Q.Integral from 0 to 2 of (x^2 + 1) dx as the limit of a sum. (Scores : 4)

Build the Riemann sum with n equal strips of width h = 2/n, then take the limit as n -> infinity, using the standard sum-of-squares formula.

By definition, $\displaystyle\int_a^b f(x)\,dx = \lim_{h\to0} h\big[f(a) + f(a+h) + \cdots + f(a+(n-1)h)\big]$, where $h = \dfrac{b-a}{n}$ and $n\to\infty$.

Here $a=0,\ b=2,\ f(x)=x^2+1$, so $h = \dfrac{2}{n}$.

$\displaystyle S = h\sum_{r=0}^{n-1} f(rh) = h\sum_{r=0}^{n-1}\big[(rh)^2+1\big] = h^3\sum_{r=0}^{n-1} r^2 + nh$

Using $\displaystyle\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$:

$S = h^3\cdot\dfrac{(n-1)n(2n-1)}{6} + nh$

Substitute $h=2/n$, so $nh = 2$ and $h^3 = 8/n^3$:

$S = \dfrac{8}{n^3}\cdot\dfrac{(n-1)n(2n-1)}{6} + 2 = \dfrac{4}{3}\cdot\dfrac{(n-1)(2n-1)}{n^2} + 2$

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