Chapter 5: Integrals - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises 5.2 - Page 266: 42

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Consider $f(x)=3x^2$ Here,we have $\Sigma_{i=1}^n (\dfrac{3}{n}) (k_i)^2=(\dfrac{3}{n^3}) \Sigma_{i=1}^n (i^2)$ This implies that $(\dfrac{3}{n^3})(\dfrac{n(n+1)(2n+1)}{6})=\dfrac{2n^3+3n^2+n}{2n^3}=\dfrac{2+\dfrac{3}{n}+\dfrac{1}{n^2}}{2}$ Now, $\lim\limits_{n \to \infty}\dfrac{2+\dfrac{3}{n}+\dfrac{1}{n^2}}{2}=1$