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What does it mean when you have notation like this

$$F\bigl(g(u)\bigr)\Bigr|_0^t,$$ where $F$ and $g$ are some general functions?

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4  
sounds to me like an evaluation statement for $F(g(u))$. That is, $F(g(u))|^t_0 = F(g(t)) - F(g(0))$. Without context it's hard to tell more. –  user20266 Apr 1 '12 at 12:40
    
I agree with @Thomas. if there's just a subscript, it means to substitute that value for the function argument. if there's also a suberscript, then it means to take the difference of the function evaluated at the top value and at the bottom: $$f(x)|_a=f(x)|_{x=a}=f(a)$$ $$F(x)|_a^b=\left.F(x)\right|_{x=a}^{x=b}=F(b)-F(a)$$ Using the explicit x= is good form but only necessary if it is not obvious from the context. –  bgins Apr 1 '12 at 12:47
    
Yes that looks right to me. Thank you –  user1066113 Apr 1 '12 at 12:49
    
@bgins: maybe that can be an answer? –  J. M. Apr 10 '12 at 8:31

2 Answers 2

up vote 3 down vote accepted

There are actually two equivalent notations in common use: matching square brackets, or a single vertical line on the right-hand-side of an expression; a matching vertical line on the left is not used because it would be confused with taking the absolute value. The usual situations where they are needed are:

  • evaluating the antiderivative of a definite integral at the upper and lower limits (as in the statement or application of the fundamental theorem of calculus)
  • evaluating any expression by substituting a value for a variable

In the former case, a superscript is used for the upper limit and a subscript for the lower limit (as in the fundamental theorem of calculus): $$ \int_a^b f(x)\,dx = \Bigl[ F(x) \Bigr]_a^b = F(b)-F(a) $$ $$ \int_a^b\frac{dx}{x} = \Bigl[\log|x|\Bigr]_a^b = \log|b|-\log|a| = \log\,\left|\frac{b}{a}\right| \qquad(ab>0) $$

while in the latter case, a single subscript is used (as in a simple evaluation by substition of a function/expression, or the definition of the Taylor series): $$ f(a) = f(x)\Big|_a = f(x)\Big|_{x=a} $$ $$ f(x) = \sum_{n=0}^{\infty} a_n(x-a)^n \quad\text{for}\quad a_n =\frac{f^{(n)}(a)}{n!} = \left.\frac1{n!}{f^{(n)}}(x)\right|_{x=a} $$

Note that the single subscript is simple evaluation, while in the subscript-superscript pair, the evaluation at the subscript is subtracted: $$ f(x)\Bigr|_a^b = f(x)\Bigr|_b - f(x)\Bigr|_a = f(b)-f(a) $$ and avoid common fallacies: $$ \frac{f(b)-f(a)}{b-a} = \frac{\bigl[f(x)\bigr]_a^b}{\bigl[x\bigr]_a^b} \ne \left[\frac{f(x)}{x}\right]_a^b \qquad\text{unless}\qquad f(x)=c\,x $$

The subscript must contain the value to be substituted into the expression. One can optionally (if unambiguous, as above) or one must (if otherwise unclear, as below) indicate into which variable the value is to be substituted (as in the formula for the derivative of an inverse function, integration by parts, or an integral with substitution/composition of functions): $$ y=f(x),~x=f^{-1}(y) ~\implies~ \left(f^{-1}\right)'(y)= \left.\frac{dx}{dy}\right|_{y=f(x)}= \left.\left(\frac{dy}{dx}\right)^{-1}\right|_{x=f^{-1}(y)}= \left.\frac1{f\,'(x)}\right|_{x=f^{-1}(y)} $$ $$ u=u(x),\quad v=v(x) \quad\implies\quad \int_{x=a}^{x=b} u\,dv = \Bigl[ uv \Bigr]_{x=a}^{x=b} - \int_{x=a}^{x=b} v\,du $$ $$ F\,'(x)=f(x)\quad\implies\quad \int_{a}^{b} f\left(g(u)\right)\,du =\Bigl[ F\left(g(u)\right) \Bigr]_{u=a}^{u=b} =\Bigl[ F(x) \Bigr]_{x=g(a)}^{x=g(b)} $$ the last of which is precisely your case with $a=0$ and $b=t$, with the part involving $f=F\,'$ valid if $F$ is also differentiable.

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Fundamental Theorem of Calculus (second part) :

Suppose $f(x)$ is a continuous function on $[a,b]$ and also suppose that $F(x)$ is any anti-derivative for $f(x)$ . Then,

$\int \limits_{a}^{b} f(x) \,dx =F(x) |^b_a =F(b)-F(a)$

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