Question
Evaluate $\int\limits_0^{3}$(2x2 + 3x + 5)dx as limit of a sum.
$\therefore$ I = $\lim\limits_{\text{ h} \to 0}\text{h}\cdot[\text{f(0) + f(h) + f(2h) + f(3h) +.......+ f}\left\{\overline{\text{(n - 1)}}\text{h}\right\}]$
=$\lim\limits_{\text{h} \to 0}\\_{\text{h} \to \infty}$ $\frac{3}{\text{n}}\cdot$[(5) + (2h2 + 3h + 5) + (2.22h2 + 3.2h + 5) + .........+ {2(n - 1)2 h2 + 3 (n - 1) h + 5 }] =$\lim\limits_{\text{n} \to \infty}$ $\frac{3}{\text{n}}\cdot$ [(5 + 5 + 5 +.........n terms) + 2h2 {12 + 22 + 32 + ........+ (n - 1)2} + 3h {1 + 2 + 3 + ..... (n - 1)}] = $\lim\limits_{\text{n} \to \infty}$ $\frac{3}{\text{n}}\cdot$ $\Bigg[\text{5n+2}\cdot\frac{9}{\text{n}^{2}}\cdot\text{n}\frac{\text{(n - 1)(2n - 1)}}{6}+\frac{3.3}{\text{n}}\cdot\frac{\text{n(n - 1)}}{2}\Bigg]$ = $\lim\limits_{\text{n} \to \infty}$ $3\Bigg[5+3\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Bigg]$ = $3\Big[5+6+\frac{9}{2}\Big]=\frac{93}{2}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.