# How do you find the total area between the curve f(x)=cos x and the x-axis on the interval [0,2pi ]?

Feb 4, 2015

The answer is: $A = 4$

There two possibily questions about the definitive integral.

The first is simply the calculus of an integral without any meaning of "area". So the definition of definite integral has to be used:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$.

BUT if the request is expressly an area, we have to be careful, because if the graph of the function, or part of it, lies under the x-axis, some areas would be consider, from the "engine" of integral, as negative!

So, before making the integral we have to make the graph of the function, that, in this case, is:

graph{cosx [-1, 7, -5, 5]}.

We can notice that in $\left(0 , \frac{\pi}{2}\right)$and in $\left(\frac{3}{2} \pi , 2 \pi\right)$ the graph lies over the x-axis, but in $\left(\frac{\pi}{2} , \frac{3}{2} \pi\right)$, the graph is under.

So, to calculate the definite integral, we have to do:

$A = {\int}_{0}^{\frac{\pi}{2}} \cos x \mathrm{dx} - {\int}_{\frac{\pi}{2}}^{\frac{3}{2} \pi} \cos x \mathrm{dx} + {\int}_{\frac{3}{2} \pi}^{2 \pi} \cos x \mathrm{dx}$

or, remembering the property: ${\int}_{a}^{b} = - {\int}_{b}^{a} \Rightarrow$

$A = {\int}_{0}^{\frac{\pi}{2}} \cos x \mathrm{dx} + {\int}_{\frac{3}{2} \pi}^{\frac{\pi}{2}} \cos x \mathrm{dx} + {\int}_{\frac{3}{2} \pi}^{2 \pi} \cos x \mathrm{dx} \Rightarrow$

$A = {\left[\sin x\right]}_{0}^{\frac{\pi}{2}} + {\left[\sin x\right]}_{\frac{3}{2} \pi}^{\frac{\pi}{2}} + {\left[\sin x\right]}_{\frac{3}{2} \pi}^{2 \pi} =$

$= \sin \left(\frac{\pi}{2}\right) - \sin 0 + \sin \left(\frac{\pi}{2}\right) - \sin \left(\frac{3}{2} \pi\right) + \sin \left(2 \pi\right) - \sin \left(\frac{3}{2} \pi\right) =$

$= 1 - 0 + 1 - \left(- 1\right) + 0 - \left(- 1\right) = 4$.