Question

$\underset{0}{\overset{3}{\int }}\left(2x^2+3x+5\right)dx$

Answer 1

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{\lim }h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)...............+f\left\{a+\left(n-1\right)h\right\}\right] \\ \mathrm{where}h=\frac{b-a}{n} \end{gathered}$$

$$\begin{gathered} \mathrm{Here}a=0,b=3,f\left(x\right)=2x^2+3x+5,h=\frac{3-0}{n}=\frac{3}{n} \\ \mathrm{Therefore}, \\ I={\int }_0^3\left(2x^2+3x+5\right)dx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+....................+f\left\{0+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[\left(0+0+5\right)+\left(2h^2+3h+5\right)+...............+\left\{2{\left(n-1\right)}^2{h}^2+3\left(n-1\right)h+5\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[5n+2h^2\left(1^2+2^2+3^2.........+{\left(n-1\right)}^2\right)+3h\left\{1+2+.......+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[5n+2h^2\frac{n\left(n-1\right)\left(2n-1\right)}{6}+3h\frac{n\left(n-1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }\frac{3}{n}\left[5n+\frac{3\left(n-1\right)\left(2n-1\right)}{n}+\frac{9\left(n-1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }3\left[5+3\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{9}{2}\left(1-\frac{1}{n}\right)\right] \\ =15+18+\frac{27}{2} \\ =\frac{93}{3} \end{gathered}$$