Given $f(x)=big(frac{1}{x}big)^{1/3}$. Is the area bounded by the function and the $x$ axis finite?

Consider the function $f(x)=Big(frac{1}{x}Big)^{1/3}$ with $xin[-1,1]$.

I want to find out wether the area bounded by the function and $x$ axis is finite?

Using simple strategy (i.e integrating $f(x)$ from $-1$ to $0$ and then from $0$ to $1$ and taking the absolute values) results in a finite area.

But the doubt is: Since the function $f(x)$ is discontinuous at $0$, the function doesn’t even exist at $0$. So how can the area be finite?

Answer

As long as $x in [-1, 0) cup (0,1]$ you calculate

$$lim_{t to 0^{+}}int_{t}^{1} frac{1}{x^{1/3}} dx = lim_{t to 0^{+}} frac{3x^{2/3}}{2}Bigg|_t^1 = frac{3}{2} – frac{3}{2}lim_{t to 0^{+}} x^{2/3} = color{#05f}{frac{3}{2}}$$

Similar approach to $t to 0^{-}$.

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