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Answer 1 · 2018-03-30T00:51:32

#11/28#

The Riemann sum for an integral #int _a^b f(x)dx# is

#int _a^b f(x)dx = lim_{n to infty } [h sum_{i=1}^n f(a+ih)],qquad h = (b-a)/n#

Our sum is

# lim_{n to infty } sum_{i=1}^n [i^6/n^7+i^3/n^4] = lim_{n to infty } 1/n sum_{i=1}^n [(i/n)^6+(i/n)^3] #

This is in the form of the Riemann sum if we identify

#h = 1/n, implies b-a = 1#

and

#f(x) = x^6+x^3, qquad a = 0#

Thus

# lim_{n to infty } sum_{i=1}^n [i^6/n^7+i^3/n^4] = int_0^1 (x^6+x^3)dx = (x^7/7+x^4/4)_0^1 #
#qquad = 1/7+1/4 = 11/28#