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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 47 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 47 Maths textbook solution.

Answers (1)

Answer:

A= \frac{33}{2}\; \text{sq. units}

Hints:

Use concept.

Given:

The curves 3x^2+5y =32 and y=\left | x-2 \right | .

Solution:

To find area enclosed by

3x^2+5y=32

3x^2=-5\left ( y-\frac{32}{5} \right )   ……. (1)

And

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

Equation (1) represents a downward parabola with vertex \left (0,\frac{32}{5} \right )  and equation (2) represents lines. A rough sketch of curves is given as:

Required area = Region ABECDA

A = Region ABEA + Region AECDA

=\int_{2}^{3}\left(y_{3}-y_{4}\right) d x+\int_{-2}^{2}\left(y-y_{2}\right) d x

\begin{aligned} &=\int_{2}^{3}\left(\frac{32-3 x^{2}}{5}-x+2\right) d x+\int_{-2}^{2}\left(\frac{32-3 x^{2}}{5}-2+x\right) d x \\ &=\frac{1}{5} \int_{2}^{3}\left(\frac{32-3 x^{2}-5 x+10}{5}\right) d x+\int_{-2}^{2}\left(\frac{32-3 x^{2}-10+5 x}{5}\right) d x \\ &=\frac{1}{5}\left[\int_{2}^{3}\left(42-3 x^{2}-5 x\right) d x+\int_{-2}^{2}\left(22-3 x^{2}+5 x\right) d x\right] \end{aligned}

\begin{aligned} &=\frac{1}{5}\left[\left(42 x-x^{3}-\frac{5 x^{2}}{2}\right)_{2}^{3}+\left(22 x-x^{3}+\frac{5 x^{2}}{2}\right)_{-2}^{2}\right] \\ &=\frac{1}{5}\left[\left\{\left(126-27-\frac{45}{2}\right)-(84-8-10)\right\}+\{(44-8+10)-(-44+8+10)\}\right] \\ &=\frac{1}{5}\left[\left\{\frac{153}{2}-66\right\}+\{46+26\}\right] \\ &=\frac{1}{5}\left[\frac{21}{2}+72\right] \end{aligned}

A= \frac{33}{2}\; \text{sq. units}

  

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