Question

$\underset{0}{\overset{2}{\int }}\left(x^2+4\right)dx$

Answer 1

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

$$\begin{gathered} \mathrm{Here}a=0,b=2,f\left(x\right)=x^2+4,h=\frac{2-0}{n}=\frac{2}{n} \\ \mathrm{Therefore}, \\ I={\int }_0^2\left(x^2+4\right)dx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+....................+f\left\{0+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[\left(0+4\right)+\left(h^2+4\right)+...............+\left\{{\left(n-1\right)}^2{h}^2+4\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[4n+h^2\left\{1^2+2^2+3^2.........+{\left(n-1\right)}^2\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[4n+h^2\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right] \\ =\underset{n\to \infty }{\lim }\frac{2}{n}\left[4n+\frac{2\left(n-1\right)\left(2n-1\right)}{3n}\right] \\ =\underset{n\to \infty }{\lim }2\left\{4+\frac{2}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right\} \\ =8+\frac{8}{3} \\ =\frac{32}{3} \end{gathered}$$