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# Bản mẫu:Intmath

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This template generates integral symbols using unicode, for inline {{math}} formulae as an alternative to LaTeX generated in [itex].

## Các tham số

The template has three parameters, applicable one by one:

1. Integral sign: Choose one of:
These match the LaTeX definitions exactly: please do not modify them for something more "convenient", thank you.
2. Subscript: Enter the subscript (symbol or short expression), for the lower limit or denoting an n-dimensional space or the (n − 1)- dimensional boundary.
3. Superscript: Enter the superscript (symbol or short expression) for the upper limit.

NB:

• Applying font-style: italic; or font-style: oblique; to the integral symbol has no effect in Firefox, it remains upright. E.g.
<span style="font-style: italic;">∫</span> yields ;
<span style="font-style: oblique;">∫</span> yields .
• This template already includes {{su}}.

## Các ví dụ

### Không {{math}}

Γ(z) =
0
ettz − 1dt
Γ(''z'') = {{intmath|int|0|∞}} ''e''<sup>−''t''</sup>''t''<sup>''z'' − 1</sup>''dt''

C
F(x) ∙ dx = −
C
F(x) ∙ dx
{{intmath|varointclockwise|''C''}} ''F''('''x''') ∙ ''d'''''x''' = −{{intmath|ointctrclockwise|''C''}} ''F''('''x''') ∙ ''d'''''x'''

V
EdS = 1/ε0
V
ρ dV

V
BdS = 0

S
Edx = −
S
B/tdS

S
Bdx =
S
(μ0J + 1/c2E/t) ∙ dS
{{intmath|oiint|∂''V''}} '''E''' ∙ ''d'''''S''' = {{sfrac|1|''ε''<sub>0</sub>}}{{intmath|iiint|''V''}} ''ρ'' ''dV''
{{intmath|oiint|∂''V''}} '''B''' ∙ ''d'''''S''' = 0
{{intmath|oint|∂''S''}} '''E''' ∙ ''d'''''x''' = −{{intmath|iint|''S''}} {{sfrac|∂'''B'''|∂''t''}} ∙ ''d'''''S'''
{{intmath|oint|∂''S''}} '''B''' ∙ ''d'''''x''' = {{intmath|iint|''S''}} (''μ''<sub>0</sub>'''J''' + {{sfrac|1|''c''<sup>2</sup>}}{{sfrac|∂'''E'''|∂''t''}}) ∙ ''d'''''S''' 

### {{math}}

Γ(z) =
0
ettz − 1dt
{{math|Γ(''z'') {{=}} {{intmath|int|0|∞}} ''e''<sup>−''t''</sup>''t''<sup>''z'' − 1</sup>''dt''}}

C
F(x) ∙ dx = −
C
F(x) ∙ dx
{{math|{{intmath|varointclockwise|''C''}} ''F''('''x''') ∙ ''d'''''x''' {{=}} −{{intmath|ointctrclockwise|''C''}} ''F''('''x''') ∙ ''d'''''x'''}}

V
EdS = 1/ε0
V
ρ dV

V
BdS = 0

S
Edx = −
S
B/tdS

S
Bdx =
S
(μ0J + 1/c2E/t) ∙ dS
{{math|{{intmath|oiint|∂''V''}} '''E''' ∙ ''d'''''S''' {{=}} {{sfrac|1|''ε''<sub>0</sub>}}{{intmath|iiint|''V''}} ''ρ'' ''dV''}}
{{math|{{intmath|oiint|∂''V''}} '''B''' ∙ ''d'''''S''' {{=}} 0}}
{{math|{{intmath|oint|∂''S''}} '''E''' ∙ ''d'''''x''' {{=}} −{{intmath|iint|''S''}} {{sfrac|∂'''B'''|∂''t''}} ∙ ''d'''''S'''}}
{{math|{{intmath|oint|∂''S''}} '''B''' ∙ ''d'''''x''' {{=}} {{intmath|iint|''S''}} (''μ''<sub>0</sub>'''J''' + {{sfrac|1|''c''<sup>2</sup>}}{{sfrac|∂'''E'''|∂''t''}}) ∙ ''d'''''S'''}}