definite intregral?

Question

Determine the definite integral that represents the area of the region bounded by the graphs of x=0, y=0 , and the graph of y=8-(x-2)^3 .

Answer 1

We need to find out what the graph of y = 8 - (x - 2)^3 looks like, so that we can figure out where the bounds are of this graph. We can start by trying to find zeroes of the equation:

0 = 8 - (x - 2)^3
8 = (x - 2)^3
2 = x - 2
x = 4

This tells us that the graph hits the x-axis at x = 4, so our bounds for our integral are 0 <= x <= 4.

As for setting up the integral, we need to know where the rest of the curve between 0 and 4 are. We find the area by subtracting the bottom bound from the top bound. In our case, one bound is the given function and the other bounds is x = 0. If the curve is positive from 0 to 4, then the integral is simply of the given function (given - 0). But if it is negative, then we need to take the integral of the opposite of the function (0 - given). (Graphing this may help visualize this concept more.)

An easy way to see what the function is doing between 0 and 4 is to plug in a test number. 1 is easy, so we'll try that:

y = 8 - (x - 2)^3
y = 8 - (1 - 2)^3
y = 8 - (-1)^3
y = 8 - (-1)
y = 9

So we find that the function is positive for our given interval, so our definite integral is (I'm using S as the integral symbol):

A = S [8 - (x - 2)^3] dx, bounded at x = 0 and x = 4