Question

$\underset{0}{\overset{3}{\int }}\left(x^2+1\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Here},a=0,b=3,f\left(x\right)=x^2+1,h=\frac{3-0}{n}=\frac{3}{n} \\ \mathrm{Therefore}, \\ {\int }_0^3\left(x^2+1\right)dx=\underset{h\to 0}{\lim }h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+............+f\left(a+\left(n-1\right)h\right)\right] \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+..........+f\left(0+\left(n-1\right)h\right)\right] \\ =\underset{h\to 0}{\lim }h\left[0+1+h^2+1+{\left(2h\right)}^2+1+.........+{\left(\left(n-1\right)h\right)}^2+1\right] \\ =\underset{h\to 0}{\lim }h\left[n+h^2\left(1^2+2^2+..............{\left(n-1\right)}^2\right)\right] \\ =\underset{h\to 0}{\lim }h\left[n+h^2\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right] \\ =\underset{h\to 0}{\lim }\left[3+\frac{9}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right] \\ =3+9=12 \end{gathered}$$