Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

sketch the graph of y=|x+3| and evaluate $\int_{-6}^0|x+3| dx.$

Can you answer this question?

1 Answer

0 votes
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
Now applying the limits we get,
Hence the value of $\int_{-6}^0|x+3|dx=\frac{9}{2}.$
answered Jan 21, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App