# Thread: Integral of a function as limit of a sequence

1. ## Integral of a function as limit of a sequence

Hey people.
Given a continuous function f(x) in the interval [a,b] , I need to prove that
$\lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) = \int_a^b f(x)dx$

I think I should do it somehow through Riemann's Integral defnition and/or use Riemann's sums , but I tried to play with it for a while and I haven't gotten anywhere....

Any ideas how should I start proving this?

2. Originally Posted by Gok2
Hey people.
Given a continuous function f(x) in the interval [a,b] , I need to prove that
$\lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n}) = \int_a^b f(x)dx$

I think I should do it somehow through Riemann's Integral defnition and/or use Riemann's sums , but I tried to play with it for a while and I haven't gotten anywhere....
I am not sure what is meant by “to prove” in this case.
You have given a Riemann approximating sum for the integral of $f\text{ on }[a,b]$.
$\lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n})$ in that expression $\Delta x = \frac{{b - a}}
{n}\,\& \,t_k = a + t_k \Delta x\,,\,k = 1,2, \cdots ,n$
.

Because $f\text{ is continuous on }[a,b]$ it is Riemann integrable and all valid approximating sums converge to the integral.

So I don’t know what there is to prove.

I see.

Thanks a lot .

4. Hey again,
In my task , after this question, I need to calculate a few limits using the formula
$\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) =\int_0^1 f(x)dx$

I managed to calculate all the limits in the task except of this limit

$\lim_{n\to \infty} \frac{\ln(2n+1)+\ln(2n+2)+...+\ln 3n}{n}-\ln n$

I tried many times but I couldn't write this expression as a sum of f(k/n)...
The problem is what's inside the ln function , i.e 2n+1 , 2n+2 , ... , 2n+k , can't be written as an expression of k/n , so I am kinda stuck....

Any ideas people ?