00:01
So in this question, they say i want to use a definite interval to find the area between the x -axis and f -of -x over the indicated interval.
00:10
I'm going to have to check first to see if the graph crosses the x -axis in the given interval.
00:17
Here i have f -of -x equals 2 times e to the x -power minus 1, and i'm on the closed interval from negative 2 to 5.
00:26
First, i'm going to check to see if we cross the x -axis in the interval.
00:30
To do that, i'm setting my y equal to 0.
00:34
I'm setting 2 e to the x minus 1 equal to 0.
00:42
2 e to the x equals 1, e to the x equals 1 1ā2 .0.
00:49
Now i am getting x equals the natural log of a half.
00:53
And so we do cross the x -axis in our interval.
00:58
To the left of ln of 1 1 -half, f of x is negative.
01:03
To the right of x equals ln of 1 half, f of x is positive.
01:10
So first, i'm going to need an integral from negative 2 to the ln of 1 half of 2 e to the x minus 1 dx.
01:23
So let's see, my antiderivative is 2e to the x minus x.
01:30
This gets evaluated on the interval from negative 2 to the l.
01:36
Of one half.
01:39
I plug in my l .n of a half.
01:42
I'm getting two times e to the power of ln of a half minus the ln of a half.
01:50
And from this, i subtract what i get when i plug in negative two.
01:54
2 e to the negative second minus negative 2 is plus 2...