$dynamic range$ is defined as the lowest concentration at which quantitative measurements can be obtained ($LOQ$) to the concentration at which the calibration curve departs from linearity ($LOL$).
Let $\alpha = LOL$
Let $\beta = LOQ$
I have to two different approach on finding the value of the $dynamic range$. First I use subtraction,
$dynamic range$ = $\alpha$ - $\beta$.
However, the value of the $dynamic range$ results on a negative number because $\beta$ is always greater than $\alpha$.
Second, I use distance formula,
Let $(x - xo) = \alpha = LOL$
Let $(y - yo) = \beta = LOQ$
$d = \sqrt{(x - xo)^2 + (y - yo)^2}$
$d = \sqrt{(\alpha)^2 + (\beta)^2}$
I also found a value using distance formula, yet it doesn't make sense because it is outside the values of $\alpha$ and $\beta$.
For example: $\alpha = 4.219434974$
$\beta = 6.027764249$
$d = \sqrt{(\alpha)^2 + (\beta)^2}$
$d = \sqrt{(4.219434974)^2 + (6.027764249)^2}$
$d = 7.357823954$
Is it applicable to use the area under the curve to find the value of the $dynamic range$?
I post this question here at Math Stack Exchange because the question is more of mathematical concept rather than chemistry concept. Any help is highly appreciated. Thank You.
I don't have the relevant chemistry background, but after carefully reading your description, and it seems that the simple
$$\text{Dynamic Range}= \beta - \alpha = \text{LOL} - \text{LOQ}$$
suits your needs. So basically your first guess is no wrong, just that the roles are switched (when you denoted $\alpha$ as LOL erroneously.
Actually, it is also indicated in the diagram as the range of the horizontal double-arrow. Thus just keep it simple and stay horizontal. There's no need to use Pythagorean Theorem or whatever.