# Make Integral Sign Larger

by S_David
Tags: integral, sign
 P: 600 Hello, I know that I am asking too many questions these days, but I really need these helps, and I think this forum can do that. My problem is: I have equations that when I wrote using Latex, the integral sign seems very small relative to other terms. How can I make it larger? Thanks in advance
 P: 74 The best answer I can give you so far is to look into the \DeclareMathSizes command. This needs to go in your preamble, and lets you redefine the size of the object when in display mode, inline text mode, sub/superscript, and nested sub/superscripts. %put this in your preamble, and fiddle with the last three parameters for each line \DeclareMathSizes{10}{18}{12}{8} % For size 10 text \DeclareMathSizes{11}{19}{13}{9} % For size 11 text \DeclareMathSizes{12}{20}{14}{10} % For size 12 text For more info, check out http://en.wikibooks.org/wiki/LaTeX/A...ging_font_size
 Sci Advisor P: 1,498 What are you using for your integral? When I typically use it, it displays a fine, normal size: $$\int \sin(x)\,dx = -\cos(x) \int \frac{1}{x}\,dx = \ln(x)$$
P: 600
Make Integral Sign Larger

 Quote by minger What are you using for your integral? When I typically use it, it displays a fine, normal size: $$\int \sin(x)\,dx = -\cos(x) \int \frac{1}{x}\,dx = \ln(x)$$

I am using the \int control sequence.

Regards
Attached Thumbnails

 P: 74 It looks perfectly legible to me, both in the forums and in your attachment. You would prefer to see the integral extend above and below the fraction? I was pondering this, and maybe there's a way to custom design the rubber length parenthesis. You know the commands \left(, \left\[, left\{, \left<, \left., and their \right counterparts? These expand to fit the contents. Maybe if you could custom define a \left\int, the \int sign will rubberband to expand to the right size. Every time you use a \left command, you need to close it with a \right command. The . is great for doing this if you don't want something to appear on the opposite side. $$\begin{split} g(x) &= \int_0^\infty \left(\frac{ \frac{ \frac{1}{1 + x} }{ 2 + x} }{3 + x}\right) dx\\ &= \int_0^\infty \frac{df(x)}{dx}dx\\ &= \left. f(x) \right|_0^\infty \end{split}$$ The first line looks kinda bad, so if the \int could expand in the same way the parenthesis are, it would be great.
 Sci Advisor P: 1,498 One problem which I think makes it "look" smaller is the limits. $$f(x) = \int_0^1 \sin(x)\,dx g(x) = \int\limits_0^1 \cos(x)\,dx$$ Try using the \limits on your \int to put the limits above the integral sign, makes it look better IMHO. $$f(x) = \int_0^1 \sin(x)\,dx$$ $$g(x) = \int\limits_0^1 \cos(x)\,dx$$
P: 600
 Quote by minger One problem which I think makes it "look" smaller is the limits. $$f(x) = \int_0^1 \sin(x)\,dx g(x) = \int\limits_0^1 \cos(x)\,dx$$ Try using the \limits on your \int to put the limits above the integral sign, makes it look better IMHO. $$f(x) = \int_0^1 \sin(x)\,dx$$ $$g(x) = \int\limits_0^1 \cos(x)\,dx$$
Good to know that, it is useful in some cases for sure. For my case, it still small even though. Anyway, thanks minger for this information.

Regards

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