00:01
The question asked to find the centroid of the region bounded by the graphs of y equals x plus cosine x and y equals zero for zero for x is greater than or equal to zero and less than or equal to two pi.
00:21
Essentially what we're doing is finding the centroid between lines f of x and g of x, which is the center between them.
00:33
And to find that is by looking for the coordinates in terms of x bar and y bar.
00:45
So what we can do first is affine the area between f of x and g of x between boundaries a through b.
00:58
So the formula for that is a equals the integral from a to b of f of x minus g of x, d x.
01:11
And using the area, we can find x bar and y bar.
01:16
The formula for x bar would be one over the area times the integral from a to b of x times f of x minus g of x d x and for y bar we'll have one over area times the integral from a to b of one half times f of x squared minus g of x squared d we can first start by finding the area let's just separate this.
02:04
We can find the area by putting in the boundaries.
02:10
Bineries is from 0 to 2 pi.
02:15
And plugging in f of x and g of x.
02:19
F of x in this case would be x plus cosine x and g of x would be y equals 0.
02:28
So we'll have minus 0 d x.
02:32
Simplifying that, we'll just have zero, the integral from 0 to pi of x plus cosine x d x.
02:42
And then integrating this, we'll have x squared over 2 minus cosine x from 0 to pi.
02:55
So now we'll have 2 pi squared over 2 minus sine of 2 pine.
03:05
Minus 0 squared over 2 minus sign of 0.
03:14
2 pi squared over 2 would just be, oops, would just be 4 pi squared over 2.
03:28
Sign 2 pi would just be 0 over 2.
03:33
Would just be 0 and sign of 0 would also just be 0.
03:40
In the end we'll just have 4 pi squared over 2 or 2 pi squared as our area now that we know our area we can find what x bar and y bar is let's start with x bar so let's just let's just delete this for a little more space and put a sign out a equals to find x bar we'll have 1 over the area, which is 1 over 2 pi squared, times the integral from a to b, or an integral from 0 to 2 pi, times x times f of x, which is x plus cosine x minus g of x or 0.
04:42
But since it's 0, i'm not going to include it.
04:46
And then we're just going to simplify this.
04:49
We'll have 1 over 2x squared times the integral from 0 to 2 pi of x squared, plus x cosine x d x.
05:03
We know the integral of x squared, which is just x q over 3, but to take the integral of x cosine x, we have to use integration by parts.
05:22
So we'll have u b equal x and dv equals cosine x d x.
05:30
So this means v has to be sine x and d u has to be 1 d x.
05:38
So putting it all together, we'll have the integral of x cosine x x cosine x equals x sine x minus the integral of sine x v x.
05:55
And i just want to remind you that where i got this equation from, is from the integration by parts equation, which is u times d u times d equals u v minus the integral of b d u.
06:15
So going back, we'll then have x sine x minus the integral of x d x would just would just be negative cosine x but then the negative times the negative is a positive so we'll have sine x plus cosine so going back to x bar we'll then have one over two pi squared times integral of x squared which is x q or three plus the integral of x cosine x which we just found was x sine x plus so now we'll just subtract the 2.
07:14
We'll have 1 times 2 pi squared times 2 pi cube over 3 plus 2 pi sine 2 pi plus cosine 2 pi plus cosine 2 pi minus, well just 0 over 3 plus 0 plus 0 plus .0 times anything would just be 0, plus.
07:41
Cosine of 0.
07:44
So 2 pi cube over 3 would just then be 8 pi cubed over 3.
07:52
Um, sign, oops, this would be 2 pi.
07:56
Uh, 2 pi sine 2 pi, sign of 2 pi would be 0.
08:00
So this would be 0.
08:02
So this would be 0 and cosine 2 pi will be 1.
08:05
This would be 0, this is 0, cosine of 0 is 1.
08:11
So simplifying it.
08:13
Further, we'll have 8 pi cubed over 3 plus 1 minus 1.
08:22
So this will end up canceling out each other.
08:27
And now we can just multiply 1 over 2 pi by 8 pi cube over 3, which is just 8 pi cubed over 6 pi squared.
08:44
We can simplify this, this would be gone, this would just be pi, this would just be three, this would be four, so x bar would just be four -thirds pi.
09:00
So i will just make a note of that x -bar would be four -thirds pi.
09:09
Now that we found a and x -bar, we can find y bar.
09:14
So i'll just erase.
09:18
Yes.
09:31
Y bar.
09:32
All right...