Question

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

Answer 1

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

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