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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} $

1 Answer

  • (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,
On applying limits,


answered Feb 14, 2013 by meena.p

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