RD Sharma Class 12 Exercise 20.2 Areas Of Bounded Region Solutions Maths - Download PDF Free Online

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-20.3

• Chapter 20- Areas Of Bounded Region - Ex-20.4

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

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

Areas of Bounded Regions Exercise:20.2

Area of bounded region exercise 20.2 question 1

Answer:

$\frac{7}{3}$ sq.units
Hint:
$y=4x^{2}$ is an upward parabola with vertex (0,0)
Given:
$y=4x^{2}$
Solution:

x=0 is the line with points (0,1),(0,0) etc.,
y=1 is the line with points (0,1),(1,1) etc.,
y=4 is the line with points (0,4),(1,4) etc.,
Parabola meets at points (0,0) with line x=0
$\left ( \frac{1}{2},1 \right )$ with liney=1, (1,4) with line y=4
Here we need to do along y-axis as we don’t know equation of line with respect to x
Required area (Shaded one)=Area under parabola from y=1 to y=4
$\begin{aligned} &\int_{1}^{4}|x| d y=\int_{1}^{4} \sqrt{\frac{y}{4}} d y \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int_{1}^{4} \sqrt{y} d y \\ &=\left[\frac{1}{2} \frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4} \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \times \frac{2}{3}\left[y^{\frac{3}{2}}\right]_{1}^{4} \\ &=\frac{1}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right] \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{3}(8-1) \\ &=\frac{7}{3} s q \cdot u n \text { its } \end{aligned}$

Area of bounded region exercise 20.2 question 2

Answer:

$\frac{56}{3}sq.units$
Hint:
To find the area under two or more than two curves, the first crucial step is to find the intersection points of the curves.
Given:
$x^{2}=16y$, $y=1$, $y=4$and y-axis in the first quadrant
Solution:

$y=\frac{x^{2}}{16}$ (Curve)
$y=4$ (Line )
$y=1$ (Line)
Between curve A and line C
$16=x^{2}\\ x=4$
Between curve A and line B
$4=\frac{x^{2}}{16}\\ x=\sqrt{64}\\ x=8$
Required area can be calculated by breaking the problem into two parts

1. Calculate area under the curve A and Line B
2. Subtract the area enclosed by curve A and line C from the above area.

1. $\int_{0}^{8}\left(4-\frac{x^{2}}{16}\right) d x=$ Area under B and A
$\begin{aligned} &=[4 x]_{0}^{8}-\left[\frac{x^{3}}{16(3)}\right]_{0}^{8} \\ &=\frac{64}{3} s q \cdot \text { units } \end{aligned}$
2. $\int_{0}^{4}\left(1-\frac{x^{2}}{16}\right) d x=$ Area under C and A
$\begin{aligned} &=[x]_{0}^{4}-\left[\frac{x^{3}}{16(3)}\right]_{0}^{4} \\ &=\frac{8}{3} s q \cdot \text { units } \end{aligned}$
The required area under the curve
$\frac{64}{3}-\frac{8}{3} =\frac{56}{3}$ sq.units

Area of bounded region exercise 20.2 question 3

Answer:

$\frac{1}{24a}$
Hint:
Find the latus rectum and its intersection points with the parabola and then integrate the expression to find the area enclosed by the curves.
Given:
$x^{2}=4ay$
Solution:

First, we find the latus rectum and its intersection points with the parabola
$x^{2}=4ay$ … (i)
$y=\frac{x^{2}}{4a}$ … (ii)
Comparing it with the standard from of a parabola
$Y^{2}=4aX^{2}$
Where (0,A) is the coordinate of the focus of the parabola and the latus rectum passes through this point and is perpendicular to the axis of symmetry
Therefore, the equation of the latus rectum is y=a
Comparing equation (i) and (ii)
$\frac{1}{4a}=4A$
$A=\frac{1}{16a}$
Intersection point
$\begin{aligned} &\frac{1}{16 a}=\frac{x^{2}}{4 a} \\ &x^{2}=\frac{1}{4} \\ &x=\pm \frac{1}{2} \end{aligned}$
Now, Integrating the expression to find the area enclosed by the curves
Since, the latus rectum is above the parabola in the Cartesian plane, the expression will be
$\begin{aligned} &\int_{-\frac{1}{2}}^{\frac{1}{2}}\left[\frac{1}{16 a}-\frac{x^{2}}{4 a}\right] d x \\ &=\left[\frac{x}{16 a}\right]_{-\frac{1}{2}}^{\frac{1}{2}}-\left[\frac{x^{3}}{4 a(3)}\right]_{-\frac{1}{2}}^{\frac{1}{2}} \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{16 a}-\frac{1}{48 a} \\ &=\frac{1}{24 a} \end{aligned}$

Area of bounded region exercise 20.2 question 4

Answer:

$\frac{1}{96}$Hint:
Find the latus rectum and its intersection points with the parabola and then integrate the expression to find the area enclosed by the curves.
Given:
$x^{2}+16y=0$
Solution:

Curve is $x^{2}+16y=0$
To find the enclosed area
$\begin{aligned} &2 \times \int_{0}^{8}\left(\frac{-x^{2}}{16}-(-4)\right) d x \\ \end{aligned}$
$\begin{aligned} &=2 \times\left[\frac{-x^{3}}{16(3)}+4 x\right]_{0}^{8} \\ &=2 \times\left[\frac{8^{3}-0^{3}}{48}-4(8-0)\right] \\ \end{aligned}$
$\begin{aligned} &=2 \times\left[\frac{512}{48}-32\right] \\ &=2 \times\left[\frac{32}{3}-32\right] \\ \end{aligned}$
$\begin{aligned} &=2 \times \frac{64}{3} \\ &=\frac{128}{3} \end{aligned}$
So, the answer is $\begin{aligned} &\frac{128}{3} \end{aligned}$ sq. Units

Area of bounded region exercise 20.2 question 5

Answer:

$\frac{3}{5}a^{2}\left ( 2^{\frac{5}{3}} -1\right )$
Hint:
Area of the bounded region = Area under (y=2a) and $(ay^{2}=x^{3})$- Area under $(y=2a)$ and $(ay^{2}=x^{2})$
Given: $ay^{2}=x^{3}, y=a, y=2a$
Solution:


Area under $(y=2a)$ and $(ay^{2}=x^{2})$

$\begin{aligned} &=\int_{0}^{1.58 a}\left(2 a-\sqrt{\frac{x^{3}}{a}}\right) d x \\ \end{aligned}$
$\begin{aligned} &=\left[2 a x-\frac{2 x^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]_{0}^{1.58 a} \\ \end{aligned}$
$\begin{aligned} &=2 a(1.58 a)-\frac{2(1.58 a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}} \end{aligned}$
Area under $(y=a)$ and $(ay^{2}=x^{2})$
$\begin{aligned} &=\int_{0}^{a}\left(a-\sqrt{\frac{x^{2}}{a}}\right) d x \\ \end{aligned}$
$\begin{aligned} &=\left[a x-\frac{2 x^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]_{0}^{a} \\ \end{aligned}$
$\begin{aligned} &=a(a)-\frac{2(a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}} \end{aligned}$
Required area
$2 a(1.58 a)-\frac{2(1.58 a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}-\left[a^{2}-\frac{2(a)^{\frac{5}{2}}}{5 a^{\frac{1}{2}}}\right]$
Solving further,

$\begin{aligned} &\frac{3}{5} a^{2}\left(2(2)^{\frac{2}{3}}-1\right) \text { sq. Units } \\ \end{aligned}$
$\begin{aligned} &=\frac{3}{5} a^{2}\left(2^{1+\frac{2}{3}}-1\right) \text { sq. Units } \\ \end{aligned}$$\begin{aligned} &=\frac{3}{5} a^{2}\left(2^{\frac{5}{3}}-1\right) \text { sq. Units } \end{aligned}$

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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