find the definite integral that is equivalent to lim n->infinity of sum over i=(1,n) n/(n^2+i^2)
since definite integral of f(x) dx over (a,b) = lim n->infinity sum over i=(1,n) f(a + (b-a)i/n) * (b-a)/n
then: f(a + (b-a)i/n) * (b-a)/n = n/(n^2+i^2)
rearrange to: so f(a + (b-a)i/n) = 1/(b-a) * n^2/(n^2 + i^2)
if (b-a)=1, then this simplifies to: f(a + i/n) = n^2/(n^2 + i^2)
I'd guess a=1, so that: f((n + i)/n) = n^2/(n^2 + i^2)
This almost works where f(x) = 1/x^2, but not quite. I'm not sure what else to do here.
To find the definite integral that is equivalent to the given limit, we can start by recognizing the form of the sum as a Riemann sum:
∫[a,b] f(x) dx ≈ lim n→∞ ∑[i=1,n] f(xi) Δx,
where Δx = (b-a)/n is the width of each subinterval and xi is a sample point in the ith subinterval.
In this case, we have f(x) = n/(n^2 + x^2) and we want to determine the equivalent definite integral with limits [a,b].
Let's set (b-a) = 1 for simplicity. Then we need to find a value of a for which f(x) matches our given function.
From the rearranged expression f(a + i/n) = n^2/(n^2 + i^2), we can see that f(1 + i/n) = n^2/(n^2 + i^2).
So, we can consider a = 1. Then our function becomes f((n + i)/n) = n^2/(n^2 + i^2).
However, as you noticed, this function does not match the form of f(x) = 1/x^2. Therefore, there may be an error or misunderstanding in the initial given problem.
To proceed further, we would need clarification on the original problem or any additional information. Could you please provide more context or clarify the question?