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

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

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Answer 1 · 2017-11-03T13:56:14

#arctan(3/4)#

#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)#