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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integrals: $\int\limits_1^4 [\; |x-1|+|x-2|+|x-3|\;]dx$

This question has appeared in modelpaper 2012

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Toolbox:
  • $\int |x+a| dx$ where $-(x+a) \geq 0\; for\; -a \leq x \leq 0\qquad (x+a) \leq 0\;for \; -a \geq x\geq 0$
  • $ \int\limits_a^b f(x)dx=\int \limits_a^b f(a-x)dx$
Given $I=\int\limits_1^4[|x-1|+|x-2|+|x-3|]dx$
 
On seperating the terms we get
 
$I=\int \limits_1^4 |x-1|\;dx+\int \limits_1^4 |x+2|\;dx+\int \limits _1^4 |x-3|\;dx$
 
(ie)$ I=I_1+I_2+I_3$
 
Consider $I_1=\int \limits_1^4 (x-1)dx$
 
where $(x+1) \geq 0\; for\; 1 \leq x \leq 4$
 
Hence the limit is 1 to 4
 
Therefore $I_1=\int \limits_1^4 (x-1)dx$
 
On integrating we get
 
$[\frac{x^2}{2}]_1^4-[x]_1^4$
 
On applying limits
 
$ [\frac{16}{2}-\frac{1}{2}]-[4-1]$
 
$=8-\frac{1}{2}-3$
 
$=\frac{9}{2}$
 
Consider $I_2=\int \limits_1^4 |x-2|dx$
 
$(x-2) \geq 0\; =>\; 2 \leq x \leq 4$
 
$(x-2) \leq 0\; => \; 1 \leq x \leq 2$
 
Hence $I-2=\int \limits_1^2 -(x-2) dx +\int \limits_2^4(x-2) dx$
 
On integrating we get
 
$-[\frac{x^2}{2}-2x]_1^2-[\frac{x^2}{2}-2x]_4^2$
 
On applying limits we get,
 
$-[\frac{4}{2}-4-\frac{1}{2}+2]+[\frac{16}{2}-8-\frac{4}{2}+4]$
 
$=-(\frac{-1}{2})+2 \qquad = \frac{1}{2}+2\qquad=\frac{5}{2}$
 
Consider $I_3=\int \limits_1^3 |x-3|dx$
 
$(x-3) \geq 0\;for\; 3 \leq x \leq 4$
 
$(x-3) \leq 0\;for \; 1 \leq x \leq 3$
 
Hence $I-3=\int \limits_1^3 -(x-3) dx +\int \limits_3^4(x-3) dx$
 
On integrating we get
 
$-[\frac{x^2}{2}-3x]_1^3+[\frac{x^2}{2}-3x]_3^4$
 
On applying limits,
 
$-[\frac{9}{2}-9-\frac{1}{2}+3]+[\frac{16}{2}-12-\frac{9}{2}-9]$
 
$=-[-2]+[\frac{1}{2}]$
 
$=2+\frac{1}{2}=\frac{5}{2}$
 
Therefore $ I=I_1+I_2+I_3$
 
$=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}$
 
$=\frac{9}{2}+5$
 
$I=\frac{19}{2}$

 

 

answered Feb 19, 2013 by meena.p
 
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