Calculating area between two zero points I have the following function $$y=\cos(2x-1)$$ and I am supposed to find the area between two zero points. 

I have calculated the first zero points which is $$x_1=0.285$$, but I do not know how to calculate the second point, $$x_2=?$$
If by ,,zero point'' you mean $x$ such that $f(x)=\cos{(2x-1)}=0$, then:

We know that $\cos(t)=t$ for $t=k\frac{\pi}{2}$, where $k$ is odd. And so we need $2x-1=k\frac{\pi}{2}$, and from this $x=k\frac{\pi}{4}+\frac{1}{2}$. You'd like to calculate the area between 
$$x_1=\frac{\pi}{4}+\frac{1}{2}\text{ and }3\frac{\pi}{4}+\frac{1}{2}$$.

Then we have that area: 
$$A=\int_{x_1}^{x_2}\cos{(2x-1)}\,\mathrm{d}x $$

You can for example substitute $u=2x-1$, then $du=2dx$ and $dx=\frac{du}{2}$ and our integral becomes:

$$ \frac{1}{2}\int_{x_1}^{x_2}\cos{u}du=\frac{1}{2}\left[\sin{u}\right]_{x_1}^{x_2}=\frac{1}{2}\left[\sin{(2x-1)}\right]_{x_1}^{x_2}$$

and so:
$$A=\frac{1}{2}\left(\sin(\frac{\pi}{2})-\sin(3\frac{\pi}{2})\right)=\frac{1}{2}(1-(-1))=\frac{1}{2}\cdot 2=1$$.
Through this MATLAB commands:

evalin(symengine,'solve(cos(2*x-1),x)')


you find all the roots:

Dom::ImageSet(pi/4 + (pi*k)/2 + 1/2, k, Z_)


$$
x=\frac{\pi}{4}+k\frac{\pi}{2}+1/2, k \in \mathbf{Z}
$$

Over the $cos$ function, you only then need to integrate the first two of them for getting the area:

eval(int(cos(2*x-1),pi/4+1/2,pi/4+pi/2+1/2))


Resulting in:

A=1