Question

The curve C has equation y = -x^3 +2x^2 +8x. the curve crosses the x-axis at the origin O and at points A and B. (see question? below)

a) using an appropriate algebraic method, find the coordinates of A and B.
b the finite region is bounded by the curve C and the x axis. use calculus to find the total area of the region.

Answer 1

a) (-2, 0) & (4, 0)
b) 148/3

Explanation:

First thing is first: let's factor this polynomial! Since this is a cubic, we expect three roots. Zero is already given to us, so we can factor that out
`y = -x(x^2 - 2x -8)`
And we notice that the second part is a quadratic so we can either use the quadratic formula or just factor by hand. I'm using rational roots theorem and Descartes rule of signs mentally to get
`y = -x(x-4)(x+2)`

So we know that A and B are at (-2, 0) and (4, 0).

Now we want to find out the area enclosed. Note: we can't just do an integral because a negative area doesn't make sense in this context. We can express the integral as
`int_-2^4 \ |y|\ dx`
but we need to find out where it is negative and where it is positive.

We have two sections: -4 to 0 and 0 to 2. We can plug in any values within these ranges and find their sign or we can think about the values at `pm infty`. Because of that negative sign and the fact that it is a cubic function, the function will be positive on the left and negative on the right. Since it crosses three times, that means it is negative from -4 to 0 and positive from 0 to 2.

Therefore, we can integrate using power rule:
#int_-2^4\ |y|\ dx = -int_-2^0\ y\ dx + int_0^4 ydx =[1/4 x^4 - 2/3 x^3 - 4x^2]_-2^0 - [1/4 x^4 - 2/3 x^3 - 4x^2]_0^4#
`=-[1/4 2^4 + 2/3 2^3 - 4(2)^2] - [1/4 4^4 - 2/3 4^3 - 4 (4)^2]`
`= -(4 - 16 + 16/3) - (4^3 - 4^3 - 2^7/3)`
`= (12 - 16/3) + 128/3 = 148/3`