Find the area between the curves:?

Question

Find the area between the curves:
y= x^3 - 10x^2 +16x
y= -x^3 + 10x^2 -16x

I found that y = 0, 8, and 2, but after that I'm lost

Answer 1

Set them equal to each other and solve for x
2x^3-20x^2+32x=0
2x(x^2-10x+16)=0
2x(x-8)(x-2)=0

x=0,2,8 (not y)

Now integrate from 0 to 2 and then integrate from 2 to 8.

integral 0 to 2 of
(x^3-10x^2+16x) - (-x^3+10x^2-16x) dx

integral 0 to 2 of 2x^3-20x^2+32x dx

Do the same for 2 to 8

If you get a negative value, then take the absolute value of the integral and then add up the two parts.

Answer 2

First thing to point out is that the two curves are the negatives of each other. That means that the figure is symmetrical in the x-axis. For every point above the y-axis on the first curve, there's a point an equal distance below the y-axis on the second curve.

You're right that the roots are 0, 2, and 8, only it should be x, not y, that is equal to 0, 2, and 8. That means that the first curve crosses the x-axis at x=0, x=2, and x=8. (At x=0 it's heading upward, at x=2 it's heading downward, and at x=8 it's heading upward again, although the problem would work out the same if it were the other way around.) The second curve is symmetrical with the first, remember, so it crosses the x-axis at the same three points. If you can visualize this, the area between the curves will form two lens shapes with their corners on the x-axis, one between x=0 and x=2, and the other between x=2 and x=8.

To compute the area between two curves, you integrate the function which represents the higher curve minus the lower one (higher means greater y-values). From x=0 to x=2, the first curve is higher; from x=2 to x=8, the second curve is higher. So you have to do it in two pieces.

From x=0 to x=2, the higher curve minus the lower is the first one minus the second, which is

(x^3 - 10x^2 +16x) - (-x^3 + 10x^2 -16x) = 2x^3 - 20x^2 + 32x.

The integral of that is 2x^4/4 - 20x^3/3 + 32x^2/2 = x^4/2 - (20/3)x^3 + 16x^2. To take the definite integral from x=0 to x=2, replace x by 2 in that formula and get a number; replace x by 0 and get a second number; subtract the second from the first. That's the area of the first lens-shaped figure. The answer I get is 18 2/3.

From x=2 to x=8, the second curve is higher than the first, so subtract in the opposite direction. It just makes the sign change. The integral will be -x^4/2 + (20/3)x^3 - 16x^2. Replace x by 8 and get a number; replace x by 2 and get another number; then subtract the second from the first. That's the area of the second lens-shaped figure.

Add the two areas to get the final answer.

Answer 3

If you don't have some bounds on that, it's going to be infinite.