Question

$\underset{-1}{\overset{1}{\int }}\left(x+3\right)dx$

Answer 1

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

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