00:01
In this question, where is to find the area enclosed by the graphs of f and g? to do that, we need to calculate the integral from a to b with the absolute value of f of x minus g of x x x.
00:19
Now we need to study the sign of the difference f of x minus g of x.
00:26
F of x minus g of x equals to 3x cube, minus x squared, minus 10x, minus negative x squared, plus 2x.
00:44
This equals to 3x cube.
00:47
Negative x squared is cancelled.
00:49
So we're going to get minus 2x, minus 10 x minus 2x is going to be minus 12x.
00:55
After factoring out 3x, we are going to get 3x times x squared minus 4.
01:01
And we can write this as 3x times x minus 2 times x plus 2.
01:10
So you can see that this difference has zeros at x equals 0, x equals 2, and x equals negative 2.
01:26
Now what happens when x is greater than 2? when x is greater than 2, everything is positive.
01:35
So the difference is going to be positive when x is greater than 2.
01:42
Next, between 0 and 2, this difference is going to be negative.
01:49
For example, plug in x equals 1.
01:52
Then from negative 2 to 0, it's going to be positive.
01:56
And then it's going to be negative again.
02:00
However, the part we're interested in is from negative 2 to 2 because that's where our region is enclosed, right? our region is enclosed between x equals negative 2 and x equals 2.
02:16
So a is going to be negative 2 and b is going to be 2 in this formula for the area.
02:23
Now, the difference is positive between negative 2 and 0 therefore, we can rewrite this integral as the integral from negative 2 to 0, f of x minus g of x dx.
02:37
We can open absolute value sign as just f of x minus g of x, because the difference is positive there, here in this interval...