x dx = +C

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- March 21st 2009, 05:44 PMmanyarrowstrying out LATEX
x dx = +C

- March 23rd 2009, 06:03 PMlmk866
This is what must be typed: [tex]\int x ~dx = \frac{x^2} {2} + C[/tex]

the "~" adds a space. alternatively, you can use "\," or "\;" - March 23rd 2009, 06:40 PMReckoner
- April 27th 2009, 10:21 PMfardeen_genTest
\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}

This isn't working! Why?

EDIT:

Thanks Opalg! Also, eight shouldn't be there. - April 28th 2009, 02:58 AMOpalg
I don't know why, but for some reason the compiler doesn't like the unnecessary braces that you have used to wrap around some sections of the formula. Remove the red ones in this code

\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}and you get

(And it looks as though the 8 probably shouldn't be there at all?)

Also note that you don't need to use \limits with \sum. They are supplied automatically. - April 29th 2009, 04:56 PMlmk866
i am not exactly sure about the equation you are trying to write but here are the syntax problems

correct form

\sum_{n = 0}^{\infty} {\frac{1}{n!} \left[ \sum_{k = 0}^{n} {(k+1)\left( \int_{0}^{1} 2^{-(k + 1)x} dx \right)} \right]}

problems

1 - \dx

2 - the location of closed }

note: you can eliminate the \limits. \sum_{lowerbound}^{upperbound} works

i hope this helps.. - April 29th 2009, 05:00 PMchella182
The .pdf file in this thread is really helpful for LaTex codes I find :)

- August 31st 2009, 10:01 AMedutabacman
how about this here, or

like this? - August 31st 2009, 10:02 AMedutabacman
well, \n \n

eh? - August 31st 2009, 10:05 AMedutabacman
or, better

like this? - August 31st 2009, 10:08 AMedutabacman
but

- August 31st 2009, 10:09 AMedutabacman
- August 31st 2009, 10:40 AMedutabacman
but <latex>$\left[ {\begin{array}{*{20}{c}}

{\sum\limits_{i = 1}^n {{X_i}{Y_i}} } & {\frac{{x - \mu }}{\sigma }} \\

{\frac{{{\partial ^2}\Omega }}{{\partial u\partial v}}} & {\sqrt {{b^2} - 4ac} } \\

\end{array}} \right]$</latex> - August 31st 2009, 10:41 AMedutabacman
I know

right?