RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise

• Chapter 20 - Areas Of Bounded Region - Ex-20.1

• Chapter 20- Areas Of Bounded Region - Ex-20.2

• Chapter 20- Areas Of Bounded Region - Ex-20.3

• Chapter 20 - Areas Of Bounded Region - Ex-MCQ

• Chapter 20 - Areas Of Bounded Region - Ex-FBQ

Areas of Bounded Regions Excercise:20.4

Area of bounded Region exercise 20.4 question 1

Answer:

$\frac{32}{3}$ sq.units
Hint:
Find the point of intersection of parabola and the given line and then find the required region.
Given:
$x=4y-y^{2}$and $x=2y-3$
Solution:
To find the point at intersection of the parabola $x=4y-y^{2}$and the line $x=2y-3$

Let us substitute $x=2y-3$ in equation of the parabola
$\begin{aligned} &2 y-3=4 y-y^{2} \\ &y^{2}-2 y-3=0 \\ &(y+1)(y-3)=0 \\ &y=-1,3 \end{aligned}$
Therefore, the point of intersection are D(-1,-5) and A(3,3)
The area of the required region ABCDOA,
$A=\int_{-1}^{3}\left | x_{1}-x_{2} \right |dy$where $x=4y-y^{2}$and $x=2y-3$
$\begin{aligned} &=\int_{-1}^{3}\left(x_{1}-x_{2}\right) d y \quad\left[\because x_{1}>x_{2}\right] \\ &=\int_{-1}^{3}\left[\left(4 y-y^{2}\right)-(2 y-3)\right] d y \\ \end{aligned}$
$\begin{aligned} &=\int_{-1}^{3}\left[4 y-y^{2}-2 y+3\right] d y \\ &=\int_{-1}^{3}\left[-y^{2}+2 y+3\right] d y \\ \end{aligned}$
$\begin{aligned} &=\left[\frac{-y^{3}}{3}+y^{2}+3 y\right]_{-1}^{3} \\ &=\left[\frac{-3^{3}}{3}+3^{2}+3(3)-\left(\frac{1}{3}+1-3\right)\right] \\ \end{aligned}$
$\begin{aligned} &=\left[-3^{2}+3^{2}+9-\frac{1}{3}-1+3\right] \\ &=11-\frac{1}{3} \\ &=\frac{32}{3} s q \cdot \text { units } \end{aligned}$

Area of bounded Region exercise 20.4 question 3 sub question 1

Answer:

9 sq.units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
$y=2x-4$ and $y^{2}=4x$
Solution:

Substitute $y=2x-4$ in $y^{2}=4x$
$\begin{aligned} &(2 x-4)^{2}=4 x \\ &4 x^{2}+16-16 x=4 x \\ &4 x^{2}-20 x+16=0 \\ \end{aligned}$
$\begin{aligned} &x^{2}-5 x+4=0 \\ &(x-1)(x-4)=0 \\ &x=1,4 \\ &y=-2,4 \end{aligned}$
Therefore point of intersection are (1,-2) and (4,4)
Using horizontal strips:
The required area is the shaded region
$\begin{aligned} A &=\int_{-2}^{4}\left(x_{1}-x_{2}\right) d y \end{aligned}$ where $\begin{aligned} x_{1}=\frac{y+4}{2} \end{aligned}$ and $\begin{aligned} x_{2}=\frac{y^{2}}{4} \\ \end{aligned}$
$\begin{aligned} &=\int_{-2}^{4}\left[\left(\frac{y+4}{2}\right)-\left(\frac{y^{2}}{4}\right)\right] d y \\ \end{aligned}$
$\begin{aligned} &=\left[\frac{y^{2}}{4}+2 y-\frac{y^{3}}{12}\right]_{-2}^{4} \\ \end{aligned}$
$\begin{aligned} &=\left[\frac{4^{2}}{4}+2(4)-\frac{4^{3}}{12}\right]-\left[\frac{(-2)^{2}}{4}+2(-2)-\frac{(-2)^{3}}{12}\right] \\ \end{aligned}$
$\begin{aligned} &=12-\frac{16}{3}+3-\frac{2}{3} \\ &=9 \mathrm{sq} \cdot \text { units } \end{aligned}$

Area of bounded Region exercise 20.4 question 4

Answer:

18 sq.units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
$x-y=4$ and $y^{2}=2x$
Solution:


The parabola $y^{2}=2x$ opens towards the positive x-axis and its focus is $\left ( \frac{1}{2},0 \right )$
The straight line$x-y=4$ passes through (4,0) and (0,-4)
Solving $x-y=4$ and $y^{2}=2x$, we get
$\begin{aligned} &y^{2}=2(y+4) \\ &y^{2}-2 y-8=0 \\ &(y-4)(y+2)=0 \\ &y=4,-2 \end{aligned}$
So, the points of intersection of given parabola and the line are(8,4) and (2,-2)
Required area,
$\begin{aligned} &=\int_{-2}^{4} x_{\text {line }} d y-\int_{-2}^{4} x_{\text {parabola }} d y \\ \end{aligned}$
$\begin{aligned} &=\int_{-2}^{4}(y+4) d y-\int_{-2}^{4} \frac{y^{2}}{2} d y \\ \end{aligned}$
9$\begin{aligned} &=\frac{(y+4)^{2}}{2}-\left[\frac{1}{2}\left(\frac{y^{3}}{3}\right)\right]_{-2}^{4} \\ \end{aligned}$$\begin{aligned} &=\frac{1}{2}(64-6)-\frac{1}{6}[64-(-8)] \\ &=30-12 \\ &=18 \mathrm{sq} \cdot \text { units } \end{aligned}$

The RD Sharma class 12th exercise 20.4 consists of a total of 5 questions that are short and easy, covering up the effectively important concepts of this chapter that are profitable for board exams. The concepts covered in the RD Sharma class 12 solution of Area of bounded region exercise 20.4 are-

• The line segment of the curve

• Area of the special bounded region

• Area of the region that is bounded by the curve

• Area of the region that is bounded by a parabola

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RD Sharma Chapter-wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable