Question

Express the following integral as limit of sum and hence evaluate, ${\int }_0^2(3x^2+5)dx$

Answer 1

${\int }_0^2(3x^2+5)dx$ $h=\frac{b-a}{x}=\frac{2-0}{x}=\frac{2}{x}$

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

$I={\int }_0^2(3x^2+5)dx=\underset{h\to 0}{lim}h\left[f\right(0)+f(0+h)+f(0+2h)+....+f\{0+(n-1)h\left\}\right]$

$=\underset{h\to 0}{lim}h\left[\right\{3(0{)}^2+5\}+\left\{3\right(0{)}^{2}h{)}^2+5\}+\{3(0+2h{)}^2+5\}+...$

$=\underset{h\to 0}{lim}h[5+3h^2+5+6h^2+5+9h^2+5+...+3(n-1)h^2+5]$

$=\underset{h\to 0}{lim}h[5n+3h^2[1+2+3+....+(n-1)\left]\right]$

$=\underset{h\to 0}{lim}h[5n+3h^2\times \frac{n(n-1)}{2}]$

$=\underset{n\to \infty }{lim}\frac{2}{n}[5n+3\times \frac{4}{n^2}\times \frac{n(n-1)}{2}]$

$=\underset{n\to \infty }{lim}\frac{2}{n}[5n+\frac{6(n-1)}{n}]$

$=\underset{n\to \infty }{lim}[10+\frac{12(n-1)}{n^2}]$

$=[10+0]$

$=10$