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

1 Answer

  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c.$
Step 1:
Given curve is $y=1+\mid x+1\mid,x=-3,x=3$ and $y=0$
$y=1+\mid x+1\mid\Rightarrow y=1+x+1$,if $x+1\geq 0$ and $y=1-(x+1)$ if $x+1\leq 0$
Thus the equations of the given curves are $y=x+2$ is a straight line cutting the $x$-axes and $y$-axes at (-2,0) and (0,2) respectively.
Hence $y=x+2$,for $x >-1$ represents the part of the line which is on the right side of $x=-1$
The equation $y=-x$ for $x < -1$ represents that the part of the line passing through the origin and making an angle of $135^\circ$ with $x$-axis which is on the left side of $x=-1$
The region bounded by the given curves is the shaded portion shown in the fig.
Step 2:
Area of the shaded portion is $\int_{-3}^{-1}y_1dx+\int_{-1}^3y_2dx.$
Where $y_1=-x$ and $y_2=x+2$
Hence $A=\int_{-3}^{-1}-xdx+\int_{-1}^3(x+2)dx.$
On integrating we get,
$A=-\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_{-3}^{-1}+\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 2x\end{bmatrix}_{-1}^3$
On applying limits we get,
$A=-\begin{bmatrix}\large\frac{(-1)^2}{2}-\frac{(-3)^2}{2}\end{bmatrix}+\begin{bmatrix}\large\frac{3^2}{2}-\frac{(-1)^2}{2}+\normalsize 2(3)-2(-1)\end{bmatrix}$
Hence the required area is 16 sq. units.
answered May 6, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai

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