logo

Ask Questions, Get Answers

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

By using the properties of definite integrals,evaluate the integral\[\int\limits_0^4\mid x-1\mid dx\]

$\begin{array}{1 1} 5 \\ 4\\ 3\\ 0 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
  • (iii)$\int \limits_a^c |x-a|=> (x-a) \leq 0\;when\;0 \leq x \leq a\;and (x-a) \geq0,\;when a \leq x \leq c$
Given $I=\int\limits_0^4\mid x-1\mid dx$
 
Here $(x-1) \leq 0\;when\;0 \leq x \leq 1\;and (x-1) \geq0,\;when 1 \leq x \leq4$
 
$I=\int \limits_0^1 |x+1|dx+\int \limits_1^4 |x-1|dx$
 
$=\int \limits_0^1 -(x+1)dx+\int \limits_0^4 (x-1)dx$
 
On integrating we get,
 
$-\bigg[\frac{x^2}{2}-x\bigg]_0^1+\bigg[\frac{x^2}{2}-x\bigg]_1^4$
 
On applying limits,
 
$I=-\bigg[\frac{1}{2}-1\bigg]+\bigg[(\frac{4^2}{2}-4)-(\frac{1^2}{2}-1)\bigg]$
 
$=\frac{1}{2}+4+\frac{1}{2}=5$
 
$I=5$

 

answered Feb 14, 2013 by meena.p
 

Related questions

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
...