Question

$$\underset{1}{\overset{3}{\int }}(2x+3)dx$$

Answer 1

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

$$\text{Here, }a=1,b=3,f\left(x\right)=2x+3,h=\frac{3-1}{n}=\frac{2}{n}$$

Therefore,

$$I={\int }_1^3(2x+3)dx$$
$$=\underset{h\to 0}{lim}h[f\left(1\right)+f(1+h)+....................+f\{1+(n-1)h\}]$$
$$=\underset{h\to 0}{lim}h[(2+3)+(2+2h+3)+...............+\{2+2(n-1)h+3\}]$$
$$=\underset{h\to 0}{lim}h[5n+2h\{1+2+3.........+(n-1)\}]$$
$$=\underset{h\to 0}{lim}h[5n+2h\frac{n(n-1)}{2}]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\frac{2}{n}[5n+2n-2]$$
$$=\underset{n\to \mathrm{\infty }}{lim}2(7-\frac{2}{n})$$
$$=14$$