Question

Evaluate the following definete integrals as limit of sums.
${\int }_4^1(x^2-x)dx.$

Answer 1

We know that ${\int }_a^{b}f\left(x\right)dx={lim}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+....+f\{a+(n-1\left)\right\}h]$ where nh =b-a
Given, ${\int }_1^4(x^2-x)dx,\text{here}a=1,b=4\text{and}nh=3\text{and}f\left(x\right)=x^2-x=x(x-1)$
$\therefore {\int }_1^4(x^2-x)dx={lim}_{h\to 0}h\left[f\right(1)+f(1+h)+f(1+2h)+....+f(1+(n-1)h\left)\right]={lim}_{h\to 0}h\left[1\right(1-1)+(1+h)h+(1+2h\left)\right(2h)+....+(1+(n-1)h\left)\left\{\right(n-1\left)h\right\}\right][\because f(x)=x(x-1\left)\right]f\left(1\right)=1(1-1);f(1+h)=(1+h)(1+h-1)=(1+h)h$ and so on]
$={lim}_{h\to 0}{h}^2\left[\right(1+h)+2(1+2h)+3(1+3h)+...+(n-1\left)\right(1+(n-1)h\left)\right]={lim}_{h\to 0}{h}^2\left[\right(1+2+3+...+(n-1))+\{h+2^{2}h+3^{2}h+...+(n-1{)}^{2}h\}]={lim}_{h\to 0}{h}^2\{\frac{(n-1)n}{2}+h(1^2+2^2+....+(n-1{)}^2)\}[\because 1+2+3+....(n-1)=\sum (n-1)=\frac{n(n-1)}{2}]={lim}_{h\to 0}\{\frac{(n-1)n{h}^2}{2}+\frac{h^3(n-1)n(2n-1)}{6}\}$

$\{\because \sum n^2=\frac{n(n+1)(2n+1)}{6}\}\therefore \sum (n-1{)}^2=\frac{(n-1)n(2n-1)}{6}={lim}_{h\to 0}\{\frac{(nh-h)\left(nh\right)}{2}+\frac{(nh-h)\left(hn\right)(2nh-h)}{6}\}={lim}_{h\to 0}\{\frac{(3-h)3}{2}+\frac{(3-h)3(2\times 3-h)}{6}\}[\because nh=3]=\frac{9}{2}+\frac{3\times 3\times 6}{6}=\frac{9}{2}+9=\frac{27}{2}$

Note While determine the value of definite integrals by the method of sum of a limits, this value can be checked by simple integration method, to avoid the any chance of error.