Integrating a dirac delta 'function' on a definite domain

Question

Came across a question that requires evaluation of $\int_{-3}^{+1} \left(x^{3}-3x^{2}+2x-1\right)\delta\left(x+2\right)dx$

Here's my attempt:

Recall that:

$\int_{-\infty}^{\infty} f\left(x\right)\delta\left(x-a\right)dx=f\left(a\right)\int_{-\infty}^{+\infty} \delta\left(x-a\right)dx=f\left(a\right)$

Note that a=-2 relative to the question.

Then,$$\int_{-2-1}^{-2+3} \left((-2)^{3}-3\left(-2\right)^{2}-2\left(-2\right)-1\right)\delta\left(x+2\right)dx= (-2)^{3}-3\left(-2\right)^{2}-2\left(-2\right)-1 $$

I'm still fairly uncomfortable dealing with dirac delta function due to my sparse exposure to them. My guess is that I'm doing this question wrongly and that the domain of integration requires shifting so that the domain of integration is symmetric about the point x=0. Any help is appreciated

Answer 1

In general, in a definite domain
$ \int_b^c f(x)\delta(x-a)dx = f(a)$ if $b < a < c$,
then $\int_{-3}^{1} (x^3 - 3x^2 + 2x -1) \delta(x+2) dx = (-2)^3 - 3(-2)^2 + 2(-2) - 1$ because $-3 < -2 < 1$.

Answer 2

In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

Here, using the Unit Step Function $u(x)$ defined by

$$ u(x)= \begin{cases}1&,x>0\\\\ 0&,x<0\\\\ 1/2&,x=0 \end{cases} $$

We interpret the notation $\int_a^b f(x)\delta(x-x')\,dx$ using the unit step function and write

$$\begin{align} \mathscr{D_{x';a,b}}\{f\}&=\int_a^b f(x)\delta(x-x')\,dx\\\\ &=\int_{-\infty}^{\infty}f(x)\left(u(x-a)-u(x-b)\right)\delta(x-x')\,dx\\\\ &=f(x')\left(u(x'-a)-u(x'-b)\right) \end{align}$$

Now, depending on $x'$ relative to $a$ and $b$, we have

$$\begin{align} \mathscr{D_{x';a,b}}\{f\}&= \begin{cases} f(x')&,a<x'<b\\\\ \frac12 f(x')&,x'=a\,\,\text{or}\,\,x'=b\\\\ 0&, \text{otherwise} \end{cases} \end{align}$$