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As we are evaluating from 0 to 4, x will be evaluated at 4i/n for i = 1 to n. f(x i) = (4i/n) 2 /3 - 6. ∑ 1 n 4/n ((4i/n) 2 /3 - 6) =. 4/n ((16/(3n 2)n(n+1)(2n+1)/6 - 6n) = 64/9 + 32/(3n)+ 32/(9n 2)- 24= -152/9 + 32/(3n) + 32/(9n 2). The limit as n goes to infinity is -152/9

Since f(π) = π^2 < 100, f(10π) = 100π^2 > 100, and f is continuous on the closed interval [π, 10π], the intermediate value theorem ensures the existence of a number c in (π, 10π) such that f(c) = 100.

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