Question

Evaluate the following integral by expressing as a limit of sum
${\int }_1^2(2x+5)dx$

Answer 1

$weknow$
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx={lim}_{h\to 0}h[f\left(a\right)+f(a+h)+f(a+2h)+.......+f(a+(n-1)h)]$
$h=\frac{b-a}{n}$
$\underset{1}{\overset{2}{\int }}(2x+5)dx={lim}_{h\to 0}h[f\left(1\right)+f(1+h)+f(1+2h)+.......+f(1+(n-1)h)].......\left(i\right)$
$h=\frac{2-1}{n}=\frac{1}{n}$
$f\left(1\right)=2\left(1\right)+5=7$
$f(1+h)=2(1+h)+5=7+2h$
$f(1+2h)=2(1+2h)+5=7+4h$
$f(1+(n-1)h)=2(1+(n-1)h)+5=7+2nh-2h$
$pluggingin\left(i\right)$
$\underset{1}{\overset{2}{\int }}(2x+5)dx={lim}_{h\to 0}h[7+(7+2h)+(7+4h)+.....+(7+2nh-2h)]$
$={lim}_{h\to 0}h[7n+(2h+4h+2nh)]$
$={lim}_{h\to 0}h[7n+2h(1+2+........n)]$
$={lim}_{h\to 0}h[7n+2h\left(\frac{n(n+1)}{2}\right)]$
$={lim}_{n\to \propto }\frac{1}{n}[7n+\frac{1}{n}\left(\frac{n(n+1)}{1}\right)]$
$={lim}_{n\to \propto }\frac{1}{n}[7n+(n+1)]$
$={lim}_{n\to \propto }\frac{1}{n}[7n+n(1+\frac{1}{n})]$
$={lim}_{n\to \propto }\frac{1}{n}\left[n(7+(1+\frac{1}{n}))\right]$
$={lim}_{n\to \propto }[7+(1+\frac{1}{n})]$
$=7+1+0=8$