Area between two curves

May 2010
4
0
For this problem there are two functions
f(x) = cos x
g(x) = 2-cos x
where 0<x<2π
I know this is probably fairly simple, but we just started learning this stuff
I know the equation ∫f(x) - g(x) or ∫top-bottom curve, but if you graph it, you can see that it's not that simple (I think)
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
For this problem there are two functions
f(x) = cos x
g(x) = 2-cos x
where 0<x<2π
I know this is probably fairly simple, but we just started learning this stuff
I know the equation ∫f(x) - g(x) or ∫top-bottom curve, but if you graph it, you can see that it's not that simple (I think)
yes, it's that simple ...

\(\displaystyle A = \int_0^{2\pi} (2-\cos{x}) - \cos{x} \, dx\)
 
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Reactions: Breadordecide
May 2010
4
0
Thank you! Yes, I tried that, but the answer I got (plugging into 2x - 2sinx) was 4π, a really big number. It just didn't seem right to me
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Thank you! Yes, I tried that, but the answer I got (plugging into 2x - 2sinx) was 4π, a really big number. It just didn't seem right to me
Not that big a number. When \(\displaystyle x= \pi\), cos(x)= -1 so 2- cos(x) is 3 and cos(x)= -1. If you put a rectangle about that figure, it would have height 4 and width \(\displaystyle 2\pi\), so area \(\displaystyle 8\pi\). It looks reasonable that the area between the two curves is half the area of the rectangle containing them.