# Area for cosx on interval [0,2pi]

1. Apr 22, 2010

### tjohn101

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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"?

2. Apr 22, 2010

### Stratosphere

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

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

now you solve it.

3. Apr 22, 2010

### 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?

4. Apr 22, 2010

### Stratosphere

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

5. Apr 22, 2010

### 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.