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  • Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 39 Maths.

Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 39 Maths.

Answers (1)

Answer: The Area of the region \left\{(x, y): 0 \leq y \leq x^{2}+3 ; 0 \leq y \leq 2 x+3 ; 0 \leq x \leq 3\right\} is \frac{38}{3} sq.units.

Hints:

Given: \left\{(x, y): 0 \leq y \leq x^{2}+3 ; 0 \leq y \leq 2 x+3 ; 0 \leq x \leq 3\right\}

Solution: To find area given equation are

y= x^3+3...... (i)

y= 2x+3....... (ii)

 And x=3    .......(iii)

Solving the above three equations to get the intersectons points,

\begin{aligned} &x^{2}+3=2 x+3 \\ &\text { Or } x^{2}-2 x=0 \\ &\text { Or } x(x-2)=0 \\ &\text { And } x=0 \text { or } x=2 \\ &y=3 \text { or } y=7 \end{aligned}

Equation (i) represents a parabola with vertices (3,0) and axis as y - axis

Equation (ii) represents a line passing through (0,3) and \left (-\frac{3}{2},0 \right )

The points of intersection are A(0,3) and B(2,7)

These are shown in the graph below:

Required area =

\begin{aligned} =& \int_{2}^{3} y_{1} d x+\int_{0}^{2} y_{2} d x \\ =& \int_{2}^{3}(2 x+3) d x+\int_{0}^{2}\left(x^{2}+3\right) d x \\ &=\left(x^{3}+3 x\right)_{2}^{3}+\left(\frac{x^{2}}{3}+x\right)_{0}^{2} \\ &=[(9+9)-(4+6)]+\left[\left(\frac{8}{3}+2\right)-(0)\right] \\ &=[18-10]+\left[\frac{14}{3}\right] \\ &=8+\frac{14}{3} \end{aligned}

=\frac{38}{3} sq.units

Therefore, the area of the region is \frac{38}{3}  sq.units

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