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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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