I know that $\delta(0)=1$, and $\delta(x)=0$ otherwise. So for the integral $\int_{-\infty}^{\infty}\delta(6-2x)x^2$, why can't you say that $\delta(6-2x)= \delta(0)$ at $x=3$, and therefore evaluating $x^2$ at $x=3$ you get 9? I know you have to do it by substitution. But why??
And how do you tackle a problem where the limits aren't $-\infty$ to $\infty$ and where there's a quadratic in the delta function? So
$\int_{0}^{\infty} \delta(x^2+x-6)x^2 = 0$?
On
Notice:
$$\delta(6-2x)x^2=\frac{x^2\delta(x-3)}{2}=\frac{9\delta(x-3)}{2}$$
So:
$$\int_{-\infty}^{\infty}\delta(6-2x)x^2\space\text{d}x=\lim_{n\to\infty}\int_{-n}^{n}\frac{9\delta(x-3)}{2}\space\text{d}x=\frac{9}{2}\lim_{n\to\infty}\int_{-n}^{n}\delta(x-3)\space\text{d}x=$$ $$\frac{9}{2}\lim_{n\to\infty}\left[\theta(x-3)\right]_{-n}^{n}=\frac{9}{2}\lim_{n\to\infty}\left(\theta(n-3)-\theta(-n-3)\right)=$$ $$\frac{9}{2}\left(\lim_{n\to\infty}\theta(n-3)-\lim_{n\to\infty}\theta(-n-3)\right)=\frac{9}{2}\left(1-0\right)=\frac{9}{2}$$
On
Can I give you a stupid one, but it fits with your intuition....
$\delta(x)$ is a curve that is 0 everywhere except in an infinatessimal area where it is infinitely tall and: $\int_a^b \delta(x)\,dx = 1$ if $a<0<b.$
$\delta(2x)$ his half a wide and just as tall, and
$\int_a^b \delta(2x)\,dx = 1/2$ if $a<0<b$
$\int_a^b \delta(6-2x)x^2 \,dx= (1/2) 3^2$ if $a<0<b.$
$\int_0^\infty \delta(x^2 + x -6)x^2\,dx$
$\int_0^\infty \delta((x-2)(x+3))x^2\,dx$
Clearly x = -3 is out of the interval... at x = 2
$\lim_\limits{h\to0}\int_{2-h}^{2+h}\delta((x-2)(x+3))x^2\,dx$
$\lim_\limits{h\to0}\int_{2-h}^{2+h}\delta((x-2)5)4\,dx = 4/5$
The delta function of a function $f(x)$ is given by $$\delta(f(x))=\sum_{i=1}^n\frac{\delta(x-x_i)}{|f'(x_i)|}$$ where the $x_i$ are the roots of $f(x)$. So for $f(x)=x^2+x-6=(x+3)(x-2)$ we have $$ \delta(f(x))=\frac{\delta(x-2)}{5}+\frac{\delta(x+3)}{5} $$ $$ \begin{align} \int_{0}^{\infty} \delta(x^2+x-6)x^2 \mathrm d x &=\int_{0}^{\infty}\left(\frac{\delta(x-2)}{5}+\frac{\delta(x+3)}{5}\right)x^2\mathrm d x\\ &=\int_{0}^{\infty}\frac{\delta(x-2)}{5}x^2\mathrm d x\\ &=\int_{0}^{\infty}\frac{\delta(x-2)}{5}(2)^2\mathrm d x\\ &=\frac{4}{5}\int_{0}^{\infty}\delta(x-2)\mathrm d x\\ &=\frac{4}{5} \end{align} $$ observing that the point $-3$ is outside of interval of integration and then $\int_{0}^{\infty}\frac{\delta(x+3)}{5}x^2\mathrm d x=0$.
For $g(x)=6-2x$ we have $$ \begin{align} \int_{0}^{\infty} \delta(6-2x)x^2 \mathrm d x =\int_{0}^{\infty}\frac{\delta(x-3)}{2}x^2\mathrm d x =\int_{0}^{\infty}\frac{\delta(x-3)}{2}3^2\mathrm d x =\frac{9}{2}\int_{0}^{\infty}\delta(x-3)\mathrm d x &=\frac{9}{2} \end{align} $$