Question

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

Answer 1

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

$$Herea=0,b=2,f\left(x\right)=x^2+2x+1,h=\frac{2-0}{n}=\frac{2}{n}$$
$$Therefore,$$
$$I={\int }_0^2(x^2+2x+1)dx$$
$$=\underset{h\to 0}{lim}h[f\left(0\right)+f(0+h)+....................+f\{0+(n-1)h\}]$$
$$=\underset{h\to 0}{lim}h[(0+0+1)+(h^2+2h+1)+...............+\{{(n-1)}^2{h}^2+2(n-1)h+1\}]$$
$$=\underset{h\to 0}{lim}h[n+h^2(1^2+2^2+3^2.........+{(n-1)}^2)+2h\{1+2+.........+(n-1)h\}]$$
$$=\underset{h\to 0}{lim}h[n+h^2\frac{n(n-1)(2n-1)}{6}+2h\frac{n(n-1)}{2}]$$
$$=\underset{n\to \mathrm{\infty }}{lim}\frac{2}{n}[n+\frac{2(n-1)(2n-1)}{3n}+2n-2]$$
$$=\underset{n\to \mathrm{\infty }}{lim}2\{3+\frac{2}{3}(1-\frac{1}{n})(2-\frac{1}{n})-\frac{2}{n}\}$$
$$=6+\frac{8}{3}$$
$$=\frac{26}{3}$$