How can I split a long equation in eqnarray?

Question

All equations fit in a single line, but the 3rd line (equation) is too long, so I need to divided it into 2 lines. I have tried \\ but it doesn't work. This is what I have (I added more space in between so you can see the 3rd equation):

\begin{eqnarray}
&I&= \displaystyle \int_{x_0=0}^{x_2=2h} \left[ \left(\dfrac{x^2-3hx+2h^2}{2h^2}\right)f(x_0) + \left(\dfrac{4hx-2x^2}{2h^2}\right)f(x_1) + \left(\dfrac{x^2-hx}{2h^2}\right)f(x_2)\right] \ dx \nonumber\
&=& \dfrac{1}{2h^2}\left[ \left(\dfrac{x^3}{3}-\dfrac{3hx^2}{2}+2h^2x\right)f(x_0) + \left(\dfrac{4hx^2}{2}-\dfrac{2x^3}{3}\right)f(x_1) + \left(\dfrac{x^3}{3}-\dfrac{hx^2}{2}\right)f(x_2)\right]_0^{2h} \nonumber\
&=& \dfrac{1}{2h^2}\left[ \left(\dfrac{(2h)^3}{3}-\dfrac{3h(2h)^2}{2}+2h^2(2h)\right)f(x_0) + \left(\dfrac{4h(2h)^2}{2}-\dfrac{2(2h)^3}{3}\right)f(x_1) +\left(\dfrac{(2h)^3}{3}-\dfrac{h(2h)^2}{2}\right)f(x_2) + 0\right] \nonumber\
&=&\dfrac{1}{2h^2}\left[ \left(\dfrac{8h^3}{3}-6h^3+4h^3\right)f(x_0) + \left(8h^3-\dfrac{16h^3}{3}\right)f(x_1) + \left(\dfrac{8h^3}{3}-2h^3\right)f(x_2) \right]\nonumber \
&=&\dfrac{1}{2h^2}\left[ \left(\dfrac{2h^3}{3}\right)f(x_0) + \left(\dfrac{8h^3}{3}\right)f(x_1) + \left(\dfrac{2h^3}{3}\right)f(x_2) \right] \nonumber\
&=&\dfrac{1}{2h^2}\cdot \dfrac{2h^3}{3}\left[ f(x_0) + 4f(x_1) + f(x_2) \right] \nonumber\
&=&\dfrac{h}{3}\left[ f(x_0) + 4f(x_1) + f(x_2) \right] \
\end{eqnarray}

You shouldn't use eqnarray, it often gives inconsistent spacing. Instead, you should use align from amsmath. — Skillmon, Dec 04 '20 at 15:57
@Skillmon Thanks! That worked but "program@e..." appeared just at the end of the 3rd equation, and don't know how to remove it. — Anna, Dec 04 '20 at 16:12
Off topic: Blank lines aren't allowed in math; this should have resulted in error messages on that account. — barbara beeton, Dec 04 '20 at 16:23

egreg · Accepted Answer · 2020-12-04T17:17:39.337

Never ever use eqnarray. In this case, equation and split are the best.

\documentclass{article}
\usepackage{geometry} % more generous text width
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{split}
I&= \int_{x_0=0}^{x_2=2h} \left[
       \left(\frac{x^2-3hx+2h^2}{2h^2}\right)f(x_0)
     + \left(\frac{4hx-2x^2}{2h^2}\right)f(x_1)
     + \left(\frac{x^2-hx}{2h^2}\right)f(x_2)
    \right] , dx
\[1ex]
 &= \frac{1}{2h^2}\left[
       \left(\frac{x^3}{3}-\frac{3hx^2}{2}+2h^2x\right)f(x_0)
     + \left(\frac{4hx^2}{2}-\frac{2x^3}{3}\right)f(x_1)
     + \left(\frac{x^3}{3}-\frac{hx^2}{2}\right)f(x_2)
    \right]_0^{2h}
\[1ex]
 &= \frac{1}{2h^2}\biggl[
       \begin{aligned}[t]
       & \left(\frac{(2h)^3}{3}-\frac{3h(2h)^2}{2}+2h^2(2h)\right)f(x_0)
        + \left(\frac{4h(2h)^2}{2}-\frac{2(2h)^3}{3}\right)f(x_1) \
       &+ \left(\frac{(2h)^3}{3}-\frac{h(2h)^2}{2}\right)f(x_2) + 0
    \biggr]
       \end{aligned}
\[1ex]
 &=\frac{1}{2h^2}\left[
      \left(\frac{8h^3}{3}-6h^3+4h^3\right)f(x_0)
    + \left(8h^3-\frac{16h^3}{3}\right)f(x_1)
    + \left(\frac{8h^3}{3}-2h^3\right)f(x_2) \right]
\[1ex]
 &=\frac{1}{2h^2}\left[
      \left(\frac{2h^3}{3}\right)f(x_0)
    + \left(\frac{8h^3}{3}\right)f(x_1)
    + \left(\frac{2h^3}{3}\right)f(x_2) \right]
\[1ex]
 &=\frac{1}{2h^2}\cdot \frac{2h^3}{3}\left[ f(x_0) + 4f(x_1) + f(x_2) \right]
\[1ex]
 &=\frac{h}{3}\left[ f(x_0) + 4f(x_1) + f(x_2) \right]
\end{split}
\end{equation}
\end{document}

I'd consider also splitting the computations (if at all necessary, as they're quite elementary):

\documentclass{article}
\usepackage{geometry} % more generous text width
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{split}
I&= \int_{x_0=0}^{x_2=2h} \left[
       \left(\frac{x^2-3hx+2h^2}{2h^2}\right)f(x_0)
     + \left(\frac{4hx-2x^2}{2h^2}\right)f(x_1)
     + \left(\frac{x^2-hx}{2h^2}\right)f(x_2)
    \right] , dx
\
 &= \frac{1}{2h^2}\bigl[f(x_0)I_0 + f(x_1)I_1 + f(x_2)I_2\bigr]
\[2ex]
I_0 &= \left[\frac{x^3}{3}-\frac{3hx^2}{2}+2h^2x\right]_0^{2h}
     = \left(\frac{(2h)^3}{3}-\frac{3h(2h)^2}{2}+2h^2(2h)\right) \
    &= \left(\frac{8h^3}{3}-6h^3+4h^3\right)
     = \frac{2h^3}{3}
\[1ex]
I_1 &= \left[\frac{4hx^2}{2}-\frac{2x^3}{3}\right]_0^{2h}
     = \left(\frac{4h(2h)^2}{2}-\frac{2(2h)^3}{3}\right) \
    &= \left(8h^3-\frac{16h^3}{3}\right)
     = \frac{8h^3}{3}
\[1ex]
I_2 &= \left[\frac{x^3}{3}-\frac{hx^2}{2}\right]_0^{2h}
     = \left(\frac{(2h)^3}{3}-\frac{h(2h)^2}{2}\right) \
    &= \left(\frac{8h^3}{3}-2h^3\right)
     = \frac{2h^3}{3}
\[2ex]
I &= \frac{h}{3}[f(x_0) + 4f(x_1) + f(x_2)]
\end{split}
\end{equation}
\end{document}

Uhm, given the amount of whitespace introduced by \right] this is one of the few situations where I wouldn't necessarily put \, before the differential. — campa, Dec 04 '20 at 17:05

How can I split a long equation in eqnarray?

1 Answers1