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January 26, 2015

January 26, 2015

Posted by **Sarah** on Saturday, April 21, 2007 at 6:30pm.

f(x) = sinx

g(x) = cosx

x = 0

x = 2pi

I set my calculator to graph on the x-axis as a 2pi scale. The two functions appear to cross three times between x = 0 and 2pi. (including 2pi) Now, from top to bottom, I can't distinguish which equation is which. I know to look from left to right, and it appears that the first intersection has cosine on top, and the second has sin on top, but I'm not sure about the intersection at 2pi. Further, I know this requires the use of values on the unit circle that I'm not entirely sure how to use.

After I've found each top-bottom equation and can add them all together to get the integral from 0-2pi OF (3 different sets of cos and sin subtractions) how do I determine the values? Will it be for quadrant 1, quadrant 4, then quadrant 1 again? (In order of intersection)

Thank you in advance. These problems have a tendency to get very involved and I lose myself in them.

You need to integrate the function

ABS[f(x) - g(x)] from zero to 2 pi.

Here ABS denotes the absolute value.

Note that:

cos(x) - sin(x) = sqrt(2)cos(x + pi/4)

Use the formula:

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

to derive this formula.

So, you have to integrate the function:

sqrt(2) ABS[cos(x + pi/4)]

from zero to 2 pi. Since the integral is over one period you can forget about the pi/4 and just integrate

sqrt(2) ABS[cos(x)] or perhaps slightly easier the function sqrt(2) ABS[sin(x)] from zero to 2 pi.

Integral of sqrt(2) ABS[sin(x)] dx

from zero to 2 pi is th same as twice the integral of sqrt(2) sin(x) from zero to pi.

Thank you so much for your help!

I'm having trouble putting that into a calculator using FnInt.

FnInt: (sqrt2) (ABS (cos(x))), x, 0, 2pi

Returns and error.

Is that the correct function etc? or am I missing something else?

Thanks again,

Sarah

Maybe you get the error because your calculator doesn't know how to integrate the absolute value of cosine.

Because we are integrating over an entire period, you can also integrate the absolute value of the sinus from zero to 2 pi, but that's twice the integral of sin(x) from zero to pi. The integral of sin(x) from zero to pi is 2. The answer is thus 4 sqrt(2).

If you sketch the two curves it is easy to see that they intersect at pi/4 and again at 5pi/4.

(Sarah, they don't intersect at 2pi because

sin(2pi)= 0, cos(2pi)=1 )

total area =

integral(cosx-sinx)dx from 0 to pi/4

+

integral(sinx-cosx)dx from pi/4 to 5pi/4

+ integral(cosx-sinx)dx from 5pi/4 to 2pi.

the integral of cosx-sinx is (sinx+cosx), the angles are all special angles and can be found without any calculators.

Thank you both again! Which values am I supposed to be using? I set up the integrals and FnInted the first one, which gave me a decimal. I figured I should get the same answer if I use the values from the unit circle for pi/4.

I had cos [(sqrt 2)/2]- sin [(sqrt 2)/2] - 1. This gave me a very different answer than the first decimal I got.

Thank you,

Sarah

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