Integral | symbol format

In this case I'd just use \Big|_{-1}^{1}.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\int_{-1}^{1}8x^3-5x^2+4dx=\tfrac{8}{4}x^4-\tfrac{5}{3}x^3+4x\Big|_{-1}^{1}
\]
\end{document}

And here is why I don't recommend using anything similar to \left. \int_{-1}^{1}....\right|_{-1}^{1} as then the | with limits is actually taller than the integral which is unnecessary

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\left.\int_{-1}^{1}\right|_{1}^{-1} 
\]
\end{document}

I'm not a fan of this notation, because it doesn't make clear what the evaluation should apply to.

\documentclass{article}
\usepackage{amsmath}

\newcommand{\evalint}{%
  \left.\kern-\nulldelimiterspace
  \vphantom{\int}\right|%
}

\begin{document}

\[
\int_{-1}^{1}(8x^3-5x^2+4)\,dx=
\frac{8}{4}x^4-\frac{5}{3}x^3+4x\evalint_{-1}^{1}
\]

$
\int_{-1}^{1}(8x^3-5x^2+4)\,dx=
\frac{8}{4}x^4-\frac{5}{3}x^3+4x\evalint_{-1}^{1}
$

\end{document}

A different implementation that guarantees the limits to be (quite accurately, although not completely exact) at the same height as in the integral.

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\makeatletter
\NewDocumentCommand{\evalint}{e{_^}}{%
  \mathpalette\eval@int{{#1}{#2}}%
}
\newcommand{\eval@int}[2]{\eval@@int#1#2}
\newcommand{\eval@@int}[3]{%
  \ifx#1\displaystyle\eval@@@int{#2}{#3}\else
    \ifx#1\textstyle\big|_{#2}^{#3}\else
      \vert_{#2}^{#3}\fi\fi
}

\newcommand{\eval@@@int}[2]{%
  \left.\kern-\nulldelimiterspace
  \sbox0{$\displaystyle\int_{#1}^{#2}$}\global\dimen1=\dimexpr\ht0+\dp0\relax
  \vphantom{\int}%
  \right|\!
  \vcenter to\dimen1{\hbox{$\scriptstyle#2$}\vfill\hbox{$\scriptstyle#1$}}%
}
\makeatother

\begin{document}

\[
\int_{-1}^{1}(8x^3-5x^2+4)\,dx=
\frac{8}{4}x^4-\frac{5}{3}x^3+4x\evalint_{-1}^{1}
\]

$
\int_{-1}^{1}(8x^3-5x^2+4)\,dx=
\frac{8}{4}x^4-\frac{5}{3}x^3+4x\evalint_{-1}^{1}
$

\end{document}

With \Big as @daleif suggested, or \bigg, and some cosmetic improvements: a correctly spaced upright d, and medium-sized fractions for numerical coefficients:

\documentclass[ a4paper]{article}
\usepackage{nccmath}

\newcommand*{\dd}{\mathop{}\!\mathrm{d}}

\begin{document}

\[ \int_{-1}^{1}(8x^3-5x^2+4)\dd x =\mfrac{8}{4}x^4-\mfrac{5}{3}x^3+4x\bigg|_{1}^{-1} \]

\[ \int_{-1}^{1}(8x^3-5x^2+4)\dd x =\mfrac{8}{4}x^4-\mfrac{5}{3}x^3+4x\Big|_{1}^{-1} \]

\end{document}