$f(x)=1/(x+3)$ over the interval [0,1]
Here the function is a monotone function. So it is Riemann integrable. But is there a way I could prove that the given function is Riemann Integrable. I tried solving the problem by using partitions but I could not prove it.
Best Answer
Even easier: the integrand function is a Lipschitz function, hence for every partition of $[0,1]$ the difference between the upper sum and the lower sum is bounded by the Lipschitz constant $L$ times the width of the largest interval of the partition. Riemann integrability readily follows.