Why does the Dirac delta function satisfy $f(x)\delta(x-a) = f(a)\delta(x-a)$? Why does the Dirac delta function have the property that 
$$ f(x)\delta(x-a)= f(a)\delta(x-a) , $$
where $\delta(x-a)$ is the Dirac delta function? Won't the Dirac delta function just stay the same even after being multiplied?
 A: The definition of the dirac delta is an object $\delta$ such that, for any (suitably well behaved) function $f$, $$\int_{a}^{b}\delta(x)f(x)dx $$
takes the value $f(0)$ if $a<0<b$, and $0$ otherwise. This isn't really a function, as there is provably no function $\delta: \mathbb{R} \to \mathbb{R}$ with this property. If it's not a function, then what the hell is it? Well, as the other answers point out, $\delta$ is formally defined as something called a distribution. However, it turns out you can manipulate $\delta$ as though it were a function, and in surprisingly many cases the sky does not fall on your head.  
Fixing an $a \in \mathbb{R}$, it is then immediate from the definition that
$$\int_{c}^{d}\delta(x-a)f(x)dx=\int_{c}^{d}f(a)\delta(x-a)dx$$
for any choice of $c$ and $d$, because both sides are equal to $f(a)$ when $a \in [c,d]$, and $0$ otherwise.
It seems that what the question setter has done is then equate the integrands, and say that the objects $$f(x)\delta(x-a)=f(a)\delta(x-a)$$
are the same. In my opinion this is a strange thing to say, because it suggests that you are treating $\delta$ as though it were a real function, which it is not.
A: Remember that 
the $\delta$ ''function'' is not really a function (proper or “improper”) but a different mathematical object ( a distribution) that live always in the shade of an integral sign.
So, when we write an ''identity'' as:
$$ f(x)\delta(x-a)= f(a)\delta(x-a) , $$
The meaning of this equations is that:

its two sides give equivalent results when used as factors in an integrand.

And this is easy verified using the definition 
$$
\int_{-\infty}^{+\infty}f(x)\delta(x)dx=f(0)$$
for $x-a=0$.
A: The only correct way to see it is from the distributional theory of the Delta function. Indeed, for a test function $\phi$, we have:
$$\langle \delta(x-a), \phi(x)\rangle = \langle \theta'(x-a), \phi(x)\rangle = - \langle \theta(x-a), \phi'(x)\rangle  = -\int_{-\infty}^{+\infty}\phi'(x)\theta(x-a)\ \text{d}x$$
Now since we have a $\theta$, we know that that $\theta(x-a) = 1$ for $x-a\geq0$ and $0$ otherwise. Hence the integral becomes
$$-\int_{a}^{+\infty}\phi'(x)\theta(x-a)\ \text{d}x = -\phi(x)\bigg|_{a}^{+\infty}$$
The test function $\phi$ goes to zero for $x\to \infty$ hence you remain with
$$\phi(a)$$
This shows you that the action of the $\delta(x-a)$ on a function $f(x)$ gives you
$$f(a)$$
Now the reason why the Dirac Delta remains, it's because the expression $f(x)\delta(x-a)$ makes sense only in the support of the Delta which is $x = a$, otherwise it's zero. To remark this, you write $f(x)\delta(x-a)$ because every other value for $x$ is meaningless. 
A: Simple answer is that cancelling the Dirac "function" on both sides is not allowed. 
Why? 
Because the function equals zero for $x$ not equal to $a$. And if $x=a$, the value becomes infinity. The rules of maths don't allow for direct cancellation of zeroes or infinities on both the sides.
A: The delta distribution is defined as 
$\delta: C^{\infty}_{0}(\mathbb{R}) \rightarrow \mathbb{R},\,\phi \mapsto \phi(0)$
i.e. a (continuous, whatever this means) functional on the smooth functions of compact support.
Now translating a delta distribution amounts to translating its argument into the opposite direction:
$\tau_{a}\delta(\phi) := \delta(\tau_{-a}\phi) = \phi(a)$
where $(\tau_{b}\phi)(x) := \phi(x-b)$.
If we now define the product of the delta distribution $\delta$ with a smooth funtion $f \in C^{\infty}(\mathbb{R})$ as merely evaluating a test function $\phi$ via the product $f\phi$, you get
$(f\cdot\delta)(\phi) := \delta(f\phi) = f(0)\phi(0)  = \delta(f(0)\phi) = (f(0)\cdot\delta)(\phi)$
and in this sense you have $f\cdot\delta = f(0)\cdot\delta$. Together with the above, the full claim follows.
A: "Physicists rush in where mathematicians fear to tread". Let's dance.
There are a few definitions of $\delta(x)$ that give intuition.
If we start with the propertyt hat $\int_a^b\delta(x)f(x)dx=f(0)$ when $a<x<b$, then we can get a definition:
$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\delta(x)e^{-ikx}dx=\frac{1}{\sqrt{2\pi}}$. The Fourier Transform of the dirac delta function is a constant, so the inverse Fourier Transform is that original function:
$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}dk$
In many cases, a suitable curve with finite area will do.
$\delta(x)=\lim _{\alpha\to \infty}\int_{-\infty}^\infty \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2}dx$. This holds for arbitrary positive values of $\alpha$. As alpha increases, the height increases, but the width shrinks, maintaining unit area. We essentially have an infinitely high, infinitely narrow curve and this is one characterization of the dirac-delta function.
Given one of these definitions, we can plug it into the integral as we would the diract delta function directly:
$\lim_{\alpha \to \infty} \int_a^b \sqrt{\frac{\alpha}{\pi}}e^{-\alpha(x-c)^2}f(x)dx$
Then with $u$ substitution $u=x-c, du=dx$:
$\lim_{\alpha \to \infty} \int_{a-c}^{b-c}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha u^2}f(u+c)du=f(c)$ for suitable $f$ and associated interval.
