Question:

By definition $\int_{a}^{b} f(x) dx = (b-a) \lim_{n \to \infty} \frac{1}{n} [f(a) + f(a+h) + \dots + f(a+(n-1)h)]$, where $h = \frac{b-a}{n}$.<br>Here $a = 0, b = 2, f(x) = x^2 + 1, h = \frac{2-0}{n} = \frac{2}{n}$.<br>Therefore, $\int_{0}^{2} (x^2 + 1) dx = 2 \lim_{n \to \infty} \frac{1}{n} \left[ f(0) + f\left(\frac{2}{n}\right) + f\left(\frac{4}{n}\right) + \dots + f\left(\frac{2(n-1)}{n}\right) \right]$<br>$= 2 \lim_{n \to \infty} \frac{1}{n} \left[ 1 + \left(\frac{2^2}{n^2} + 1\right) + \left(\frac{4^2}{n^2} + 1\right) + \dots + \left(\frac{(2n-2)^2}{n^2} + 1\right) \right]$<br>$= 2 \lim_{n \to \infty} \frac{1}{n} \left[ \underbrace{(1 + 1 + \dots + 1)}_{n\text{-terms}} + \frac{1}{n^2} (2^2 + 4^2 + \dots + (2n-2)^2) \right]$<br>$= 2 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{2^2}{n^2} (1^2 + 2^2 + \dots + (n-1)^2) \right]$<br>$= 2 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{2^2}{n^2} \frac{(n-1)n(2n-1)}{6} \right] = 2 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{2}{3} \frac{(n-1)(2n-1)}{n} \right]$<br>$= 2 \lim_{n \to \infty} \left[ 1 + \frac{2}{3} \left(1 - \frac{1}{n}\right) \left(2 - \frac{1}{n}\right) \right] = 2 \left[ 1 + \frac{4}{3} \right] = \frac{14}{3}$.