Show that limit is a definite integral

Show that the given limit is a definite integral $\displaystyle \int_a^b f(x) \, dx$ for a suitable interval $\displaystyle [a,b]$ and function $\displaystyle f$

The limit is:

$\displaystyle \lim \limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n}{n^2 + i^2}$

That's the problem I can't solve. Here is how far I've made it:

Since a definite integral can be defined as

$\displaystyle \int_a^b f(x) \, dx = \lim\limits_{n \to \infty} \sum\limits_{i=1}^n f\left(a + (b-a)\frac{i}{n}\right) \cdot \frac{b-a}{n}$

Then:

$\displaystyle f\left(a + (b-a)\frac{i}{n}\right) \cdot \frac{b-a}{n} = \frac{n}{n^2 + i^2}$

which simplifies to:

$\displaystyle f\left(a + (b-a)\frac{i}{n}\right)= \frac{1}{b-a} \frac{n^2}{n^2 + i^2}$

I'm not sure how to solve, but, I can guess that $\displaystyle b-a=1$ which would simplify things to:

$\displaystyle f\left(a + \frac{i}{n}\right)= \frac{n^2}{n^2 + i^2}$

I can also guess that $\displaystyle a=1$ so that:

$\displaystyle f\left(\frac{n + i}{n}\right)= \frac{n^2}{n^2 + i^2}$

Now, I can't find a function $\displaystyle f$ that satisfies this. $\displaystyle f(x)=\frac{1}{x^2}$ almost works but not quite.

Re: Show that limit is a definite integral

Quote:

Originally Posted by

**VinceW** Show that the given limit is a definite integral $\displaystyle \int_a^b f(x) \, dx$ for a suitable interval $\displaystyle [a,b]$ and function $\displaystyle f$

The limit is:

$\displaystyle \lim \limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n}{n^2 + i^2}$

Let $\displaystyle a=0,~b=1~\&~f(x)=\frac{1}{1+x^2}$.

Re: Show that limit is a definite integral

that works. I take it that was just intuition? Not sure how I would have figured that out by myself. thanks!