Question

Find the value of definite integrals, as the limit of a sum (by first principle).
${\int }_0^2(x+4)dx$

Answer 1

${\int }_0^2(x+4)dx$
Here $f\left(x\right)=x+4,a=0,b=2$
and $nh=b-a=2-0=2$
$\therefore {\int }_0^2(x+4)dx$
$=\underset{h\to 0}{lim}h\left[f\right(0+h)+f(0+2h)+...+f(0+nh\left)\right]$
$=\underset{h\to 0}{lim}h\left[f\right(h)+f(2h)+...+f(nh\left)\right]$
$=\underset{h\to 0}{lim}h\left[\right(h+4)+(2h+4)+...+(nh+4\left)\right]$
$=\underset{h\to 0}{lim}h\left[h\right(1+2+...+n)+(4+4+...+4\left)\right]$
$=\underset{h\to 0}{lim}h[h\left(\frac{n(n+1)}{2}\right)+4n]$
$=\underset{h\to 0}{lim}[\frac{nh(nh+h)}{2}+4nh]$
$=\underset{h\to 0}{lim}[\frac{2(2+h)}{2}+4\times 2]$
$=(2+0)-8=10$