Function diverges but area under curve is finite? I have a curve given by $f(x)=\frac{x}{b^2}(1-\frac{x^2}{b^2})^{-1/2}$ for $x \in (0,b)$. Upon integration, I get that the area under the curve on the interval $(t, b)$ is given by $F(t) =\sqrt{1-\frac{t^2}{b^2}}$.
The value of $f(x)$ is $0$ for $x=0$ but grows and diverges to positive infinity as $x \to b$, but the area under the curve  on the interval $(0, b)$ gives a unit area. How can a diverging curve have a finite area? Any help?
Ps: its not a homework question.
 A: To answer your title's question, that negative sign in $f(x)=-\sqrt{1-\frac{x^2}{b^2}}$ is perfectly alright. Technically there should be a $+C$ appended. If $C\geq1$, then $f(x)\geq 0$ when $0 \leq x \leq 1$.
Also, since $f$ is increasing from $0\leq x<1$. This means any measuring of area under $f'$ for $0 \leq x \leq 1$ (for any constant $C$) will be positive and the world makes sense.
It is indeed strange that in mathematics (and in nature), things can have infinite perimeter but finite area (take fractals for instance).
Or instead, take geometric series' an example. You're adding an infinite number of terms (a diverging number of terms!) but the whole sum can converge.
In other words, even though $f$ approaches $+\infty$ as $x$ aproaches $b$, the graph attempts to hug the line $x=b$ so tightly that the area essentially vanishes as fast (or faster) than the terms of a converging geometric series do.
A: "How can a diverging curve have a finite area?" Because it can.
Let's say, more generally, that $g(x)$ is a differentiable function that has a point of divergence $b>0$.
$g(x) \rightarrow \infty$, as $x \rightarrow b$ implies the area under $g$ between $a$ and $b$ is either finite or infinite. In order to figure out which one for the specific function $g$, you have to carry out the integration between $a$ and $b$ of that function.
For example, the function you give has finite area, whereas $\tan(x)$ between $0$ and $\frac{\pi}{2}$ has infinite area.
As $x$ tends to $b$ from below(left), whether or not the area of $g(x)$ between $0$ and $x$ increases by a small amount vs a large amount, depends on how quickly the gradient of $g(x)$ increases as $x$ tends to $b$.
You could instead think of it as a sum of rectangles with width $x_i$ with $x_i \rightarrow 0$ as $i \rightarrow \infty$, and height(length) $y_i$ with $y_i \rightarrow \infty$ as $i \rightarrow \infty$. Note that this is not the usual way we do integration, but you can get the exact area this way.
Then the area is the limit of $ \sum x_i \cdot y_i$
which boils down to an indeterminate of the form $0 \times \infty$, which can be finite or infinite.
See: https://en.wikipedia.org/wiki/Indeterminate_form
