Question

Evaluate the given definite integrals as limit of sums:
${\int }_0^5(x+1)dx$

Answer 1

Let $I={\int }_0^5(x+1)dx$
It is known that,
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\left[f\right(a)+f(a+h)....f(a+(n-1)h]$, where $h=\frac{b-a}{n}$
Here, $a=0,b=5$, and $f\left(x\right)=(x+1)$
$\Rightarrow h=\frac{5-0}{n}=\frac{5}{n}$
$\therefore {\int }_0^5(x+1)dx=(5-0)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f\left(\frac{5}{n}\right)+...+f\left(\right(n-1\left)\frac{5}{n}\right)]$
$=5\underset{n\to \infty }{lim}\frac{1}{n}[1+(\frac{5}{n}+1)+...\{1+\left(\frac{5(n+1)}{n}\right)\}]$
$=5\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(1+\underset{ntimes}{1+1}.....1)+[\frac{5}{n}+2\cdot \frac{5}{n}+3\cdot \frac{5}{n}+...(n-1\left)\frac{5}{n}\right]]$
$=5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5}{n}\{1+2+3....(n-1\left)\right\}]$
$=5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5}{n}\cdot \frac{(n-1)n}{2}]$
$=5\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{5(n-1)}{2}]$
$=5\underset{n\to \infty }{lim}[1+\frac{5}{2}(1+\frac{1}{n})]$
$=5[1+\frac{5}{2}]$
$=5\left[\frac{7}{2}\right]=\frac{35}{2}$