Let [tex]f(x)=\frac{1}{x} , x \geqq 0[/tex](adsbygoogle = window.adsbygoogle || []).push({});

[tex]\therefore \int_{0}^{\infty} f(x) dx = \infty[/tex] (1)

However the function f(x) has an inverse function of f(x), ie. the function f(x) when m,irrored around y=x is the same shape. Does this imply that when x<1 f(x) approaches [itex]\infty[/itex] as fast as f(x) approaches 0 when x>1?

If so wouldn't:

[tex]\int_{0}^{\infty} f(x) dx = 2 \times \int_{1}^{\infty} f(x) dx (2)[/tex]

I understand mathematically that (1) is correct, but when I think about the areas logically I can't understand how lines that approach the respective axis at what appears to me to be the same rate can have one part (x>1) with a non-infinite area and yet have a part ([itex]0 \leqq x < 1[\itex]) which has an infinite area.

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# Infinite area of f(x)=1/x

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