The question is $$ \int ^4 _{-4} (t-2)^2\delta'\left(-\frac13t+\frac12\right)dt $$
The solution of the text book is
\begin{align} \int ^4 _{-4} (t-2)^2\delta'\left(-\frac13t+\frac12\right)dt &=\int ^4 _{-4} 3(t-2)^2\delta'\left(t-\frac32\right)dt \\ &=\int ^4 _{-4} \left[{\frac34\delta'\left(t-\frac32\right)+3\delta\left(t-\frac32\right)}\right]dt\\ &=3 \end{align}
My solution is \begin{align*} \int ^4 _{-4} (t-2)^2\delta'\left(-\frac13t+\frac12\right)dt &=3\int ^4 _{-4} (t-2)^2\delta'\left(t-\frac32\right)dt \\ &=-3\int ^4 _{-4} \left[{\frac{d}{dt}(t-2)^2}\right]_\frac32 \delta\left(t-\frac32\right)dt\\ &=-3\cdot2\cdot\left(\frac32-2\right)\int ^4 _{-4}\delta\left(t-\frac32\right)dt\\ &=3\cdot1\\ &=3 \end{align*}
My solution is used the following equation derived by parts.
$if\ \ \ t_1 < t_0 < t_1,$ $$ \int ^{t_2} _{t_1} x(t) \delta ^ {(n)} (t-t_0) dt = (-1)^{(n)} \int ^{t_2} _{t_1} x^{(n)}(t) \delta(t-t_0) dt = (-1)^{(n)}x^{(n)} (t_0)\\ $$
The result is same as $3$. However, I cannot understand the textbook's solution.
Can someone explain how the first line became the second line in textbook's solution?
Neither seems correct: $$ \begin{align} \int_{-4}^4(t-2)^2\delta'\left(-\frac13t+\frac12\right)\,\mathrm{d}t &=-3\int_{-4}^4(t-2)^2\,\mathrm{d}\delta\left(-\frac13t+\frac12\right)\\ &=3\int_{-4}^42(t-2)\delta\left(-\frac13t+\frac12\right)\,\mathrm{d}t\\ &=9\int_{-4}^42(t-2)\cdot\frac13\delta\left(-\frac13t+\frac12\right)\,\mathrm{d}t\\ &=9\cdot2\left(\frac32-2\right)\\[6pt] &=-9\tag{1} \end{align} $$
I've also done some numerical tests with approximations of $\delta$ in Mathematica, and they support the computation above.
I used $\delta_n(x)=ne^{-\pi n^2x^2}$ and computed $$ \int_{-4}^4(x-2)^2\delta_n'\!\left(-\tfrac13x+\tfrac12\right)\,\mathrm{d}x\tag{2} $$ Using
d[x_,n_] := n Exp[-Pi n^2x^2]d1[x_,n_] := Evaluate[D[d[x,n],x]]NIntegrate[(t-2)^2 d1[-t/3+1/2,100],{t,-4,4},WorkingPrecision->20,MaxRecursion->100]for $n=100$, Mathematica 8 gives $-9.0000000000000000000$.