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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 49 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 49 Maths textbook solution.

Answers (1)

Answer:

121:4

Hints:

Use basic concepts.

Given:

The parabola y=4x-x^2  and y=x^2-x .

Solution:

The curves are

\begin{aligned} &y=4 x-x^{2}\\ &\Rightarrow-(y-4)=(x-2)^{2}\\ &\text { And } y=x^{2}-x \text {. }\\ &\Rightarrow\left(y+\frac{1}{4}\right)^{2}=\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) and \left (\frac{5}{2}, \frac{15}{4} \right ) . A rough sketch of the curves is as under:

Area of the region above x-axis

A_1  = Area of region OBACO

      = Region OBCO + Region BACB

=\int_{0}^{1} y_{1} d x+\int_{1}^{\frac{5}{2}}\left(y_{1}-y_{2}\right) d x

\begin{aligned} &=\int_{0}^{1}\left(4 x-x^{2}\right) d x+\int_{1}^{\frac{5}{2}}\left(4 x-x^{2}-x^{2}+x\right) d x \\ &=\left(\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{1}+\left[\frac{5 x^{2}}{2}-\frac{2 x^{3}}{3}\right]_{1}^{\frac{5}{2}} \\ &=\left(2-\frac{1}{3}\right)+\left[\left(\frac{125}{8}-\frac{250}{24}\right)-\left(\frac{5}{2}-\frac{2}{3}\right)\right] \\ &=\frac{5}{3}+\frac{125}{24}-\frac{11}{6} \\ &=\frac{121}{24} \text { sq. units } \end{aligned}

Area of the region below x-axis

A_2 = Area of region OPBO

     = Region OBCO + Region BACB

  \begin{aligned} &=\left|\int_{0}^{1} y_{2} d x\right| \\ &=\left|\int_{0}^{1}\left(x^{2}-x\right) d x\right| \\ &=\left|\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}\right)_{0}^{1}\right| \\ &=\left|\left(\frac{1}{3}-\frac{1}{2}\right)-(0)\right| \\ &=\left|-\frac{1}{6}\right| \\ A_{2} &=\frac{1}{6} \text { sq. units } \end{aligned}

\begin{aligned} &A_{1} \cdot A_{2}=\frac{121}{24}: \frac{4}{24} \\ &A_{1} \cdot A_{2}=121: 24 \end{aligned}

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