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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 48 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 48 Maths textbook solution.

Answers (1)

Answer:

\frac{125}{24}  sq. units                                                                                                       

Hints:

Use concept.

Given:

The parabola y=4x-x^2 and y=x^2 

Solution:

To area enclosed by

\begin{aligned} &y=4 x-x^{2} \\ &\Rightarrow-y=x^{2}-4 x+4-4 \\ &\Rightarrow-y+4=(x-2)^{2} \\ &\Rightarrow-(y-4)=(x-2)^{2} \\ &\text { And } y=x^{2}-x \\ &\left(y+\frac{1}{4}\right)=\left(x-\frac{1}{2}\right)^{2} \end{aligned}

Equation (1) represents a parabola downward with vertex at (2,4) and meets axis at (4,0), (0,0). Equation (2) represents a parabola upward whose vertex is \left (\frac{1}{2},\frac{1}{4} \right ) and meets axis at (1,0), (0,0). Points of intersection of parabola are (0,0) and \left (\frac{5}{2},\frac{15}{4} \right ) .

A rough sketch of the curves is as under:

Shaded region is required area it is sliced into rectangles with area = (y_2-y_1)x . It slides from x =0  to x=\frac{5}{2} , so

Required area = Region OQAP

\begin{aligned} \mathrm{A} &=\int_{0}^{\frac{{5}}{2}}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{\frac{5}{2}}\left[4 x-x^{2}-x^{2}+x\right] d x \\ &=\int_{0}^{\frac{5}{2}}\left[5 x-2 x^{2}\right] d x \\ &=\left[\frac{5 x^{2}}{2}-\frac{2}{3} x^{2}\right]_{0}^{\frac{5}{2}} \\ &=\left[\left(\frac{125}{8}-\frac{250}{24}\right)-(0)\right] \\ & A=\frac{125}{24} \text { sq. units } \end{aligned}

Therefore, the area enclosed by the parabola y=4x-x^2 and y=x^2 is \frac{125}{24}  sq. units.

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