Question

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

Answer 1

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

$$\begin{gathered} \mathrm{Here}a=0,b=2,f\left(x\right)=x^2+x,h=\frac{2-0}{n}=\frac{2}{n} \\ \mathrm{Therefore}, \\ I={\int }_0^2\left(x^2+x\right)dx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+....................+f\left\{0+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[\left(0+0\right)+\left(h^2+h\right)+...............+\left\{{\left(n-1\right)}^2{h}^2+h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[h^2\left(1^2+2^2+3^2.........+{\left(n-1\right)}^2\right)+h\left\{1+2+3........+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[h^2\frac{n\left(n-1\right)\left(2n-1\right)}{6}+h\frac{n\left(n-1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }\frac{2}{n}\left[\frac{2\left(n-1\right)\left(2n-1\right)}{3n}+n-1\right] \\ =\underset{n\to \infty }{\lim }2\left[\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+1-\frac{1}{n}\right] \\ =\frac{8}{3}+2 \\ =\frac{14}{3} \end{gathered}$$