# Strange result regarding riemann sums. [closed]

Given a curve $$y = \frac{1}{t}$$, for $$t>0$$, show that for $$x>0$$, $$0<\frac {1}{3\ln(x)}<\sqrt[3]{x}$$.

I know that $$\int _1^{\sqrt[3]{x}}\frac{dt}{t}\:=\ln \:\sqrt[3]{x}$$ but the width of the rectangle for $$1 is $$\sqrt[3]{x}-1$$

So how can $$0<\frac {1}{3\ln(x)}<\sqrt[3]{x}$$?

It should be $$0<\frac {1}{3\ln(x)}<\sqrt[3]{x}-1$$

We have $$\sqrt[3]x-1 < \sqrt[3]x$$.
If you have $$0 < 1/3 \ln (x)< \sqrt[3]x-1$$, then we must have
$$0 < 1/3 \ln (x)< \sqrt[3]x$$