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This problems is a straight application of the right hand sum approximation, use
x[i] = a + (b-a)/n * i = 0 + (4-0)/n * i = 4i/n
where i = 1...n for the right handed limit. Notice that dx = delta x = (b-a)/n = 4/n. Substituting these results into the integrand without writing all the limit and summation pieces we have
(x^2/3+2) dx = [ (4i/n)^2 *(1/3) + 2 ] (4/n) = 64/[3 n^3] * i^2 + 8/n
Plugging this result into the summation, without writing the limit piece yet we have using the relationships
sum(i^2) = (1/6) [ n (n+1) (2n+1)] and sum(8) = 8 * n
where the indices are from i = 1 to n. Substuting this result and our integrand result we have
(8/9)*(1/n^2) * [17 n^2 + 12n + 4]
Notice in the limit as n goes to infinity the only surviving term is the term with n^2 over n^2, which yields
(8/9) * (17/1) = 136/9
I hope that answers the question
x[i] = a + (b-a)/n * i = 0 + (4-0)/n * i = 4i/n
where i = 1...n for the right handed limit. Notice that dx = delta x = (b-a)/n = 4/n. Substituting these results into the integrand without writing all the limit and summation pieces we have
(x^2/3+2) dx = [ (4i/n)^2 *(1/3) + 2 ] (4/n) = 64/[3 n^3] * i^2 + 8/n
Plugging this result into the summation, without writing the limit piece yet we have using the relationships
sum(i^2) = (1/6) [ n (n+1) (2n+1)] and sum(8) = 8 * n
where the indices are from i = 1 to n. Substuting this result and our integrand result we have
(8/9)*(1/n^2) * [17 n^2 + 12n + 4]
Notice in the limit as n goes to infinity the only surviving term is the term with n^2 over n^2, which yields
(8/9) * (17/1) = 136/9
I hope that answers the question