If I have an arbitrary positive monotonically decreasing function $f(x), x \in [0,\infty]$, is there an 'efficient' method for finding $y$ in:

$r = \int\limits_0^y f(x) dx$

for a known $r \in [0, \int\limits_0^\infty f(x) dx]$. By efficient I guess I mean more efficient than doing numerical integration until one lands in within a given distance from $r$.

I particularly care about the cases where the integral of $f$ has no closed form.

Best answer so far (rewrite of below), doubt it gets much better than this:

Start with $s_0 = r$, $y_0 = 0$, then

$y_{i+1} = y_i + \frac{s_i}{f(y_i)}$

$s_{i+1} = s_i - \int\limits_{y_i}^{y_{i+1}}f(x)dx$

3 added 87 characters in body

If I have an arbitrary positive monotonically decreasing function $f(x), x \in [0,\infty]$, is there an 'efficient' method for finding $y$ in:

$r = \int\limits_0^y f(x) dx$

for a known $r \in [0, \int\limits_0^\infty f(x) dx]$. By efficient I guess I mean more efficient than doing numerical integration until one lands in within a given distance from $r$.

I particularly care about the cases where the integral of $f$ has no closed form.

If I have an arbitrary positive monotonically decreasing function $f(x), x \in [0,\infty]$, is there an 'efficient' method for finding $y$ in:
$r = \int\limits_0^y f(x) dx$
for a known $r \in [0, f(0)]$\int\limits_0^\infty f(x) dx]$. By efficient I guess I mean more efficient than doing numerical integration until one lands in within a given distance from$r\$.