RD Sharma Class 12 Exercise 20.2 Areas of Bounded Region Solutions Maths

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

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 03:19 PM IST

Maths is a tricky subject to master because of the abundance of formulae and theorems. To get a good hold on the subject, students need to refer to explicit materials covering all concepts. This is where RD Sharma books come to help with their explicit material and easy-to-understand images.

RD Sharma Class 12th Exercise 20.2 is a relatively short exercise covering concepts like finding areas of curves through integration. It contains four Level 1 questions and 1 Level 2 question. RD Sharma solutions This exercise is based on the area of the region bounded by parabola and lines, parabola and its latus rectum. However, students can refer to the material provided by Career360 to complete this exercise in a breeze.

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RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise
Areas of Bounded Regions Exercise:20.2
RD Sharma Chapter wise Solutions

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

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

Calculate area under the curve A and Line B
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

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5. What is integration using limits?

Limits are the values that are substituted after integrating the function. For example, there are upper and lower limits for an integral, which must be subtracted in the end.