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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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sketch the graph of y=|x+3| and evaluate $\int_{-6}^0|x+3| dx.$

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y=|x+3| can have two possibilities
$\;\;(i)\;y=x$ where $x\geq 0$
$\;\;(ii)\;\;y=-x$ where $x\leq 0$
y=x;$x\geq 0$ represents the portion that lies on the right side of x=0
y=-x;$x\leq 0$ represents the portion that lies on the left side of x=0
We have y=|x+3|=$\left\{\begin{array}{1 1}x+3; & if\;x\geq -3\\-x-3; & if\;x\leq -3\end{array} \right.$
Clearly y=x+3 is a straight line cutting the x and y axes at(-3,0) and (0,3) respectively.
Hence y=x+3,$x\geq -3$ represents that portion of the line which lies on the right side of x=-3.
Now $\int_{-6}^0|x+3|dx$ is the given function to be which has to be evaluated
$\;\;\;=\int_{-6}^{-3}|x+1|dx+\int_{-3}^0|x+1|dx$
$\;\;\;=\int_{-6}^{-3}(x+1)dx+\int_{-3}^0(x+1)dx$
$\;\;\;=-\begin{bmatrix}\frac{(x+3)^2}{2}\end{bmatrix}_{-6}^{-3}+\begin{bmatrix}\frac{(x+3)^2}{2}\end{bmatrix}_{-3}^0$
$\;\;\;=\frac{1}{2}\{-[(x^2+6x+9)]_{-6}^{-3}+\frac{1}{2}\{-[(x^2+6x+9)]_{-3}^0\}$
Now applying the limits we get,
$\;\;\;=\frac{1}{2}\{-[(9-18+9)-(36-36+9)]+[(0)-(9-18+9)]\}$
$\;\;\;=\frac{1}{2}[-(0-9)-0]$
$\;\;\;=\frac{9}{2}$
Hence the value of $\int_{-6}^0|x+3|dx=\frac{9}{2}.$
answered Jan 21, 2013 by sreemathi.v
 

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