INTEGRALS — Maths STD 12 Science — Question

$\therefore \text{I} = \lim\limits_{ h\rightarrow 0} \bigg[ \text{f (0)} + \text{f (h)} + \text{f (2h)} +\dots\dots\dots\dots\text{+} \text{f} \overline{(\text{n} - 1)}\text{h}\bigg]$

$= \lim\limits_{ h\rightarrow 0} \text{h} \bigg[1 + (h^{2} + 2h + 1) + (2^{2}h^{2} + 2.2h + 1) + \dots\dots\dots(n- 1^{2}) h^{2} + 2 (n - 1) h + 1)\bigg]$

$= \lim\limits^{n \rightarrow \infty}_{h \rightarrow 0} \frac{2}{n} \Big[(1+1+1+1+1+\ ........)+\text{h}^2\big(1+2^2+3^2+\ ....\big)\Big]$

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

$= \lim\limits_{n \rightarrow \infty} \Big[2 + \frac{8}{6} \frac{n - 1}{n} \frac{2n - 1}{n} + 4\frac{n - 1}{n}\Big]$

$= 2 + 4 +\frac{8}{3} = \frac{26}{3}$