Closed-form representation of area below a function

Question

Let $f:[0,1]\rightarrow[0,1]$. $f$ is strictly increasing in $x$ if $x\in[0,x^*]$ and strictly decreases otherwise.

Suppose that I'm interested in finding the area of the domain where $f(x)\leq t$. That is, if $$t=f(x)=f(x+y)$$ for some $t$, $x\leq x^*$ and $x+y\geq x^*$, I want to find $y$.

Is there any closed form representation of $y$ in terms of $f$?

Further, is there any way that I can find $\frac{dy}{dt}$? Any suggestions?

score 1 · Accepted Answer · answered Dec 11 '18 at 18:39

1

Let's write $$f(x)=\begin{cases}f_1(x),x\in[0,x^*]\\f_2(x),x\in(x^*,1]\end{cases}$$ We know that both these branches are invertible (one is strictly increasing, one of them is strictly decreasing), so we can calculate $f_{1,2}^{-1}(y)$. If you sketch some graph of $f$, draw the line parallel to $x$ axis through $t$. You notice that it intersects $f_1$ at $(f_1^{-1}(t),t)$ and $f_2$ at $(f_2^{-1}(t),t)$. From these $$y=f_2^{-1}(t)-f_1^{-1}(t)$$ You can now take the derivative to get $\frac{dy}{dt}$.

In the same sketch of $f$, you can write the area under the curve in terms of integral along $y$ axis as well: $$A=\int_0^t(f_2^{-1}(y)-f_1^{-1}(y))dy-yt$$

answered Dec 11 '18 at 18:39

Andrei

37,370

Thanks @Andrei. Can we define $\frac{dy}{dt}$ at $t^=f(x^)$? If not, can we define a limit that is approached from above, say $\lim_{t\rightarrow t^*}y'(t)$? – Greenteamaniac Dec 12 '18 at 01:55
Note that $t^*$ is a maximum, so you can only approach it from below. You can define the limit, but there is no guarantee that it exists – Andrei Dec 12 '18 at 02:08
Alright, thanks! – Greenteamaniac Dec 12 '18 at 02:48

Closed-form representation of area below a function

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