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Evaluate:$\int \limits_{-5}^ 8 |x+3|dx$

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  • $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $ \int \limits_a^b |x+a| dx \qquad (x+a) \geq 0\; if\; x \geq -a$
  • $ -(x+a) \; if\; x < -a$
  • Therefroe $ \int \limits_a^b |x+a| dx = \int \limits_a^c -(x+a)dx+\int \limits_c^b (x+a) dx$
Step 1:
Given $I=\int \limits_{-5}^8 |x+3|dx$
Here we need to break the intervals $[8,-5]$
we know $|(x-3)|=(x-3)$ if $x \geq 3 \; and\; -(x-3)\; if \;x < 3$
Therefore $\int \limits_{-5}^8 |x-3| dx= \int \limits_{-5}^{-3} -(x-3) dx+\int \limits_{-3}^8 (x-3) dx$
Step 2:
On integrating we get
$-\bigg[\large\frac{x^2}{2}-$$3x\bigg]_{-5}^{-3} +\bigg[\large\frac{x^2}{2}-$$3x\bigg]_{-3}^8$
On applying the limits,
$- \bigg [\bigg( \large\frac{(-3)^2}{2}-$$3 (-3)\bigg)-\bigg( \large\frac{(-3)^2}{2}-$$3(-5)\bigg)\bigg]+\bigg[\bigg(\large\frac{8^2}{2}-$$3 \times 8\bigg)-\bigg(\large\frac{(-3)^2}{2}-$$3(-3)\bigg)\bigg]$
On simplifying,
$-\bigg\{\bigg[\large\frac{9}{2}$$+9 \bigg]-\bigg(\large\frac{25}{2}$$+15\bigg)\bigg\}+\bigg\{\bigg(\large\frac{64}{2}$$-24\bigg)-\bigg(\large\frac{9}{2} $$+9\bigg)\bigg\}$
$=-14+\bigg( \large\frac{-11}{2} \bigg)$
$=-14- \large\frac{11}{2} $
answered Apr 26, 2013 by meena.p
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