00:01
In this question, we are asked to find the area of the region under the graph of f over the interval from negative 2 to 5.
00:08
By the formula, by the area formula, area is equal to the interval from negative 2 to 5 of f of x d x.
00:18
However, the problem is, in our case, f of x is a piecewise defined function.
00:24
But that's not a big deal because according to the definition of f, we can rewrite the interval, as a sum of two interals.
00:35
One integral is going to be the interval from negative 2 to 3 of f of x d x plus the integral from 3 to 5 of f of x d x.
00:52
We can do that by using the properties of interals.
00:55
The reason we chose the middle point to be 3, because that's where our function changes its definition at x equals 3.
01:04
Now, on each of the sub -intervals, f of x is defined uniquely.
01:12
So, for example, on the interval from negative 2 to 3, x is less equal than 3, which means we can replace a function f by x squared plus 6.
01:23
So the interval from negative 2 to 3 of f becomes the interval from negative 2 to 3 of x squared plus 6, plus the interval from 3 to 5 of f of x d x d x...