Am I correct in interpreting the integral of this delta-function graph? I have $f(x)$ defined as $\delta(x)-\delta(x-\frac{1}{2})$ $\forall x\in(-\frac{1}{6},\frac{3}{4})$. Outside this boundary it is periodically repeating.
From the definition of the $\delta$-function, when $x=\frac{1}{2}$, $f(x)=-1$ and when $x=0$, $f(x)=1$. 
Visually, I interpret this as an upward blip to $\infty$ at $x=0$ and a downward blip to $-\infty$ at $x=\frac{1}{2}$. This repeats as we go along the x-axis. 
Is this a correct interpretation?
If so, I go to my next question. 
From this it can be seen that $$\int_{-1}^1f(x)dx=1$$ because between the points $x=-1$ and $x=1$ we have 3 upward blips at $x=-1,0,1$ and 2 downward blips at $x=-\frac{1}{2},\frac{1}{2}$. Since the area of each upward blip is 1 and the area of each downward blip is -1 we have the total area and hence the integral to be $3-2=1$.
However, if I split the integral into four parts my answer seems to be different: $$\int_{-1}^{1}f(x)dx=\int_{-1}^{-\frac{1}{2}}f(x)dx+\int_{-\frac{1}{2}}^{0}f(x)dx+\int_{0}^{\frac{1}{2}}f(x)dx+\int_{\frac{1}{2}}^{1}f(x)dx.$$
This would imply that for the first integral, the upward blip at $-1$ and the downward blip and $-\frac{1}{2}$ will cancel, making the integral zero. The same would then occur for all subsequent integrals. For example, for the second integral the downward blip at $-\frac{1}{2}$ and the upward blip at $0$ will cancel, rendering the integral zero. 
Obviously I have misunderstanding somewhere, since this doesn't make sense. Please don't explicitly answer my question, just give me hints as to where I am going wrong or where I have a misunderstanding. 

For your first question, the dirac delta function $\delta(x)$ is undefined at $x=0$, you can think of it as being 'infinite' there, but this isn't strictly true. Thus $f(x)$ is undefined at $x=\frac{1}{2}$. Your intuition in terms of 'blips' is correct however.
For your second question, note that:
$$\int_{-\infty}^{+\infty}\delta(x)dx=1,$$
but
$$\int_{-\infty}^{0}\delta(x)dx=\frac{1}{2}.$$
One way to think of this is to view $\delta(x)$ as the limit of a series of Gaussians, as on the Wikipedia page. In that case if you are integrating to $0$ you will only have half of the area of the Gaussian.
