$E(x) = 1.5x \frac{N}{C}$ between $2.5\hspace{0.1cm}cm$ and $4.0\hspace{0.1cm} cm$ Determine the electrical potential difference for the fields in the x direction:
$E(x) = 1.5x \frac{N}{C}$ between $2.5\hspace{0.1cm}cm$ and $4.0\hspace{0.1cm} cm$
\begin{eqnarray*}
  \Delta V =-\int_{.025}^{.04}1.5x\hspace{0.1cm}dx&=& \bigg[-\frac{1.5}{2}\left(x^{2}\right) \bigg|_{.025}^{.04}\hspace{0.2cm}\bigg]\text{ V}  \\
&= &-0.75 \cdot (0.04)^{2}-(0.025)^{2} \text{ V} \\ 
& =& -0.75 \cdot (0.0016-0.000625)\text{ V}  \\ 
& =&(-0.75 \cdot 0.000975)\text{ V}  \\   
& =& (-0.75 \cdot 9.75 \cdot 10^{-4})\text{ V}  \\  
& =& (-7.31 \cdot 10^{-4})\text{ V}  \\  
& =&  (-0.731 \cdot 10^{-3})\text{ V} \\   
&=& -0.731\text{ mV}  \hspace{0.3cm}\mbox{Milivolts} 
 \end{eqnarray*}

Questions:
Is there a systematic approach to finding a solution to this problem?
Is there a simpler elementary function with required properties?

Your final answer is correct! Just be careful in your first integral. You have written
$$ \Delta V =-\int_{2.5}^{4.0}1.5x\hspace{0.1cm}dx $$ when it should be
$$ \Delta V =-\int_{.025}^{.04}1.5x\hspace{0.1cm}dx $$
so that you stay with SI units. You convert to mV after evaluating the integral if you wish. Using a consistent unit system tends to cut down on errors!
