RD Sharma Solutions Class 12 Mathematics Chapter 20 FBQ

The Class 12 RD Sharma chapter 20 exercise FBQ solution deals with the chapter of 'Area of bounded region,' which brings out the concepts of a bounded region, which means any flat, curved, or irregular expanse of a surface or the extent of a two-dimensional surface enclosed within a specified boundary or geometric figure. The RD Sharma class 12 exercise FBQ collects the most critical questions important in this chapter.

RD Sharma Class 12 Solutions Chapter20 FBQ 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-20.4

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

Areas of Bounded Regions Excercise:FBQ

Areas Of Bounded Region exercise Fill in the blanks question 1

Answer:

$\frac{37}{3}$ Sq. units
Hint:
Use indefinite integral formula then put limits to solve this integral.
Given:
$x=y^2,y-axis$
Explanation:
$$\begin{gathered} y=3,x=9[\because x=y^2] \\ y=4,x=16 \end{gathered}$$

$y^2=x$ denotes the curve in the graph.
Area, $OABC={\int }_3^4{y}^{2}dy$
$\begin{array}{rl}& ={\left[\frac{y^3}{3}\right]}_3^4\\ & =\frac{1}{3}[(4{)}^3-(3{)}^3]& & & & & & & & [\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]\\ & =\frac{1}{3}[64-27]\\ & =\frac{37}{3}\mathrm{sq}\cdot \text{ units }\end{array}$

Areas Of Bounded Region exercise Fill in the blanks question 4

Answer:

4 sq.units
Hint:
Use indefinite integral formula then put limits to solve this integral
Given:
$$\begin{gathered} y=\sin x,x-axis \\ x=0,x=2\pi \end{gathered}$$
Explanation:
Required area=$2{\int }_0^{\pi }\sin xdx$
$$\begin{gathered} ={[-2\cos x]}_0^{\pi } \\ =-2[\cos \pi -\cos 0] \\ =-2[-1-1] \\ =4sq.units\left[\begin{array}{c}\because {\int }_a^b\sin xdx=[-cosx{]}_a^b\\ \because \cos \pi =-1\\ \because \cos 0=1\end{array}\right] \end{gathered}$$

Areas Of Bounded Region exercise Fill in the blanks question 9

Answer:

$\frac{8a^2}{3}$ sq.units
Hint:
Use this formula to integrate : ${\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b$
Given:$y^2=4ac$, latus-rectum
Explanation:
Latus-rectum(x=a)

Area $\begin{array}{r}=2{\int }_0^{a}2\sqrt{ax}dx\end{array}$
$\begin{array}{rl}& =2\times 2{\left[\sqrt{a}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a\end{array}$
$\begin{array}{rl}& =2\times \frac{4}{3}\sqrt{a}[a^{\frac{3}{2}}-0][\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]\end{array}$
$=\frac{8a^2}{3}$ sq.units

Areas Of Bounded Region exercise Fill in the blanks question 10

Answer:

$\frac{1}{\sqrt{3}}$
Hint:
Use this formula to integrate : ${\int }_b^a{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b$
Given:
$y=a{x}^2,x=a{y}^2,a>0$
Area=1sq. units
Explanation:
Intersection point
$$\begin{gathered} y=a{x}^2=a(a{y}^2{)}^2[\because x=a{y}^2] \\ y=a^3{y}^4 \\ {a}^3{y}^4-y=0 \\ y(a^3{y}^3-1)=0 \\ y=0y=\frac{1}{a} \\ x=a{\left[\frac{1}{a}\right]}^2=\frac{1}{a} \end{gathered}$$


Required area= ${\int }_1^{\frac{1}{a}}(\sqrt{\frac{x}{a}}-a{x}^2)dx$
$\begin{array}{rl}& =\frac{1}{\sqrt{a}}{\int }_0^{\frac{1}{a}}\sqrt{x}dx-a{\int }_0^{\frac{1}{a}}{x}^{2}dx\end{array}$
$\begin{array}{rl}& =\frac{1}{\sqrt{a}}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{1}{a}}-a{\left[\frac{x^3}{3}\right]}_0^{\frac{1}{a}}\end{array}$
$\begin{array}{rl}& =\frac{2}{3\sqrt{a}}[{\left(\frac{1}{a}\right)}^{\frac{3}{2}}-0]-\frac{a}{3}[{\left(\frac{1}{a}\right)}^3-0][\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]\end{array}$
$\begin{array}{rl}& =\frac{2}{3\sqrt{a}}\left(\frac{1}{a\sqrt{a}}\right)-\frac{1}{3a^2}\\ & =\frac{1}{3a^2}\end{array}$
Given, Area=1
$\begin{array}{rl}& \frac{1}{3a^2}=1\\ & a^2=\frac{1}{3}\\ & a=\pm \frac{1}{\sqrt{3}}\\ & \text{ As }a>0,\\ \therefore a=\frac{1}{\sqrt{3}}\end{array}$

Areas Of Bounded Region exercise Fill in the blanks question 12

Answer:

2 sq.units
Hint:
Use indefinite integral formula then put limits to solve this integral.
Given:
$y=\cos x,x=0,x=\pi$
Explanation:
Required area,

$$\begin{gathered} =2{\int }_0^{\frac{\pi }{2}}\cos xdx \\ ={[2\sin x]}_0^{\frac{\pi }{2}} \\ =2[\sin \frac{\pi }{2}-\sin 0] \\ =2sq.units[\because \int \cos xdx=\sin x+c] \end{gathered}$$

Areas Of Bounded Region exercise Fill in the blanks question 13

Answer:

$\pi$ sq. units
Hint:
Use this formula to integrate : $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}$
Given:
$x^2+y^2=1$
Explanation:

Area $\begin{array}{rl}& \text{ Area }ABCD=4(\text{ Area }OAB)\end{array}$
$\begin{array}{rl}& =4{\int }_0^1\sqrt{1-x^2}dx\\ & =4{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\frac{x}{1}]}_0^1\end{array}$$\begin{array}{rl}& =4[\frac{1}{2}\sqrt{1-1}+\frac{1}{2}{\sin }^{-1}1-\frac{0}{2}\sqrt{1-0}-\frac{1}{2}{\sin }^{-1}0][\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{1}{2}{\sin }^{-1}\frac{x}{a}]\end{array}$
$\begin{array}{rl}& =4[\frac{1}{2}\times \frac{\pi }{2}]\\ & =4\times \frac{\pi }{4}\\ & =\pi \text{ sq.units }\end{array}$

Areas Of Bounded Region exercise Fill in the blanks question 14

Answer:

$20\pi$sq.units
Hint:
Use this formula to integrate: $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}$
Given:
$x^2+y^2=1$
Explanation:

Area ABCD=4(Area OAB) … (i)
Now, $\frac{x^2}{25}+\frac{y^2}{16}=1$
$$\begin{gathered} \frac{y^2}{16}=1-\frac{x^2}{25} \\ \frac{y^2}{16}=\frac{25-x^2}{25} \\ y=\frac{4}{5}\sqrt{25-x^2} \end{gathered}$$
Now, from (i) we have
Required area
$$\begin{gathered} =4{\int }_5^0\frac{4}{5}\sqrt{25-x^2}dx \\ =\frac{16}{5}{[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}]}_0^5 \\ =\frac{16}{5}[\frac{5}{2}\sqrt{25-5^2}+\frac{25}{2}{\sin }^{-1}1]-\frac{16}{5}[\frac{0}{2}\sqrt{25-0}+\frac{25}{2}{\sin }^{-1}0] \\ =\frac{16}{5}[\frac{25}{2}\times \frac{\pi }{2}] \\ =20\pi sq.units \end{gathered}$$

The RD Sharma class 12th exercise FBQ consists of a total of 15 questions that are short and precise, covering up the essential concepts of this chapter that are important for board exams. The concepts covered in the RD Sharma class 12 solution of Area of bounded region exercise FBQ are mentioned below-

• Method to find the area between two curves

• The area between two curves using vertical and horizontal stripes

• Area of the region bounded by ellipse

• Area of the region bounded by the curve and the line

• Area of the region bounded by a parabola and latus ractum

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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