Question

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

Answer 1

We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+......+f((a+(n-1\left)h\right)]$, where nh=b-a
Here, a=0, b=5 and nh =5-0=5
and $f\left(x\right)=(x+1)\Rightarrow f\left(0\right)=1$
$\therefore {\int }_0^5(x+1)dx={lim}_{h\to 0}h[1+(1+h)+(1+2h)+....+(1+(n-1)h\left)\right]$

$={lim}_{h\to 0}h[n+h\frac{n(n-1)}{2}](\because \sum n=\frac{n(n+1)}{2})and\sum 1=n\therefore \sum (n-1)=\frac{(n-1)n}{2}={lim}_{h\to 0}[nh+h^2\frac{n(n-1)}{2}]={lim}_{h\to 0}[nh+\frac{(h^2{n}^2-h^{2}n)}{2}]={lim}_{h\to 0}[hn+\frac{(hn-h)\left(hn\right)}{2}]$ $={lim}_{h\to 0}[5+\frac{(5-h)\left(5\right)}{2}][Here,nh=5]=5+\frac{5\times 5}{2}=\frac{35}{2}$