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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.

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• 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}$$+c. Step 1: Given curve is y=1+\mid x+1\mid,x=-3,x=3 and y=0 y=1+\mid x+1\mid\Rightarrow y=1+x+1,if x+1\geq 0 and y=1-(x+1) if x+1\leq 0 Thus the equations of the given curves are y=x+2 is a straight line cutting the x-axes and y-axes at (-2,0) and (0,2) respectively. Hence y=x+2,for x >-1 represents the part of the line which is on the right side of x=-1 The equation y=-x for x < -1 represents that the part of the line passing through the origin and making an angle of 135^\circ with x-axis which is on the left side of x=-1 The region bounded by the given curves is the shaded portion shown in the fig. Step 2: Area of the shaded portion is \int_{-3}^{-1}y_1dx+\int_{-1}^3y_2dx. Where y_1=-x and y_2=x+2 Hence A=\int_{-3}^{-1}-xdx+\int_{-1}^3(x+2)dx. On integrating we get, A=-\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_{-3}^{-1}+\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 2x\end{bmatrix}_{-1}^3 On applying limits we get, A=-\begin{bmatrix}\large\frac{(-1)^2}{2}-\frac{(-3)^2}{2}\end{bmatrix}+\begin{bmatrix}\large\frac{3^2}{2}-\frac{(-1)^2}{2}+\normalsize 2(3)-2(-1)\end{bmatrix} \;\;\;=[\large\frac{1}{2}-\frac{9}{2}]\normalsize+4+8 \;\;\;=\large\frac{8}{2}$$+4+8=16$sq.units.
Hence the required area is 16 sq. units.
edited Dec 22, 2013