You need to think about what your delta function is defined to be right from the start. You are right that the definition you are using does not naturally extend to an obvious answer, but its only the bad integral notation for the delta combined with ones intuition with manipulating integrals that causes the confusion!
Usually the formal mathematical way of defining the delta is as a distribution.
On a regular domain $\Omega \subset \mathbb{R}^n$ a distribution is a linear functional defined on $C^\infty_c(\Omega)$, where $C^\infty_c(\Omega)$ denotes the space of smooth functions whose support is a compact set. The Dirac delta about the point $x_0\in\Omega$ is then defined via
$$ \delta_{x_0}\colon C^\infty_c(\Omega)\to\mathbb{R} $$
$$ \delta_{x_0}(f)=f(x_0) .$$
If this is the first time you are seeing this, note that this is precisely the same definition you were using. You just used the notation of the integral sign for some convenience, but the integrand was never actually an integrable function (it was never a function at all!) so you were kidding yourself that the operation being performed was a classical integral :p
Note that we only have here the delta defined on functions which vanish near the boundary $\partial \Omega$! (One can of course then extend the definition to spaces in which $C^\infty_c(\Omega)$ is appropriately dense, however the condition of vanishing at the boundary is crucial to the whole theory of distributions as it relies on defining duality by analogy with integration by parts). In the end, in your situation the answer depends whether the underlying space you are working over is in fact $[a,b]$ or if you are working on $\mathbb{R}$. If it is the former, then your answer won't be defined unless $f(a)=f(b)=0$. If it is the latter, then the answer is $f(x_0)$, straight from the definition!
I would guess the case is probably the latter, if you are dealing with functions on $\mathbb{R}$. To avoid confusion with the integral notation you should perhaps try and avoid this functional $I[a,b]$ that you are trying to define, and instead always define the dirac with 'limits' over the whole line, writing
$$ \delta_{x_0}(f)= \int_{-\infty}^\infty f(x) \delta_{x_0}(x) dx = \int_{-\infty}^\infty f(x) \delta(x-x_0) dx = f(x_0) .$$
Then you will safely never make mistakes. But I hope this motivates you to read a bit about distributions anyway :)
The usual nascent delta function is $\frac{1}{\epsilon}\phi(\frac{x}{\epsilon})$, from which the substitution $y=\frac{x}{\epsilon}$ reveals the limiting behaviour. Your example is more involved. We have$$\phi\bigg(\frac{x+\epsilon}{\sqrt{2\epsilon}}\bigg)=\frac{1}{\sqrt{2\pi}}\exp\bigg(-\frac{x^2}{4\epsilon}-\frac{x}{2}-\frac{\epsilon}{4}\bigg),$$so$$f(x)\phi\bigg(\frac{x+\epsilon}{\sqrt{2\epsilon}}\bigg)=\sqrt{2\epsilon}\exp-\frac{\epsilon}{4}\frac{f(x)\exp -\frac{x}{2}}{\sqrt{4\pi\epsilon}}\exp\bigg(-\frac{x^2}{4\epsilon}\bigg).$$Why did I write it in that complicated way? Well,$$\lim_{\epsilon\to 0^+}\int_{-2}^\infty f(x)\phi\bigg(\frac{x+\epsilon}{\sqrt{2\epsilon}}\bigg)=\lim_{\epsilon\to 0^+}\sqrt{2\epsilon}\exp-\frac{\epsilon}{4}\int_{-2}^\infty\frac{f(x)\exp -\frac{x}{2}}{\sqrt{4\pi\epsilon}}\exp\bigg(-\frac{x^2}{4\epsilon}\bigg).$$On the right-hand side, the integral $\to f(0)\exp-\frac{0}{2}=f(0)$, but its coefficient $\sqrt{2\epsilon}\exp-\frac{\epsilon}{4}\to 0$.
Best Answer
The sequence $f_n$ of piecewise-linear tent functions of height $n$ and base width $2/n$, but centered at $1/n$ rather than $0$, approximates $\delta$, but they all have integral $\int_0^\infty$ equal to $1$. Symmetrically, centering the tents at $-1/n$ have those integrals all $0$. If we interleave the two sequences, they approach $\delta$ distributionally but their integrals oscillate between $0$ and $1$, so have no limit.