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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Sketch the graph of y=| x+3 | and evaluate$\Large \int\limits_{-6}^0\normalsize| x+3 |dx$

This question has appeared in model paper 2012

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$y=\mid x+3\mid=\left\{\begin{array}{1 1}-(x+3)\;for\; x<-3\\x+3\;for\; x\geq -3\end{array}\right.$
When $x<-3$
$y=-x-3$
When x=-4,
$y=4-3=1$
When x=-5,
$y=5-3=2$
When x=-6,
$y=6-3=3$
Step 2:
When $x\geq -3$
$x=-1\Rightarrow y=2$
$x=-2\Rightarrow y=1$
$x=-3\Rightarrow y=0$
Step 3:
Required Area=Area of region ABC+Area of region OAD
$\qquad\qquad\quad=\int\limits_{-6}^{-3}\mid x+3\mid dx+\int \limits_{-3}^{0}\mid x+3\mid dx$
$\qquad\qquad\quad=\int\limits_{-6}^{-3}(-x-3)dx+\int \limits_{-3}^0(x+3)dx$
$\qquad\qquad\quad=\big[\large\frac{-x^2}{2}$$-3x\big]_{-6}^{-3}+\big[\large\frac{x^2}{2}+3x\big]_{-3}^0$
Step 4:
On applying limits we get,
$\qquad\qquad\quad=\big[(-\large\frac{9}{2}$$+9)-(-18+18)\big]+\big[\large\frac{9}{2}\big]$
$\qquad\qquad\quad=\large\frac{9}{2}+\frac{9}{2}$
$\qquad\qquad\qquad=9$sq.units
answered Sep 10, 2013 by sreemathi.v
 

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