Area for cosx on interval [0,2pi]

[tjohn101]

Homework Statement

f(x)=cosx and the x-axis on the interval [0,2pi]

A) Set up definite integral that represents area above
B) Find area using the fundamental theorem

Homework Equations

The Attempt at a Solution

cosxdx [0,2pi]
= sinx [0,2pi]
= sin(2pi)-sin(0)
= 0

Area= (cosxdx [0,pi/2]) - (cosxdx [pi/2,3pi/2]) + (cosxdx [3pi/2,2pi])

Area= (sinx [0,pi/2]) - (sinx [pi/2,3pi/2]) + (sinx [3pi/2,2pi])

Area= (1-0) - (-1-1) + (0-1)
Area= (1) -(-2) + (1)
Area= 4 square units.

So... What do you think? Is it right? And what exactly does the question mean by "Set up definite integral that represents area above"?

 

[Stratosphere]

When it says set up the definite integral it means to do this.

\int^{2\pi}_{0}cosx dx

now you solve it.

 

[tjohn101]

I did that, but I didn't know how to type it into the forums. So I made it the cosxdx [0,2pi] and solved from there. What do you think of my answer for the area?

 

[Stratosphere]

You have the integral correct, check this again though
sin(2\pi)-sin(0)

 

[LCKurtz]

As you have noted, the area in question is not given by

\int_0^{2\pi} \cos x\ dx

because that gives zero. That formula only works for non-negative integrands. To express it as a single integral you might write:

\int_0^{2\pi} |\cos x|\ dx

which you have properly evaluated by breaking it up to subtract the negative parts.