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?
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?
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:
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)g\,'(u)\,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.
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,
$\int \limits_{a}^{b} f(x) \,dx =F(x) |^b_a =F(b)-F(a)$