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  • Draw a rough sketch of the given curve y = 1 + \vert x +1 \vert , x = –3, x = 3, y = 0 and find the area of the region bounded by them, using integration.

Draw a rough sketch of the given curve y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0 and find the area of the region bounded by them, using integration.

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Given;

Curve y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0  

|x+1|=\left\{\begin{array}{l}-x-1, x<-1 \\ x+1, x \geq-1\end{array}\right. \therefore y=1+|x+1| =\left\{\begin{array}{c}-x, x<-1 \\ x+2, x \geq-1\end{array}\right.
Area of the region bounded by the curve y=f(x) , the  x  -axis and the ordinates  x=a  and  x=b,  where  f(x)  is a continuous function defined on  [a, b],  is given by  A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x \\

 

 

 

 

Required area 

=\int_{-3}^{-1}(-\mathrm{x}) \mathrm{dx}+\int_{-1}^{3}(\mathrm{x}+2) \mathrm{d} \mathrm{x}\\ =-\left[\frac{\mathrm{x}^{2}}{2}\right]_{-3}^{-1}+\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{3}\\ =-\left(\frac{1}{2}-\frac{9}{2}\right)+\left(\frac{9}{2}+6-\frac{1}{2}+2\right)\\ =4+12=16 sq.units \\

 

 

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