Finding Areas of Regions Bounded by Trig Functions Using Integrals

[calchelpneeded9]

Homework Statement

Find the exact total of the areas bounded by the following functions:

f(x) = sinx

g(x) = cosx

x = 0

x = 2pi

Homework Equations

the integral of (top equation - bottom equation)

The Attempt at a Solution

Change the window on the graphing calculator to an x scale of 2pi?
I'm having trouble finding where x = 2pi.

the integral of (cosx - sinx)
to me, cosine appears to be the top equation

Take the anti-derivative of them both---> (-sinx - cosx) (if i remember correctly)

Finding the bounds for the integral is what I'm having problems with. Any hints or suggestions would help me out. This is the last stretch of my ap calculus class, and I guess my brain is starting to fizz out!

Thank you everyone!

 

[Dick]

If you've drawn the picture as you should, it ought to be clear that the integration limits need to be set where the curves cross. So at how many points do cos(x) and sin(x) cross between 0 and 2pi. Hint: sin(x)=cos(x) -> sin(x)/cos(x)=tan(x)=1.

 

[Gib Z]

Homework Equations

the integral of (top equation - bottom equation)

What Dick was saying is that you are right, it has to be the integral of the function on top minus the function on bottom. However, through 0 to 2pi, is cosine always on top?

 

[calchelpneeded9]

thank you! ok...

From what I can tell, there appears to be three intersections of the two functions. It looks like cos is on top for the first one, and sin on the second, but I'm not sure about the third one.

Once I know that, would I then subtract each pair of sin/cos intersections and then take the anti derivatives of them?

This would then give me the equation I plug in 2pi, find a value, and subtract the value I found using 0 from it? (top - bottom)

Thanks.. I'm still a little stumped!

 

[Gib Z]

Ok well now that you know which function is on top for each intersection, you need to find the area for each region separately.

Say for the first region before they intersect, cos is on top. The you have cos x- sin x. Find the anti derivative and then sub in the bounds of the integral as you normally do. Once you have done this for the first section, do the same for the next.

 

[calchelpneeded9]

ok

Thank you for all your help!

what I have now is:

the integral from 0-(pi/4) of: cosx-sinx +

the integral from (pi/4)-(5pi/4) of: sinx-cosx +

the integral from (5pi/4)-(2pi) of: cosx-sinx

I need the exact area..so I thought I would get the same values (for the first integral) by using unit circle values and using FnInt.

Unfortunately, I got different answers for FnInt and using:

[cos (sqrt2/2)- sin (sqrt2/2)] - 1

I thought I was supposed to get an answer in terms of pi?

 

[Gib Z]

Nope, why would you think that! The anti derivatives, sin and cos, are sort of pi eliminators! Sub in a pi thing, get back 1/sqrt 2 or 1 or 0, no more pi! The area won't be in terms of pi!

 

[calchelpneeded9]

ok

so just by adding the values returned by those 3 integrals and not rounding, I should have my exact answer! thank you. :)