Question

$\underset{0}{\overset{5}{\int }}\left(x+1\right)dx$

Answer 1

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

$$\begin{gathered} \mathrm{Here},a=0,b=5,f\left(x\right)=x+1,h=\frac{5-0}{n}=\frac{5}{n} \\ \mathrm{Therefore}, \\ I={\int }_0^5\left(x+1\right)dx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+....................+f\left\{0+\left(n-1\right)h\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[\left(0+1\right)+\left(h+1\right)+...............+\left\{\left(n-1\right)h+1\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[n+h\left\{1+2+3.........+\left(n-1\right)\right\}\right] \\ =\underset{h\to 0}{\lim }h\left[n+h\frac{n\left(n-1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }\frac{5}{n}\left[n+\frac{5n-5}{2}\right] \\ =\underset{n\to \infty }{\lim }5\left(\frac{7}{2}-\frac{5}{n}\right) \\ =\frac{35}{2} \end{gathered}$$