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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 50 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 50 Maths textbook solution.

Answers (1)

Answer:

4 sq. units.

Hints:

Use concept of definite integrals.

Given:

The curves y=\left | x-1 \right |  and y=3-\left | x \right |

Solution:

To find area bounded by the curve

 

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

And   y=3-\left | x \right |

\Rightarrow \quad y= \begin{cases}3+x, \text { if } x<0\; \; \; \; \; \; \; \; \; ......(3) \\ 3-x, \text { if } x \geq 0\; \; \; \; \; \; \; \; \; ......(4)\end{cases}

 

Drawing the rough sketch of lines (1), (2), (3) and (4) as under:

Shaded region is the required area

Required area = Region ABCDA

A = Region ABFA + Region AFCEA + Region CDEC

\begin{aligned} &=\int_{1}^{2}\left(y_{1}-y_{2}\right) d x+\int_{0}^{1}\left(y_{1}-y_{3}\right) d x+\int_{-1}^{0}\left(y_{4}-y_{3}\right) d x \\ &=\int_{1}^{2}(3-x-x+1) d x+\int_{0}^{1}(3-x-1+x) d x+\int_{-1}^{0}(3+x-1+x) d x \\ &=\int_{1}^{2}(4-2 x) d x+\int_{0}^{1} 2 d x+\int_{-1}^{0}(2+2 x) d x \\ &=\left[4-x^{2}\right]_{1}^{2}+[2 x]_{0}^{1}+\left[2 x+x^{2}\right]_{-1}^{0} \\ &=[(8-4)-(4-1)]+[2-0]+[(0)-(-2+1)] \\ &=(4-3)+2+1 \end{aligned}

A = 4 sq. units.

Therefore, the area bounded by the curve is 4 sq. units.

Posted by

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