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# Suppose that $0 < c < \frac{\pi}{2}$. For what value of $c$ is the area of the region enclosed by the curves $y = \cos x$, $y = \cos (x - c)$, and $x = 0$ equal to the area of the region enclosed by the curves $y = \cos (x - c)$, $x = \pi$, and $y = 0$.

## $\sin \frac{c}{2}=\frac{1}{2}=>\quad c=\frac{\pi}{3}$

#### Topics

Applications of Integration

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##### Catherine R.

Missouri State University

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

this problem is a little bit different from a more straightforward problem where we're finding just the area under a curve in this problem were given to functions and we have defined it. What value of C would the area be enclosed by both of these functions? So the functions were given is why equals cosine X and why equal to casa co sin of X minus c. So the first thing that I did was I graphed it so we can visualize where these curves are moving and where they're hitting on the X and Y axis. So now what we can do is we can see from this graph that the function co sign of X minus C crosses the X axis where X equals pi over two plus c. Now, this is very helpful because now we can use that to construct an integral because, remember, we're using our techniques of integration to find this value of C so we would get the integral from zero to see over to of coastline of X, minus the coastline of X minus C in de X. So then we can do a little bit of substitution to find that this integral is the same thing as saying the negative integral of pi over two, plus C two c of co sin of X minus C and D x. Then what we would do is we would simply take the anti derivative and we would get sign of X minus sign of X minus C and will evaluate that from zero to see over to. So when we do that, we'll find that this will be equivalent to negative sign of X minus C evaluated from pi over two plus C two pi and we can plug in each of these values and you'll do B minus a. And when we simplify that a little bit will find that this is equivalent to negative sign of pi minus C plus sign of pi over two plus C minus E. And for this problem, I think it's very important for us to keep our parentheses or brackets organized so we don't mess up any of the terms. Now we can use a trick identity. This trick identity says that two times the sign of C over to minus sign of C is equivalent to negative sign of C plus one so we can use as a substitution. So what we can say is that the sign of C over two is equivalent toe one half. So where are the unit circle with this statement? Be true. Well, that's when C is pi over three. So we just found the value of see where our functions are enclosed and we confined the area. So our value is C is pi over three. Let's look back at our graph very quickly and we can see that that is true based on of this graph. So I hope that this helped you understand how we confined that value of C given to functions that enclose an area and how we can specifically find that see using our techniques of integration.

University of Denver

#### Topics

Applications of Integration

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