Question

Evaluate the given definite integrals as limit of sums:
${\int }_1^4(x^2-x)dx$

Answer 1

Let $I={\int }_1^4(x^2-x)dx$
$={\int }_1^4{x}^{2}dx-{\int }_1^{4}xdx$
Let $I=I_1-I_2$, where $I_1={\int }_1^4{x}^{2}dx$ and $I_2={\int }_1^{4}xdx$ .........(1)
It is known that,
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+........f(a+(n-1)h]$, where $h=\frac{b-a}{n}$
For $I_1={\int }_1^4{x}^{2}dx$,
$a=1,b=4$, and $f\left(x\right)=x^2$
$\therefore h=\frac{4-1}{n}=\frac{3}{n}$
$I_1={\int }_1^{4}dx=(4-1)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(1)+f(1+h)+.....+f(1+(n-1)h\left)\right]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[l^2+{(1+\frac{3}{n})}^2+{(1+2\cdot \frac{3}{n})}^2+...{(1+\frac{(n-1)3}{n})}^2]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[1^2+\{1^2+{\left(\frac{3}{n}\right)}^2+2\cdot \frac{3}{n}\}+....+\{1^2+{\left(\frac{(n-1)3}{n}\right)}^2+\frac{2\cdot (n-1)\cdot 3}{n}\}]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(1^2+\underset{ntimes}{....}+1^2)+{\left(\frac{3}{n}\right)}^2\{1^2+2^2+....+(n-1{)}^2\}+2\cdot \frac{3}{n}\{1+2+....+(n-1\left)\right\}]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{9}{n^2}\left\{\frac{(n-1)\left(n\right)(2n-1)}{6}\right\}+\frac{6}{n}\left\{\frac{(n-1)\left(n\right)}{2}\right\}]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{9n}{6}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{6n-6}{2}]$
$=3\underset{n\to \infty }{lim}[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3-\frac{3}{n}]$
$=3[1+3+3]=3\left[7\right]$
$I_1=21$ ............(2)
For $I_2={\int }_1^{4}xdx$,
$a=1,b=4$, and $f\left(x\right)=x$
$\Rightarrow h=\frac{4-1}{n}=\frac{3}{n}$
$\therefore I_2=(4-1)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(1)+f(1+h)+....f(a+(n-1)h\left)\right]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[1+(1+h)+....+(1+(n-1)h]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[1+(1+\frac{3}{n})+....+\{1+(n-1\left)\frac{3}{n}\right\}]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(1+\underset{ntimes}{1+....}+1)+\frac{3}{n}(1+2+....+(n-1)\left)\right]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[n+\frac{3}{n}\left\{\frac{(n-1)n}{2}\right\}]$
$=3\underset{n\to \infty }{lim}\frac{1}{n}[1+\frac{3}{2}(1-\frac{1}{n})]$
$=3[1+\frac{3}{2}]=3\left[\frac{5}{2}\right]$
$I_2=\frac{15}{2}$ ........ (3)
From equations (2) and (3), we obtain
$I=I_1-I_2=21-\frac{15}{2}=\frac{27}{2}$