Integrating unit impulse function Given that, 
$$ \delta(t) = \begin{cases} \infty & \text{if } t = 0 \\ 
0 & \text{if } t \ne 0\\ 
\end{cases}$$
How is it that, 
(A) 
$$
\int_{-\infty}^\infty \delta(t) dt = 1
$$
(B)
$$
\int_{-\infty}^\infty f(t) \delta(t) dt = f(0)
$$
considering $f$ continuos at $t=0$
Thanks in advance

Can I assume you're a physicist and avoid the mathematical complexities a bit?
The integral finds the area under functions. $\delta(x)$ is a really skinny and really tall rectangle. In fact, it's width is $\epsilon$ and its height is $\omega$ so the area is given by $ \epsilon \cdot \omega$. The variable $\epsilon={1 \over {\omega}}$, so $ \epsilon \cdot \omega=1$. You'd just let $\omega \to \infty$ to get the integral of $\delta(x)$. Using this, I encourage you to figure out the answer to your other question.
More mathematical here. The delta 'function' is not a function in any typical sense. It's not continuous, differentiable, or integrable in the Riemann sense.
However, if you define it as a measure, you can look at it in a more rigorous way. A measure, is basically a way to assign mass, or weight, to subsets of the x axis. For instance, the way density can be integrated is a good example of a measure. There is some similarity with distributions as well.
Define the measure $\delta(dx)$ to be $1$ if $dx$ includes the value $0$ and $0$ otherwise. Using this definition, you can integrate with respect to the measure. Doing this, you get,
$$ \int_{-\infty}^{\infty} f(x) \ \delta(dx)=f(0)$$
Here's more information about the Dirac delta function, skip to the "As a Measure" section if you want.
