Find $\int\limits_{0}^{2} (x^2 + 1) \, dx$ as the limit of a sum.

$\frac{14}{3}$

The correct answer is Option (1) → $\frac{14}{3}$

By definition

$\int\limits_{a}^{b} f(x) \, dx = (b - a) \lim\limits_{n \to \infty} \frac{1}{n} [f(a) + f(a + h) + \dots + f(a + (n - 1)h)],$

where, $\displaystyle h = \frac{b - a}{n}$

In this example, $a = 0, b = 2, f(x) = x^2 + 1, h = \frac{2 - 0}{n} = \frac{2}{n}$

Therefore,

$\int\limits_{0}^{2} (x^2 + 1) \, dx = 2 \lim\limits_{n \to \infty} \frac{1}{n} [f(0) + f\left(\frac{2}{n}\right) + f\left(\frac{4}{n}\right) + \dots + f\left(\frac{2(n - 1)}{n}\right)]$

$= 2 \lim\limits_{n \to \infty} \frac{1}{n} \left[1 + \left(\frac{2^2}{n^2} + 1\right) + \left(\frac{4^2}{n^2} + 1\right) + \dots + \left(\frac{(2n - 2)^2}{n^2} + 1\right)\right]$

$= 2 \lim\limits_{n \to \infty} \frac{1}{n} [\underbrace{(1 + 1 + \dots + 1)}_{n\text{-terms}} + \frac{1}{n^2} (2^2 + 4^2 + \dots + (2n - 2)^2)]$

$= 2 \lim\limits_{n \to \infty} \frac{1}{n} [n + \frac{2^2}{n^2} (1^2 + 2^2 + \dots + (n - 1)^2)]$

$= 2 \lim\limits_{n \to \infty} \frac{1}{n} [n + \frac{4}{n^2} \frac{(n - 1)n(2n - 1)}{6}]$

$= 2 \lim\limits_{n \to \infty} [1 + \frac{2}{3} \frac{(n - 1)(2n - 1)}{n^2}]$

$= 2 \lim\limits_{n \to \infty} [1 + \frac{2}{3} (1 - \frac{1}{n})(2 - \frac{1}{n})] = 2 \left[1 + \frac{4}{3}\right] = \frac{14}{3}$