Is f Integrable and What Is the Limit of the Riemann Sum?

[skippenmydesig]

We only have the epsilon-delta definition to work with for these.

Prove that f is integrable and verify the value. On [0,1] f(x)=1 if x=1/2 else 0. \(\displaystyle \int_{0}^{1} \,f\) =0

Prove: If f is integrable on [0,1] then \(\displaystyle \lim_{{n}\to{\infty}}\ \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n})\) = \(\displaystyle \int_{0}^{1} \,f\) .

 

[chisigma]

We only have the epsilon-delta definition to work with for these.

Prove that f is integrable and verify the value. On [0,1] f(x)=1 if x=1/2 else 0. \(\displaystyle \int_{0}^{1} \,f\) =0

Prove: If f is integrable on [0,1] then \(\displaystyle \lim_{{n}\to{\infty}}\ \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n})\) = \(\displaystyle \int_{0}^{1} \,f\) .

Wellcome on MHB skippenmydesign!...

... in general the definition of Riemann Integral is...

$\displaystyle \int _{a}^{b} f(x)\ dx = \lim_{\text{max} \Delta_{k} \rightarrow 0} \sum_{k=1}^{n} f(x_{k})\ \Delta_{k}\ (1)$

... where $x_{k}$ are aritrary points in the subintervals $\Delta_{k}$ of the interval [a, b] ... in Your case all the term of the sum (1) are 0 with only one exception for the subinterval containing $x = \frac{1}{2}$ where the term is $\Delta_{k}$, so that the limit is 0...

Kind regards

$\chi$ $\sigma$