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Evaluate:$\int \limits_1^4 [\; \left |x-1 \right|+\left |x-2 \right |+\left| x-3\right |\;]\;dx$

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  • $\int \limits _a^b f(x)dx=F(b)-F(a)$
  • $\int \limits _a^b |x-c| dx$ Here we need to break the interval into two subintervals.
  • If $(x-c) \geq 0 \;(ie)\; x \geq c,$ then the interval is from a to c
  • If $(x-c) \leq 0\; (ie)\; x \leq 0 \;(ie)\; x < c $ then the interval is from c to b
Given $I=\int \limits_{-1}^2 f(x)dx$ where $f(x)=|x-1|+|x|+|x-1|$
First of all let us equate the expressions in mod to 0
Hence (x+1) gives x=-1
(x) gives x=0
(x-1) gives x=1
Therefore x=-1,x=0 and x=1.which is in acsending order
Now let us break the given interval [-1,2] into three subintervals [-1,0] [0,1] and [1,2]
case (i) on subinterval [-1,0] where $(x+1) \geq x \leq 0\;and\;(x-1) \leq 0$
Hence $f(x)=|x+1|+|x|+|x-1|$
case(ii) Let us consider the subinterval [0,1], where $(x+1) \geq 0,x \geq 0$ and |x-1|=-(x-1)
Hence $f(x)=|x+1|+|x|+|x-1|$
case(iii) Let us consider the subinterval [1,2] where $(x+1) > 0,|x| \leq 0$ and $(x-1) \geq 0$
Hence $f(x)=|x+1|+|x|+|x-1|$
Therefore $\int \limits_{-1}^2 f(x)dx=\int \limits_{-1}^0 f(x)dx+\int \limits_0^1 f(x)dx+\int \limits_{1}^2 f(x)dx$
$=\int \limits_{-1}^0 (2-x)dx+\int \limits_0^1 (x+2)dx+\int \limits_1^2 3x dx$
On integrating we get
on applying the limits we get,
Hence $\int \limits_{-1}^2 f(x)dx=\large\frac{19}{2}$
answered Apr 2, 2013 by meena.p
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