How to solve integration with Dirac Delta function? I am struggling to solve several problems in my 'Signals and Systems' textbook. However, I just met a confused problem.
Q) $\displaystyle\int_{-4}^{4}\left(t-2\right)^2\delta'\left(-\frac13t+\frac12\right)dt$
I tried to solve with the method, 'integration by substitution'.
$\displaystyle-\frac13t = x \Leftrightarrow t=-3x \Leftrightarrow dt = -3dx$
so, $\displaystyle\int_{-4}^{4}\left(t-2\right)^2\delta'\left(-\frac13t+\frac12\right)dt=\int_{\frac43}^{-\frac43}(-3x-2)^2\delta'(x+\frac12)(-3dx)$
$\displaystyle=3\int_{-\frac43}^{\frac43}(3x+2)^2\delta'(x+\frac12)dx$
Then, I found an equation on the internet,
(19) in http://mathworld.wolfram.com/DeltaFunction.html
,which is

$\displaystyle\int_{-\infty}^\infty f(x) \delta'(x-a)dx=-f'(a)$.
Actually, I failed to understand how the equation above is induced. :(
However, I can apply it.

So, I assumed that $f(x) = (3x+2)^2$ and $\displaystyle a = -\frac12$. 
$\displaystyle f'(x)=2\left(3x+2\right)\cdot3=18x+12 \Rightarrow -f'\left(-\frac12\right)=-3$
$\displaystyle\therefore3\int_{-\frac43}^{\frac43}(3x+2)^2\delta'(x+\frac12)dx=3\cdot\left(-f'\left(-\frac12\right)\right)=3\cdot(-3)=-9$
But, the answer is 3 in the solution of this book.
Are there any errors in my solving process?
Thanks a lot for the kind answers.
 A: The way to obtain
$$
\int_{-\infty}^\infty f(x)\,\delta'(x-a)\,dx = -f'(a)
$$
is to integrate by parts:
$$
\int_{-\infty}^\infty f(x)\,\delta'(x-a)\,dx = 
\int_{-\infty}^\infty \left[\, f(x)\,\delta(x-a) \,\right]'\,dx -
\int_{-\infty}^\infty f'(x)\,\delta(x-a)\,dx = 
\left[\, f(x)\,\delta(x-a) \,\right]\,\Big|_{-\infty}^{+\infty} -
f'(a) = -f'(a)
$$
A: To solve this, it's easier to go by parts (treating the delta function formally as a function, which is in fact justifiable).
$$I = \int_{-4}^4 \underbrace{(t-2)^2}_{u} \underbrace{\delta'(-\frac{1}{3} t + \frac{1}{2} )}_{d v} dt$$
$$\dfrac{du}{dt} = 2(t-2)$$
$$v = -3 \delta(-\frac{1}{3} t + \frac{1}{2})$$
Therefore $$I = \left[\underbrace{-3 \delta(-\frac{1}{3} t + \frac{1}{2})}_{v} \underbrace{(t-2)^2}_u \right]_{-4}^4 - \int_{-4}^4 \underbrace{2(t-2)}_{du} \cdot \underbrace{-3 \delta(-\frac{1}{3} t + \frac{1}{2})}_{v} dt = 6 \int_{-4}^4 (t-2) \delta(\color{blue}{-\frac{1}{3} t + \frac{1}{2}}) dt$$
Now, a substitution $u = \color{blue}{-\frac{1}{3} t + \frac{1}{2}}$, so $du = \color{red}{-\frac{1}{3}} dt$:
$$I = 6 \times \color{red}{-3} \int_{11/6}^{-5/6} (\frac{3}{2} - 3 u - 2) \delta(u) du = -18 \int_{-5/6}^{11/6} (\frac{1}{2}-3u) \delta(u) du$$
Finally, that is clearly $-18 \times \frac{1}{2} = -9$.
That is essentially the construction of the formula you found in Mathworld.
A: The Dirac Delta function $\delta(x)$ is very cool in the sense that
$$ \delta(x) = \begin{cases}
+\infty, \, & x =0 \\
0, \, & x \ne 0
\end{cases}
$$
Its unique characteristics do not end there though, because when integrating the Dirac Delta function we would get
$$ \int_{-\infty}^\infty \delta(x) dx = 1$$
Or, if we have another function $f(x)$ multiplied to the Dirac Delta function and integrating them we would get
$$ \int_\infty^\infty f(x) \delta(x) \, dx = f(x) \int_{-\infty}^\infty \delta(x) \, dx = f(0) $$ 
$$\\$$
Since
$$ \int_{-\infty}^\infty \delta(x) \, dx = \begin{cases}
0, \, & x \ne 0 \\
1, \, & x = 0 \end{cases}$$
Therefore in the previous integral we would have
$$\int_{-\infty}^\infty f(x)\delta(x) \, dx = \int_{-\infty}^\infty f(0)\delta(0) \, dx = f(0)$$.
What if we have $\delta(x-a)$? It's the same thing! The only thing here is we need to satisfy the condition $x-a = 0$ such that $\int_{-\infty}^\infty \delta(x-a) \, dx = 1$.
If the bounds of our integral though is not infinity, we need to make sure that the if we let $x = a$, $a$ would be in the bounds of the integral or else the integral would evaluate into zero.
$$\int_{-b}^b f(x)\delta(x-a) \, dx = \begin{cases}
0, & \text{if $b<a$ or $-b >a$ such that we cannot let $x \ne a$} \\
f(0), & \text{if $-b<a<b$ such that we can let $x = a$}
\end{cases}$$.
What if we have $\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx$? Let's then compute it by integration by parts.
$$\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx = f(x)\delta(x-a)\bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \delta'(x-a)f(x) \, dx = -f'(a)$$
Or in general
$$\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) \, dx = (-1)^n f^{(n)}(a)$$.
