logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the definite integral as limits of sums\[\int\limits_1^4(x^2-x)dx\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$
  • where $h=\large\frac{b-a}{n} $
  • (ii)$ \sum (n-1)=\frac{n(n-1)}{2} \qquad \sum(n-1)^2=\large\frac{n(n-1)(2n-1)}{6}$
Given $I=\int \limits_1^4(x^2-x)dx$
 
$f(x)=(x^2-x)$
 
Hence $I=\int \limits_1^4(x^2-x)dx=$
 
$lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$
 
Here $a=1\;and\;b=4,therefore\; h=\frac{b-a}{n}=\frac{4-1}{n}=\frac{3}{n}$
 
f(a)=(1-1)=0
 
$f(a+h)=[(1+h)^2-(1+h)]$
 
$=1^2+2h+h^2-1-h$
 
$=(h+h^2)$
 
$f(a+2h)=[(1+2h)^2-(1+2h)]$
 
$=1+4h+4h^2-1-2h$
 
$=2h+4h^2$
 
$f(a+(n-1)h)=[(1+(n-1)h)^2-(1+(n-1)h)]$
 
$=1+(n-1)^2 h^2+2(n-1)h-1-(n-1)h$
 
$=(n-1)^2 h^2+(n-1)h$
 
Now substituting the above values we get
 
$\int\limits_1^4 (x^2-x))dx=lim_{h->0}h[0+(h+h^2)+(2h+4h^2+.....(n-1)^2h^2+(n-1)h]$
 
We can write the above function as
 
$lim_{h->0}h[h+2h+3h+......(n-1)h+h^2+4h^2+.....(n-1)^2 h^2]$
 
On seperating the common factor,
 
$=lim_{h->0}h[h[(1+2+3+....(n-1)]+h^2(1^2+2^2+.....(n-1)^2]$
 
Now substitute $h=\frac{3}{n},\sum(n-1)=\frac{n(n-1)}{2}\; and\; \sum(n-1)^2=\frac{n(n-1)(n-2)}{6}$
 
As $ h \to 0 \qquad n \to \infty$
 
Hence $ \int \limits_1^4 (x^2-x)dx=\frac{3}{n} \lim _{n \to \infty}\bigg[\large\frac{3}{n}\frac{n(n-1)}{2}+\frac{9}{n^2}\frac{n(n-1)(2n-1)}
{6}\bigg]$
 
$\frac{3}{n} \lim _{n \to \infty}\bigg[\large\frac{3}{2}(n-1)+\frac{9}{n}\frac{n(n-1)(2n-1)}{6}\bigg]$
 
$=\lim _{n \to \infty}\large\frac{9}{2}(1-\frac{1}{n})+\frac{9}{2}(1-\frac{1}{n})(2-\frac{1}{n})$
 
Applying limits we get
 
$\large\frac{9}{2}+\frac{9}{2}(2)=\frac{9}{2}+9$
 
               $=\large\frac{27}{2}$

 

 

answered Feb 8, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...