Question

Answer 1

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

$$\begin{gathered} \mathrm{Here}a=3,b=5,f\left(x\right)=2-x,h=\frac{5-3}{n}=\frac{2}{n} \\ \mathrm{Therefore}, \\ I={\int }_3^5\left(2-x\right)dx \\ =\underset{h\to 0}{\lim }h\left[f\left(2\right)+f\left(2+h\right)+...+f\left(2+\left(n-1\right)h\right)\right] \\ =\underset{h\to 0}{\lim }h\left[\left(2-2\right)+\left(2-h-2\right)+...+\left(2-\left(n-1\right)h-2\right)\right] \\ =\underset{h\to 0}{\lim }h\left[-h\left(1+2+3+...+\left(n-1\right)\right)\right] \\ =\underset{h\to 0}{\lim }h\left[-2h\frac{n\left(n-1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }\frac{2}{n}\left[-2n+2\right] \\ =\underset{n\to \infty }{\lim }2\left(-2+\frac{2}{n}\right) \\ =-4 \end{gathered}$$