How do I convert lim_(n->oo) sum_(i=0)^n (12n)/(16n^2 + 9i^2) into a definite integral?

I'm new when it comes to Riemann sums. I also have trouble with finding Delta x_i .

1 Answer
Nov 3, 2017

arctan(3/4)

Explanation:

lim_(n->oo) sum_(i=0)^n (12n)/(16n^2 + 9i^2) = lim_(n->oo) sum_(i=0)^n (12/(16 + 9(i/n)^2))1/n

and calling now xi = i/n and d xi = 1/n

lim_(n->oo) sum_(i=0)^n (12/(16 + 9(i/n)^2))1/n=int_0^1 12/(16+9 xi^2)d xi = arctan(3/4)