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Calculate $\int_0^1f(x)dx$,where $$\ f(x) = \left\{ \begin{array}{l l} 0 & \quad \text{if $x=0$ }\\ n & \quad \text{if $x\in(\frac{1}{n+1},\frac{1}{n}]$} \end{array} \right.$$

How we can calculate this integral?

Is this simply

$$\int_0^1f(x)dx=\int_{\frac{1}{n+1}}^{\frac{1}{n}}ndx=n\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{n+1}$$?

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  • $\begingroup$ I think $n$ is intended to run through all positive integers. $\endgroup$ – Daniel Fischer Oct 21 '14 at 20:54
  • $\begingroup$ If $n$ is fixed, then yes. $\endgroup$ – Tom Oct 21 '14 at 20:58
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$$(0,1]=\bigcup_{n=1}^\infty \left(\frac1{n+1},\frac1n\right]$$ so

$$\int_0^1 f(x)dx=\sum_{n=1}^\infty\int_{1/n+1}^{1/n}ndx=\sum_{n=1}^\infty\frac1{n+1}=\infty$$

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