Question

Evaluate ${\int }_0^5(x+1)dx$ as limit of sum

Answer 1

We have, ${\int }_0^5(x+1)dx=\underset{n\to \infty }{lim}h\left[\sum _{r=0}^{n-1}f(a+r.h)\right]=\underset{n\to \infty }{lim}\frac{5}{n}\left[\sum _{r=0}^{n-1}(r\left(\frac{5}{n}\right)+1)\right]$

$\because h=\frac{5-0}{n}f\left(x\right)=x+1$

$=\frac{5}{n}$

$=\underset{n\to \infty }{lim}\left[\sum _{r=0}^{n-1}\frac{25}{n^2}r+\frac{5}{n}.n-1\right]$

$=\underset{n\to \infty }{lim}[\frac{25}{n^2}\frac{n(n-1)}{2}+5\left(\frac{n-1}{n}\right)]$

$=\frac{25}{2}+5$

$=\frac{35}{2}$