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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 46 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 46 Maths textbook solution.

Answers (1)

Answer:

\frac{1}{2} .sq. Units.

Hints:

Use concept.

Given:

The curves y=\left | x-1 \right |  and y=-\left | x-1 \right |+1

Solution:

To find area enclosed by

\begin{aligned} &y=|x-1| \\ &\Rightarrow y=\left\{\begin{array}{l} -(x-1), \text { if } x-1<0 \\ (x-1), \text { if } x-1 \geq 0 \end{array}\right. \end{aligned}

\begin{aligned} &\Rightarrow y=\left\{\begin{array}{l} 1-x, \text { if } x<1 \\ x-1, \text { if } x \geq 1 \end{array}\right. \\ &\text { And } y=-|x-1|+1 \\ &\Rightarrow y=\left\{\begin{array}{l} +(x-1)+1, \text { if } x-1<0 \\ -(x-1)+1, \text { if } x-1 \geq 0 \end{array}\right. \\ &y=\left\{\begin{array}{lr} x, \quad \text { if } x<1 \\ -x+2, \text { if } x \geq 1 \end{array}\right. \end{aligned}

A rough sketch of equation of lines (1),(2),(3),(4) is given as:

Shaded region is the required area.

Required area = Region ABCDA

Required area = Region BDCB + Region ABDA

Region BDCB is sliced into rectangles of area  =(y_2-y_1)x and it slides from x=\frac{1}{2}  to x =1

Region ABDA is sliced into rectangle of area =(y_3-y_4)x and it slides from x =1 to x=\frac{3}{2} .

So, using equation (1),

Required area = Region BDCB + Region ABDA

\begin{aligned} &=\int_{\frac{1}{2}}^{1}\left(y_{1}-y_{2}\right) d x+\int_{1}^{\frac{3}{2}}\left(y_{3}-y_{4}\right) d x \\ &=\int_{\frac{1}{2}}^{1}(x-1+x) d x+\int_{1}^{\frac{3}{2}}(-x+2-x+1) d x \\ &=\int_{\frac{1}{2}}^{1}(2 x-1) d x+\int_{1}^{\frac{3}{2}}(3-2 x) d x \\ &=\left[x^{2}-x\right]_{\frac{1}{2}}^{1}+\left[3 x-x^{2}\right]_{1}^{\frac{3}{2}} \end{aligned}

\begin{aligned} &=\left[(1-1)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]+\left[\left(\frac{9}{2}-\frac{9}{4}\right)-(3-1)\right] \\ &=\frac{1}{4}+\frac{9}{4}-2 \\ &A=\frac{1}{2} \text { sq. units } \end{aligned}

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