Question

Evaluate the following definite integrals as limit of sums.
${\int }_a^{b}xdx.$

Answer 1

We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{h\to 0}h\left[f\right(a)+f(a+h\left)\right]+f(a+2h)+...+f[a+(n-1\left)h\right]$
where, nh=b-a
Here, a=a, b=b and f(x) =x
$\therefore {\int }_a^{b}xdx={lim}_{h\to 0}h[a+(a+h)+(a+2h)+...+a+(n-1\left)h\right]={lim}_{h\to 0}h\left[\right(a+a+...+ntimes)+h(1+2+...(n-1)\left)\right]\because (\sum a=na)={lim}_{h\to 0}h[na+h(1+2+3+...+(n-1)\left)\right]={lim}_{h\to 0}[hna+h^2\frac{n(n-1)}{2}](\because \sum n=\frac{n(n+1)}{2}\therefore \sum (n-1)=\frac{(n-1)n}{2})={lim}_{h\to 0}[hna+\frac{h^2{n}^2}{2}-\frac{h^{2}n}{2}]={lim}_{h\to 0}\left[\right(b-a)a+\frac{1}{2}(b-a{)}^2-\frac{(b-a{)}^2}{n^2}.\frac{n}{2}][\therefore nh=b-a]=\left[\right(b-a)a+\frac{(b-a{)}^2}{2}][\therefore whenh\to 0,thenn\to \infty \therefore lim\frac{(b-a{)}^2}{2n}=0]=(b-a)[a+\frac{b-a}{2}]=(b-a)\left[\frac{2a+b-a}{2}\right]=(b-a)\left[\frac{a+b}{2}\right]=\frac{b^2-a^2}{2}$