sure how to solve it.Please can you help me with this part? Thank you !

Answer 1

Find where the curves intersect. Clearly, at x=2, since y(2) = 22 = 11*2

I was wondering about the "hence".

So, the area in question has two parts. That under the line and that under the curve:

a = ∫[0,2] 11x dx + ∫[2,3.7988] 10+8x+x^2-x^3 dx
= 22 + 27.2556
= 49.2556

I'm surprised they did not come up with an integer zero for y.