00:01
So we need to sketch the region enclosed by those voluering curves.
00:07
Say this is our xy plane.
00:12
And so the first function is cosine pi x.
00:19
That means x equals to 1, cosine pi equals to negative 1.
00:30
And when x equals to 1 1⁄2, cosine px equal to 0.
00:37
Then x equals 0, cosine 0 is pi, cosine 0 is 1.
00:49
This function should look like, and this is an even function, so it's a metric by y -axis.
00:59
It looks like this, here is x equals 1, this is negative 1.
01:16
This is my x equals negative 1, and y equals to negative 1.
01:20
And the second curve is a problem.
01:25
No, x equals 0, this curve across y equals to nikev 1.
01:32
So when x0, y equals 9 k1, across this point.
01:38
And when y equal to 0, x equals to positive and nai kiv 1 1⁄2.
01:44
So it crossed those 2 points.
01:47
This is 9 kiv 1 half.
01:51
So this problem looks like this, and the enclosed region is in between, this shaded area.
02:13
So the order to find this area, we write on an integral form.
02:20
The area will equal to integral with respect to we choose x because those functions can be represented by x explicitly.
02:29
So this is a equals to integral with respect, sorry, with respect to x.
02:39
And x goes from negative one half to positive one half.
02:48
And here is the upper curve minus the lower curve.
02:54
It's cosine x minus 4x square minus 1.
03:06
So it will be minus 4x squared plus 1...