Question

Find the value of definite integrals, as the limit of a sum (by first principle).
${\int }_3^5(x-2)dx$

Answer 1

${\int }_3^5(x-2)dx$
According to definition
${\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{lim}\left[f\right(a+h)+f(a+2h)+f(a+3h)+...+f(a+nh\left)\right]$
Here a = 5, b = 5, f(x) = x - 2, nh = 5 - 3 = 2
$\therefore {\int }_3^5(x-2)dx$
$=\underset{h\to 0}{lim}h\left[f\right(3+h)+f(3+2h)+f(3+3h)+...+f(3+nh\left)\right]$
$=\underset{h\to 0}{lim}h\left[\right(3+h-2)+(3+2h-2)+(3+3h-2)+...+(3+nh-2\left)\right]$
$=\underset{h\to 0}{lim}h[1+h+1+2h+1+3h+...+1+nh]$
$=\underset{h\to 0}{lim}h\left[h\right(1+2+3+...+n)+(1+1+1+...+ntimes\left)\right]$
$=\underset{h\to 0}{lim}h\left(\frac{hn(n+1)}{2}\right)+n$
$=\underset{h\to 0}{lim}\frac{h(h{n}^2+hn+2n)}{2}$
$=\underset{h\to 0}{lim}\frac{h^2{n}^2+h^{2}n+2nh}{2}$
$=\underset{h\to 0}{lim}\frac{(nh{)}^2+2(nh)+h^{2}n}{2}$
$=\frac{(2{)}^2+2\times 2+0}{2}$
$=\frac{4+4}{2}=\frac{8}{2}=4$