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  • Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 16 maths

Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 16 maths

Answers (1)

Answer:

\frac{27}{2} s q \cdot \text { units }

Hint:

y=1+|x+1|, x=-2, x=3, y=0

Given:

Using integration, find area of region bounded by following curve making a rough sketch      

y=1+|x+1|, x=-2, x=3, y=0

Solution:

We have

        y=1+|x+1|  Intersect  x=-2  at \left ( -2,2 \right ) and x=3 at (3,5)

        y=0 is x-axis

        y=1+|x+1|

        \left\{\begin{array}{cc} 1-(x+1) & x \leq-1 \\ 1+(x+1) & x \geq 1 \\ -x & x \leq-1 \\ x+2 & x \geq 1 \end{array}\right.

Let required area be A since limits on x are given we use horizontal strip to find area

        \begin{aligned} &A=\int_{-2}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}|y| d x+\int_{-1}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}-x d x+\int_{-1}^{3}(x+2) d x \end{aligned}

        \begin{aligned} &A=-\left[\frac{x^{2}}{2}\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{3} \\\\ &A=-\left[\frac{1}{2}-\frac{4}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\\\ &A=\frac{3}{2}+8+\frac{8}{2} \end{aligned}

        \begin{aligned} &A=\frac{3}{2}+8+4 \\\\ &A=\frac{27}{2} s q \cdot u n i t \end{aligned} \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

        

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