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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20 3 question 26

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 26.

Answers (1)

Answer:

\frac{16}{3 }\; \text{sq. units} .

Hint:

Use concept.

Given:

The given equations are  y^2=2x+1   and x-y-1=0

Solution:

To find area bounded by

y^2=2x+1            ...........(1)

x-y-1=0            ........(2)

 On solving the equation (i) and (ii)

\begin{aligned} &x-y=1 \\ &\text { or } y^{2}=2(y-1)+1 \\ &\text { or } y^{2}=2 y-1 \\ &\text { or }(y+1)(y-3)=0 \\ &\text { or } y=3 \text { or }-1 \\ &\therefore x=4,0 \end{aligned}

Equation (i) is a parabola with vertex \left (\frac{-1}{2},0 \right ) (−12,0) and passes through (0, 1), A (0, – 1)

Equation (ii) is a line passing through (1, 0) and (0, – 1).

 Points of intersection of parabola and line are B (4, 3) and A (0, – 1)

These are shown in the graph below:

 

 

Required area=RegionABCDA

\begin{aligned} &=\int_{-1}^{3}\left(1+y-\frac{y^{2}-1}{2}\right) d y \\ &=\frac{1}{2} \int_{-1}^{3}\left(2+2 y-y^{2}+1\right) d y \\ &=\frac{1}{2} \int_{-1}^{3}\left(3+2 y-y^{2}\right) d y \\ &=\frac{1}{2}\left[3 y+y^{2}-\frac{y^{3}}{3}\right]_{-1}^{3} \\ &=\frac{1}{2}\left[(9+9-9)-\left(-3+1+\frac{1}{3}\right)\right] \\ &=\frac{1}{2}\left[9+\frac{5}{3}\right] \\ &=\frac{16}{3} \text { sq.units } \end{aligned}

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