Question

Evaluate the definite integral as limit of sums: $∫^4_1(x^2-x)dx$

Answer 1

The correct answer is:$\frac{27}{2}$
Let $I=∫^4_1(x^2-x)dx$
$=∫^4_1x^2 dx-∫^4_1 xdx$
Let $I=I_1-I_2$,where,$I_1=∫^4_1 x^2dx$ and $I_2=∫^4_1 x\,dx...(1)$
It is known that,
$∫_a^b ƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)+ƒ(a+(n-1)h)],where\, h=\frac{b-a}{n}$
For $I_1=∫^4_1 x^2dx,$
$a=1,b=4$,and $ƒ(x)=x^2$
$∴h=\frac{4-1}{n}=\frac{3}{n}$
$I_1=∫^4_1 x^2dx=(4-1)\underset{n→∞}{lim}\frac{1}{n}[ƒ(1)+ƒ(1+h)+...+ƒ(1+(n-1)h)]$
$=3\underset{n→∞}{lim}\frac{1}{n}[1^2+(1+\frac{3}{n})^2+(1+2.\frac{3}{n})^2+...(1+(\frac{(n-1)3}{n})^2]$
$=3\underset{n→∞}{lim}\frac{1}{n}[1^2+[1^2+(\frac{3}{n})^2+2.\frac{3}{n}]+...+[1^2+(\frac{(n-1)3}{n})^2+\frac{2.(n-1).3}{n}]]$
$=3\underset{n→∞}{lim}\frac{1}{n}[(\underset{n times}{1^2+......+1^2})+(\frac{3}{n})^2[1^2+2^2+...+(n-1)^2]+2.\frac{3}{n}[1+2+...+(n-1)]]$
$=3\underset{n→∞}{lim}\frac{1}{n}[n+\frac{9}{n^2}[\frac{(n-1)(n)(2n-1)}{6}]+\frac{6}{n}[\frac{(n-1)(n)}{2}]]$
$=3\underset{n→∞}{lim}\frac{1}{n}[n+\frac{9n}{6}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{6n-6}{2}]$
$=3\underset{n→∞}{lim}\frac{1}{n}[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3-\frac{3}{n}]$
$=3[1+3+3]$
$=3[7]$
$I_1=21...(2)$
For $I_2=∫^4_1 x\,dx,$
$a=1,b=4,$and $ƒ(x)=x$
⇒h=4-1/n=3/n
$∴I_2=(4-1)\underset{n→∞}{lim}\frac{1}{n}[ƒ(1)+ƒ(1+h)+...ƒ(a+(n-1)h)]$=3limn→∞1/n[[1+(1+h)+...+(1+(n-1)h)]
$=3\underset{n→∞}{lim}\frac{1}{n}[1+(1+\frac{3}{n})+...+{1+(n-1)\frac{3}{n}}]$
$=3\underset{n→∞}{lim}\frac{1}{n}[(\underset{n times}{1+1+...+1})+\frac{3}{n}((1+2+...+(n-1))]$
$=3\underset{n→∞}{lim}\frac{1}{n}[n+\frac{3}{n}[\frac{(n-1)n}{2}]]$
$=3\underset{n→∞}{lim}\frac{1}{n}[1+\frac{3}{2}(1-\frac{1}{n})]$
$=3[1+\frac{3}{2}]$
$=3[\frac{5}{2}]$
$I_2=\frac{15}{2}...(3)$
From equation,(2)and(3),we obtain
$I=I_1+I_2=21-\frac{15}{2}=\frac{27}{2}$