RD Sharma Class 12 Exercise 20.3 Areas Of Bounded Region Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 20.3 Areas Of Bounded Region Solutions Maths - Download PDF Free Online

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

The RD Sharma class 12 solution of area of bounded region exercise 20.3 is one of the easy chapters to start with if any student is thinking of practicing the maths solutions with a less challenging chapter. The RD Sharma class 12th exercise 20.3 will give you an insight of the different types of concepts covered in the RD Sharma solutions chapter and hence improve your skills and enhance solving qualities to help you Ace the maths exam. The class 12 RD Sharma chapter 20 exercise 20.3 solution guides you so that you can get a grip on the subject of maths even if you think that you are not so good at the subject.

## Areas of Bounded Regions Excercise:20.3

Areas of Bounded Region Exercise 20.3 Question 1

12 sq.units.
Given:
$y^2=6x$
$x^2=6y$
Hint:
The intersecting points of the given parabolas are obtained by solving these equations for x and y , which are 0(0,0)and (6,6).
Solution:
The given equations are
$y^2=6x$
$y=\sqrt{6x}$ ...…(i)
And
$x=\sqrt{6y}$ (ii)
Putting x value on x = 6
When y =0 then x=6
And when y =6 then x = 6
On solving these two equations ,we get point of intersections
The points are O(0,0) and A(6,6). These are shown in the graph below

Bounded Area ,A=[Area between the curve (i) and x -axis from 0 to 6]-[Area between the curve (ii) and x -axis from 0 to 6)
\begin{aligned} &A=\int_{0}^{6}\sqrt{6 x} d x-\int_{0}^{6} \frac{x^{2}}{6} d x \\ &A=\int_{0}^{6}\left(\sqrt{6 x}-\frac{x^{2}}{6}\right) d x \end{aligned}
On integration the above definite integration
\begin{aligned} &A=\int_{0}^{6}\left(\sqrt{6 x}-\frac{x^{2}}{6}\right) d x \\ &=\left[\sqrt{6} \frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{18}\right]_{0}^{6} \\ &=\left[\sqrt{6} \frac{(6)^{3 / 2}}{3 / 2}-\frac{(6)^{3}}{18}\right] \end{aligned}
$=12 \; \text{sq units.}$

Areas of Bounded Region Exercise 20.3 Question 2

Given:
$4y^2=9x$
$3x^2=16y$
Hint:
the bounded area ,A=[Area between the curve (1) and x -axis from 0 to 4]-[Area between the curve (2)and x -axis from 0 to 4]
Solution:
The given eqation are,
$4y^2=9x$
$y=\frac{3}{2}\sqrt{x}$ .........(1)
And,
$3x^2=16y$
$y=\frac{3x^2}{16}$
Equating (1)and (2)
\begin{aligned} \frac{3}{2} \sqrt{x} &=\frac{3 x^{2}}{16} \\ x^{\frac{2}{2}} &=4^{\frac{3}{2}} \\ x &=4 \end{aligned}
When we put x =4 in eqation (I)
Then y =3 ,
When we put x =0 in eqation (I)
Then y = 3 ,
On solving these two equations,we get the point of intersections
The points are shown in the graph below

Now the bounded area ,
A= [Area between the curve (1) and x- axis from 0 to 4] - [ Area between the curve (2) and x - axis from 0 to 4]
\begin{aligned} &A=\int_{0}^{4} \frac{3}{2} \sqrt{x} d x-\int_{0}^{4} \frac{3 x^{2}}{16} d x \\ &A=\int_{0}^{4}\left(\frac{3}{2} \sqrt{x}-\frac{3 x^{2}}{16}\right) d x \end{aligned}
On integrating the above definate integration ,
The required area
\begin{aligned} A &=\int_{0}^{4}\left[\frac{3 \sqrt{x}}{2}-\frac{3 x^{2}}{16}\right] d x \\ &=\left[x^{3 / 2}-\frac{x^{3}}{16}\right]_{0}^{4} \\ &=\left[(4)^{3 / 2}-\frac{(4)^{3}}{16}\right] \\ &=\left[8-\frac{64}{16}\right] \\ &=[8-4] \\ &=4 \end{aligned}

Areas of Bounded Region Exercise 20.3 Question 3

$\frac{1}{6}$ sq.units
Given:
$y=\sqrt{x}$ and
$y=x$
Hint:
Bounded Area ,
A = [Area between curve 1 and axis from 0 to 1] - [Area between the curve 2 and axis from 0 to 1]
Solution:
The given eqation are,
$y=\sqrt{x}$ (i)and $y=x$ (ii)
Solving eqation (1) and( 2)
$y^2=x=y$
$y^2=y$
$y(y-1)=0$'
So, $y=0$
Or y = 1 and x = 0 or x = 1
On solving these two equations we get the points of intersection .
The point are O(0,0)and A(1,1) these are shown in the graph below

Now the bounded area is the required area to be calculated,
Hence,Bounded Area ,
A = [Area between curve 1 and axis from 0 to 1] - [Area between the curve 2 and axis from 0 to 1]
\begin{aligned} &A=\int_{0}^{1} \sqrt{x} d x-\int_{0}^{1} x d x \\ &A=\int_{0}^{1}(\sqrt{x}-x) d x \end{aligned}
On integrating the above definate integration,
\begin{aligned} &=\int_{0}^{1}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{1}(\sqrt{x}-x) d x \\ &=\left[\frac{2}{3} x \sqrt{x}-\frac{x^{2}}{2}\right]_{0}^{1} \\ &=\left[\frac{2}{3} 1 \sqrt{1}-\frac{(1)^{2}}{2}\right]-[0] \\ &=\left[\frac{2}{3}-\frac{1}{2}\right] \end{aligned}
=$\frac{1}{6}$ sq.units

Areas of Bounded Region Exercise 20.3 Question 4

Answer:$\frac{28}{3}$ sq.units
Given:$y=4-x^2$ and the lined y =0, y =3
Hint:
Bounded area ,A = 2 times [area between the equation 1 and y axis from y =0 to y = 3 ]
Solution:
The given equation are ,
$y=4-x^2$ …(1)
$y=0$ ..(2)
And $y=3$ …(3)
Equation 1 represents a parabola with vertex (0,4) and passes through (0,2),(0,2)
Equation 2 is x - axis and cutting the parabola at C(2,0) and D(-2,0)
Equatin 3 is a line parellel to x - axis cutting the parabola at A(3,1)and B(-3,1)
On solving these equations ,we get pont of intersection of a parabola with the other two lines are A(3,1) B(-3,1),c(2,0)and D(-2,0) these are shown in the graph below

Now the bounded area is the required area to be calculated,
Hence ,bounded area ,A = 2 times [area between the equation 1 and y axis from y = 0 to y =3 ]
$A=-2 \int_{0}^{3} \sqrt{4-y} d y$
On intregrating the above definate integration,
\begin{aligned} &=-2\left[\frac{(4-y)^{3 / 2}}{3 / 2}\right]_{0}^{3} \\ &=-2 \frac{2}{3}\left[4^{3 / 2}-1^{3 / 2}\right] \end{aligned} , as area in not negative.
=$\frac{28}{3}$ sq units
The area bounded by the curved $y=4-x^2$ and the lined y=0, y=3 is $\frac{28}{3}$ sq units

Areas of Bounded Region Exercise 20.3 Question 5

Answer:$\frac{ab}{4}[\pi -2]$ Square units.
Hint:
Given:
$\left\{(x, y): \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 \leq \frac{x}{a}+\frac{y}{b}\right\}$
Solution:

\begin{aligned} &\text { Let } R=\left\{(x, y): \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 \leq \frac{x}{a}+\frac{y}{b}\right\} \\ &R_{1}=\left\{(x, y): \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1\right\} \\ &\text { And } R_{2}=\left\{(x, y): 1 \leq \frac{x}{a}+\frac{y}{b}\right\} \\ &\text { Then } R=R_{1} \cap R_{2} \end{aligned}
considered $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ,this represent an ellipse ,symmetrical about both axis and cutting x-axis at A(a,0)and A(-a,0) and y-axis at B(0,b),B(0,-b)
$R_1=\left (\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1 \right )$ Represents the area inside the ellipse
$\frac{x}{a}+\frac{x}{b}=1$ Represent a straight line cutting x-axis at A(a,0) and y- axis at B(0,b)
$R_2=\left (\frac{x}{a}+\frac{x}{b}\geq 1 \right )$ Represent the area above the straight line
$R=R_1\cap R_2$ Represent the smaller shaded arera bounded by the line and the ellips in the shaded region, consider a vertical strip with length=$\left | y_2-y_1 \right |$ and width=dx such that $P(x,y_2)$ lies on ellipse and $(x,y_1)$ lies on the straight line area of approximating rectangle=$\left | y_2-y_1 \right |dx$
The approximating rectangle move from x = 0 to x = a
$\therefore$ Area of the shaded region$=\int_{0}^{a}\left|y_{2}-y_{1}\right| d x=\int_{0}^{a}\left(y_{2}-y_{1}\right) d x \ldots \ldots \ldots \ldots .\left[A s, y_{1}\left|y_{2}-y_{1}\right|=y_{2}-y_{1}\right]$
\begin{aligned} &A=\int_{0}^{a}\left(\frac{b}{a} \sqrt{a^{2}-x^{2}}-\frac{b}{a}(a-x)\right) d x \\ &A=\int a\left(\frac{b}{a} \sqrt{a^{2}-x^{2}}\right) d x-\int_{0} a \frac{b}{a}(a-x) d x \\ &A=\frac{b}{a}\left[\left\{\frac{x}{a} \sqrt{\left.\left.a^{2}-x^{2}+\frac{1}{2} a^{2} \sin ^{-1}\left(\frac{x}{a}\right)\right\}\right]_{0}^{a}-\frac{b}{a}\left[a x-\frac{x^{2}}{2}\right]_{0}^{a}}\right.\right. \\ &A=\frac{b}{a}\left[\frac{1}{2} a^{2} \times \frac{\pi}{2}-\frac{a^{2}}{2}\right] \\ &A=\frac{a b}{2}\left[\frac{\pi}{2}-1\right] \\ &A=\frac{a b}{4}[\pi-2] \text { Square units } \end{aligned}

Areas of Bounded Region Exercise 20.3 Question 6

4 square units
Hint:
Given:
A(2,1),B(3,4)and C(5,2)
Solution:

The equation of AB,
\begin{aligned} &y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right) \\ &y-1=\left(\frac{4-1}{3-2}\right)(x-2) \\ &y-1=\frac{3}{1}(x-2) \\ &y=3 x-5 \ldots \text { (1) } \end{aligned}
The equation of BC,
\begin{aligned} &y-4=\left(\frac{2-4}{5-3}\right)(x-3) \\ &=\frac{-2}{2}(x-3) \\ &y=-x+7 \ldots(2) \end{aligned}
The equation of AC,
\begin{aligned} &y-1=\left(\frac{2-1}{5-2}\right)(x-2) \\ &y-1=\frac{1}{3}(x-2) \\ &y=\frac{1}{3} x-\frac{2}{3}+1 \\ &y=\frac{1}{3} x+\frac{1}{3} \ldots . .(3) \end{aligned}
Now the required area (A)=[(Area between line AB and x-axis)-(Area between line AC and x-axis)from x=2 to x=3]
+[(Area between line BC and x-Axis )-(Area between line AC and x-Axis )from x=3 to x=5]
\begin{aligned} A &=\int_{2}^{3}\left(y_{1}-y_{3}\right) d x+\int_{3}^{5}\left(y_{2}-y_{1}\right) d x \\ =& \int_{2}^{3}\begin{aligned} \left[(3 x-5)-\left(\frac{1}{3} x+\frac{1}{3}\right) ]d x\right.\\ +\int_{3}^{5}\left[(-x+7)-\left(\frac{1}{3} x+\frac{1}{3}\right)\right] d x \end{aligned} \\ =& \int_{2}^{3}\left[3 x-5-\frac{1}{3} x-\frac{1}{3}\right] d x \\ +& \int_{3}^{5}\left[-x+7-\frac{1}{3} x+\frac{1}{3}\right] d x \end{aligned}
\begin{aligned} &=\int_{2}^{3}\left(\frac{8 x}{3}-\frac{16}{3}\right) d x+\int_{3}^{5}\left(-\frac{4}{3} x+\frac{20}{3}\right) d x \\ &=\frac{8}{3}\left(\frac{x^{2}}{2}-2 x\right)_{2}^{3}-\left(\frac{4 x^{2}}{6}-\frac{20}{3}\right)_{3}^{5} \\ \; \; \; \; =& \frac{8}{3}\left[\left(\frac{9}{2}-6\right)-(2-4)\right]-\left[\left(\frac{50}{3}-\frac{100}{3}\right)(6-20)\right] \\ &=\frac{8}{3}\left[-\frac{3}{2}+2\right]-\left[-\frac{50}{3}+14\right] \\ &=\frac{4}{3}-\left[-\frac{8}{3}\right] \end{aligned}
=4 square unit
The area of the region bounded by the triangle whose vertices are (2,1),(3,4) and (5,2) is 4 sq. units

Areas of Bounded Region Exercise 20.3 Question 7

$\frac{15}{2}$ sq. Units
Hint:
Given : A(-1,1),B(0,5),C(3,2)
Solution
We have to find the area of the triangle whose vertices are A(-1,1),B(0,5),C(3,2) as shown below

The equation of AB
\begin{aligned} &y=y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right) \\ &y-y_{1}=\left(\frac{5-1}{0+1}\right)(x+1) \\ &y-1=\frac{4}{1}(x+1) \\ &y-1=4 x+4 \\ &Y=4 x+5 \ldots . .(1) \end{aligned}
The equation of BC,
\begin{aligned} &y-5=\left(\frac{2-5}{3-0}\right)(x-0) \\ &=\frac{-3}{3}(x-0) \\ &y-5=-x \\ &y=5-x \ldots(2) \end{aligned}
The equation oa AC,
\begin{aligned} &y-1=\left(\frac{2-1}{3+1}\right)(x+1) \\ &y-1=\frac{1}{4}(x+1) \\ &y-1=\frac{1}{4} x+\frac{1}{4} \\ &y=\frac{1}{4}(x+5) \ldots(3) \end{aligned}
Now the required Area (A)=[(Area between line BC and x-Axis )-(Area between line AC and x-Axis) from x=0 to x=3]
Say ,Area$A =A_1+A_2$
\begin{aligned} &A_{1}=\int_{-1}^{0}\left[(4 x+5)-\frac{1}{4}(x+5)\right] d x \\ &=\int_{-1}^{0}\left[4 x+5-\frac{x}{4}-\frac{5}{4}\right] d x \\ &=\int_{-1}^{0}\left(\frac{15}{4} x+\frac{15}{4}\right) d x \\ &=\frac{15}{4}\left(\frac{x^{2}}{2}+x\right)_{-1}^{0} \\ &=\frac{15}{4}\left[(0)-\left(\frac{1}{2}-1\right)\right] \\ &=\frac{15}{8} \end{aligned}
\begin{aligned} &\text { And, } A_{2}=\int_{0}^{3}\left(y_{2}-y_{3}\right) d x \\ &=\int_{0}^{3}\left[(5-x)-\left(\frac{1}{4} x+\frac{5}{4}\right)\right] d x \\ &=\int_{0}^{3}\left[5-x-\frac{1}{4} x-\frac{5}{4}\right] d x \\ &=\int_{0}^{3}\left(-\frac{5}{4} x+\frac{15}{4}\right) d x \\ &=\frac{5}{4}\left(3 x-\frac{x^{2}}{2}\right)_{0}^{3} \\ &=\frac{5}{4}\left[9-\frac{9}{2}\right] \\ &=\frac{45}{8} \end{aligned}
So the enclosed area of the triangle is $\frac{15}{8}+\frac{45}{8}=\frac{15}{2}$ square units

Areas of Bounded Region Exercise 20.3 Question 8

8 sq units
Hint:
Given:
y=2x+1,y=3x+1 and x=4
Solution:
The given lines are
y=2x+1…..(1)
y=3x+1…..(2)
x=4…….(3)

For intersection points of (1) and( 3)
Y=2x4+1=9
Coordinates of intersecting point of 1 and 3 is (4,9) for intersection point of (2) and (3)
Y=3x4+1=13
i.e, coordinates of intersection point of (2) and (3) is (4, 3)
For intersection point of (1) and (2)
2x+1=3x+1=>x=0
Y=1
i.e., coordinates of intersection point of (1) and (2) is (0, 1)
Shaded region is required triangle region.
Required Area =Area of trapezium OABD-Area of trapezium OACD
\begin{aligned} &\int_{0}^{4}(3 x+1) d x-\int_{0}^{4}(2 x+1) d x \\ &=\left[3 \frac{x^{2}}{2}+x\right]_{0}^{4}-\left[\frac{2 x^{2}}{2}+x\right]_{0}^{4} \\ &=[(24+4)-0]-[(16+4)-0]=28-20 \end{aligned}
=8 sq units

Areas of Bounded Region Exercise 20.3 Question 9

$2\left [ \frac{\sqrt{2}}{3}+\frac{9 \pi} {4}-\frac{9}{2} \sin ^{-1} \left ( \frac{1}{3} \right ) \right ]$ square units
Hint:
Given:
$\left\{(x, y): y^{2} \leq 8 x, x^{2}+y^{2} \leq 9\right\}$
Solution:
To find area $\left\{(x, y): y^{2} \leq 8 x, x^{2}+y^{2} \leq 9\right\}$
$y^2=8x.....(1)$
$x^2+y^2=9 .....(2)$
On solving the equation (1) and (2)
\begin{aligned} &x^{2}+8 x=9 \\ &x^{2}+8 x-9=0 \\ &(x+9)(x-1)=0 \\ &x=-9 \text { or } x=1 \end{aligned}

And when x=1 then y= $\pm 2\sqrt{2}$
Equation (1) represents a parabola with vertex (0,0) and axis as x-axis equation (2) represents a circle with centre (0,0) and radius 3 units, so it meets area at $(\pm 3,0),(0,\pm 3)$
Point of intersection of parabola and circle is $(1,2\sqrt{2})$ and $(1-2\sqrt{2})$
The sketch of the curves is as below

Or, required area=(region ODCO+REGION DBCD)
\begin{aligned} &=2\left[\int_{0}^{1} \sqrt{8 x} d x+\int_{1}^{3} \sqrt{9-x^{2}} d x\right] \\ &=2\left[\left(2 \sqrt{2 \cdot \frac{2}{3} x \sqrt{x}}\right)_{0}^{1}+\left(\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right)_{1}^{3}\right] \\ &=2\left[\left(\frac{4 \sqrt{2}}{3} \cdot 1 \sqrt{1}\right)+\left\{\left(\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin ^{-1}(1)\right)-\left(\frac{1}{2} \sqrt{9-1}+\frac{9}{2} \sin ^{-1} \frac{1}{3}\right)\right\}\right] \\ &=2\left[\frac{4 \sqrt{2}}{3}+\left\{\left(\frac{9}{2} \cdot \frac{\pi}{2}\right)-\left(\frac{2 \sqrt{2}}{2}-\frac{9}{2} \sin ^{-1}\left(\frac{1}{3}\right)\right)\right\}\right] \\ &=2\left[\frac{4 \sqrt{2}}{3}+\frac{9 \pi}{4}-\sqrt{2}-\frac{9}{2} \sin ^{-1}\left(\frac{1}{3}\right)\right] \end{aligned}
Hence, the required area is $2\left [ \frac{\sqrt{2}}{3}+\frac{9 \pi} {4}-\frac{9}{2} \sin ^{-1} \left ( \frac{1}{3} \right ) \right ]$ square unit

Areas of Bounded Region Exercise 20.3 Question 10

$\left (\frac{4\sqrt{3}}{3}+\frac{16}{3}\pi \right )$ Square unit
Hint:
Given:
$x^{2}+y^{2}=16$
$y^2=6x$
Solution:

$x^{2}+y^{2}=16$ …(1)
Equation of circle with centre (0,0) and radius 4
Equation of parabola:
$y^2=6x$ ….(2)
Intersecting piont of (1) and (2)
\begin{aligned} &x^{2}+6 x-16=0 \\ &x^{2}+8 x-2 x-16=0 \\ &x(x+8)-2(x+8)=0 \\ &(x+8)(x-2)=0 \\ &x=-8,2 \\ &\text { but }, x \neq-8, x=2 \\ &y^{2}=12 \\ &y=\pm \sqrt{12} \\ &y=\pm 2 \sqrt{3} \end{aligned}
So points are $(2,2\sqrt{3})$ and $(2,-2\sqrt{3})$
\begin{aligned} &=2\left[\int_{0}^{3} \sqrt{6} \sqrt{x} d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right] \\ &=2 \sqrt{6}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{2}+2 \int_{2}^{4} \sqrt{16-x^{2}} d x \\ &=\frac{4}{3} \sqrt{6}(2)^{3 / 2}+2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2}^{4} \\ &=\frac{4}{3} \sqrt{6} \times 2 \sqrt{2}+16 \sin ^{-1} 1-2 \sqrt{12}-16 \sin ^{-1} \frac{1}{2} \\ &=\frac{8}{3} \times 2 \sqrt{3}+16 \frac{\pi}{2}-4 \sqrt{3}-16 \frac{\pi}{6} \end{aligned}
$\frac{4\sqrt{3}}{3}+8 \pi -\frac{8 \pi}{3}$
$=\left (\frac{4\sqrt{3}}{3}+\frac{16}{3}\pi \right )$ square unit

Areas of Bounded Region Exercise 20.3 question 11.

$\left (\frac{8 \pi }{3}-2\sqrt{3} \right ) \text{sq. units}$
Hint:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{x}{a}+c$ .
Given: The equation of the given curves is
$x^2+y^2=4$ ........(I)
$(x-2)^2+y^2=4$ .........(II)
Solution: The equation of the given curves is
$x^2+y^2=4$ ........(I)
$(x-2)^2+y^2=4$ .........(II)
Clearly $x^2+y^2=4$ represents a circle with center (0, 0) and radius 2. Also, $(x-2)^2+y^2=4$ represents a circle with centre (2, 0) and radius 2. To find the point of intersection of the given curves, we solve (i) and (ii). Simultaneously, we find the two curves intersect at $A(1, \sqrt{3})$ and $D(1, -\sqrt{3})$.
Since both the curves are symmetrical about x-axis, So, the required area = 2(Area OABCO) Now, we slice the area OABCO into vertical strips. We observe that the vertical strips change their character at A(1,√3). So, Area OABCO = Area OACO + Area CABC.

When area OACO is sliced in the vertical strips, we find that each strip has its upper end on the circle (x - 2)2 + (y - 0)2 = 4 and the lower end on x-axis. So, the approximating rectangle shown in figure has length = y1 width = $\Delta x$ and area = y1$\Delta x$. As it can move from x = 0 to x = 1
\begin{aligned} &\therefore \text { Area } O A C O=\int_{0}^{1} y_{1} d x \\ &\therefore \text { AreaO } A C O=\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x \end{aligned}
Similarly, approximating rectangle in the region CABC has length = y2 , width = $\Delta x$ and area = y2$\Delta x$
As it can move from x =1 to x =2
$\text { Area } C A B C=\int_{1}^{2} y_{2} d x=\int_{1}^{2} \sqrt{4-x^{2}} d x$
Hence,required area A is given by
\begin{aligned} &A=2\left[\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x+\int_{1}^{2} \sqrt{4-x^{2}} d x\right] \\ &\Rightarrow A=2\left[\left[\frac{(x-2)}{2} \sqrt{4-(x-2)^{2}}+\frac{4}{2} \sin ^{-1} \frac{(x-2)}{2}\right]_{0}^{1}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{1}^{2}\right] \\ &\Rightarrow A=2\left\{-\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(-\frac{1}{2}\right)-2 \sin ^{-1}(-1)+2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin -1 \frac{1}{2}\right\} \end{aligned}
\begin{aligned} &=2\left[-\sqrt{3}-2\left(\frac{\pi}{6}\right)+2\left(\frac{\pi}{2}\right)+2\left(\frac{\pi}{2}\right)-2\left(\frac{\pi}{6}\right)\right] \\ &=2\left(-\sqrt{3}-\frac{2 \pi}{3}+2 \pi\right) \\ &=2\left(\frac{4 \pi}{3}-\sqrt{3}\right) \\ &=\left(\frac{8 \pi}{3}-2 \sqrt{3}\right) \text { sq.units. } \end{aligned}

Areas of Bounded Region Exercise 20.3 question 12.

$\frac{9}{2}\; \text{sq. units}$ .
Hint: Using the formula
Given: The equations of the given curves are
$y^2=x$ ..........(I)
$x+y =2$ ........(II)
Solution: Plot the two curves
$y^2=x$ ..........(I)
$x+y =2$ ........(II)

Solving (I) and (II) ,we have
\begin{aligned} &y^{2}+y=2\\ &(y+2)(y-1)=0\\ &y=-2,1 \end{aligned}
We have to determine the area of shaded region.
Required Area
\begin{aligned} &=\int_{2}^{1}(2-y) d y-\int_{-2}^{1} y^{2} d y \\ &\left.=2 y-\frac{y^{2}}{2}-\frac{y^{3}}{3}\right]_{-2}^{1} \\ &=\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4-\frac{4}{2}+\frac{8}{3}\right) \\ &=\frac{9}{2} \text { square units. } \end{aligned}

Areas of Bounded Region Exercise 20.3 question 13.

\begin{aligned} &{\left[\frac{4}{\sqrt{3}} a^{3 / 2}+\left[\frac{8 \pi}{3}-a \sqrt{\frac{16}{3}-a^{2}}-\frac{16}{3} \sin ^{1}-\left(\frac{\sqrt{3 a}}{4}\right)\right]\right] \text { square unit. }} \\ &\text { where } a=\frac{-9+\sqrt{273}}{9} \end{aligned}
Hint: Using the identity formula $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -{ }^{1} \frac{x}{a}+c .$ .
Given: Here we know that $R = {(x, y): y^2 \leq 3x, 3x^2 + 3y^2\leq 16}$
Solution:

Here we know that $R = {(x, y): y^2 \leq 3x, 3x^2 + 3y^2\leq 16}$
We can write it as
$R = {(x, y): y^2 \leq 3x} \cap {(x, y): 3x^2 + 3y^2 \leq 16} = R_1 \cap R_2$
Here $R_1 = {(x, y): y^2 \leq 3x}$ which represents the region inside the parabola, y2 = 3x with vertex (0, 0) and x-axis as it axis
$R = {(x, y): 3x^2 + 3y^2 \leq 16}$ represents the interior of $3x^2 + 3y^2 = 16$ circle having (0, 0) as centre and $\frac{4}{\sqrt{3}}$ as radius
So the region R which is intersection of points R1 and R2 is shaded in the figure
$3x^2 + 3y^2 = 16$ …… (1)
$y^2 = 3x$…… (2)
Solving both the equations
3x2 + 9x – 16 = 0
So we get
$x = \frac{(-9 \pm \sqrt{2}73)}{}6$
Here $x = \frac{(-9 \pm \sqrt{2}73)}{}6$ is the rejecting negative value
Substituting y = 0 in equation (1)
$x=\frac{4}{\sqrt{3}}$
We know that the circle (1) cuts x-axis at P $\left (\frac{4}{\sqrt{3}},0 \right )$ and P’ ($\left (-\frac{4}{\sqrt{3}},0 \right )$
So the required area can be written as
Required area = 2 [area of ODPAO] = 2 [area of ODAO + area of ADPA] .
$\text { Re quired area }=2\left[\int_{0}^{a} \sqrt{3 x} d x+\int_{a}^{4, \sqrt{\beta}} \sqrt{\sqrt{3}-x^{2}} d x\right]$
Intergrating w.r.t
$=2\left(\sqrt{3}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{a}+\left[\frac{x}{2} \sqrt{\frac{16}{3}-x^{2}}+\frac{16}{3} \sin ^{-1}\left(\frac{x}{4 / \sqrt{3}}\right)\right]_{a}^{4 / \sqrt{s}}\right)$
Substituting the value of x
\begin{aligned} &{\left[\frac{4}{\sqrt{3}} a^{3 / 2}+\left[\frac{8 \pi}{3}-a \sqrt{\frac{16}{3}-a^{2}}-\frac{16}{3} \sin ^{1}-\left(\frac{\sqrt{3 a}}{4}\right)\right]\right] \text { square unit. }} \\ &\text { where } a=\frac{-9+\sqrt{273}}{9} \end{aligned}
$\frac{27}{2}$
Hint:
First we need to find the point of intersection of the given curve
Given: Equation of parabola $y=3x^2$ and equation of line $3x-y+6=0$
Solution:

Equation of parabola $y=3x^2$ ---(i)
Equation of line $3x-y+6=0$ ---(ii)
Putting the value of y in equation (ii)
\begin{aligned} &3 x-3 x^{2}+6=0 \\ &-3 x^{2}+3 x+6=0 \\ &3 x^{2}-3 x-6=0 \\ &x^{2}-x-2=0 \\ &x^{2}-2 x+x-2=0 \\ &x(x-2)+1(x-2)=0 \end{aligned}
$\begin{array}{ll} (x+1)(x-2) & =0 \\ x+1=0 & , x-2=0 \\ x=-1 & , x=2 \end{array}$
When $x=-1$ then $y=3(-1)^2=3$
When $x=2$ then $y=3(2)^2=12$
So the points of the interaction of the given curves is (1 , -3 ) and (2 , 12)
$\text{Required area}= \int_{-1}^{2}(Y_{\text{Line}}-Y_{\text{Parabola}})dx$
\begin{aligned} &=\int_{-1}^{2}\left(3 x+6-3 x^{2}\right) d x \\ &=\left[3 \frac{x^{2}}{2}+6 x \frac{-3 x^{2}}{3}\right]_{-1}^{2} \\ &=\left(\frac{3}{2} \times 4+12-8\right)-\left(\frac{3}{2}-6+1\right) \\ &=\frac{10}{1}+\frac{7}{2} \\ &=\frac{20+7}{2} \\ &=\frac{27}{2} \end{aligned}
Hence , area of the region is $\frac{27}{2}$.

Areas of Bounded Region Exercise 20.3 question 16

$\frac{16}{3}\; \text{ab sq. units}$ .
Hint:
Use basic concepts.
Given:
$y^2=4ax$ and $x^2=4by$ .
Solution: To find area enclosed by
$y^2=4ax$ ..........(1)
$x^2=4by$ .........(2)
Equation (1) represents a parabola with vertex (0,0) and axis as x-axis,
Equation (2) represents a parabola with vertex (0,0), and axis as y-axis,
Point of intersection of parabolas are (0, 0) and $(4a^{\frac{1}{3}} b^{\frac{2}{3}} , 4a^{\frac{2}{3}}b^{\frac{1}{3}})$

A rough sketch is given as:

The shaded region is required area and it is sliced into rectangle of width=$\Delta _x$ and length$(y_2-y_1)$
Area of rectangle=$(y_1-y_2)\Delta _x$
This approximation rectangle slides from $x=0$ to, $x= 4a^{\frac{1}{3}}b^{\frac{2}{3}}$ , so
Required area = Region OQAPO
\begin{aligned} &=\int_{0}^{4 a^{\frac{1}{4} b^{2}} 3}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{4 a^{\frac{1}{a} b^{2}}}\left(2 \sqrt{a} \cdot \sqrt{x}-\frac{x^{2}}{4 b}\right) d x \\ &=\left[2 \sqrt{a} \cdot \frac{2}{3} x \sqrt{x}-\frac{x^{3}}{12 b}\right]_{0}^{4 a^{\frac{1}{3} b^{2}}} \\ &=\frac{32 \sqrt{a}}{3} \cdot a^{\frac{1}{3}} b^{\frac{2}{3}} a^{\frac{1}{6}} b^{\frac{1}{3}} \\ &=\frac{64 a b^{2}}{12 b} \\ &=\frac{16}{3} a b \text { sq.units } \end{aligned}
Hence the required area is $\frac{16}{3}\; \text{ab sq. units}$ .

Areas of Bounded Region Exercise 20.3 question 17.

Answer: $\frac{\pi}{3}$
Hint:
use basic concepts.
Given: The equations of the curves are
$x^2+y^2=4$
$x=\sqrt{3}y$
Solution: Obviously $x^2+y^2=4$ is a circle having centre at (0, 0) and radius 2 units. Forgraph of line
$x=\sqrt{3}y$.
 x 0 1 y 0 0.58

For intersecting point of given circle and line
Putting $x=\sqrt{3}$ in $x^2+y^2=4$ we get
\begin{aligned} &(\sqrt{3} y)^{2}+y^{2}=4 \\ &3 y^{2}+y^{2}=4 \\ &4 y^{2}=4 \\ &y=\pm 1 \\ &x=\pm \sqrt{3} \end{aligned}
Intersecting points are $(\sqrt{3},1 ), (-\sqrt{3},-1)$

Now required area $=\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{0}^{\sqrt{3}} \sqrt{4-x^{2}}$
\begin{aligned} &=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x \sqrt{4-x^{2}}}{2}+\frac{4}{2} \sin -\frac{x}{2}\right]_{\sqrt{3}}^{2} \\ &=\frac{1}{2 \sqrt{3}}(3-0)+\left[2 \sin -{ }^{1}-\left(\frac{\sqrt{3}}{2}+2 \sin -{ }^{1} \frac{\sqrt{3}}{2}\right)\right] \end{aligned}
\begin{aligned} &=\frac{\sqrt{3}}{2}+\left[2 \frac{\pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right] \\ &=\frac{\sqrt{3}}{2}+\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3} \\ &=\pi-\frac{2 \pi}{3} \\ &=\frac{\pi}{3}(\text { proved }) \end{aligned}

Areas of Bounded Region Exercise 20.3 question 18

9 sq. units.
Hint:use basic concepts.
Given:
he given equations of the curves are
$y=\sqrt{x}$, $x=2y+3$
Solution: We have,
$y=\sqrt{x}$, $x=2y+3$
Solving we get,
\begin{aligned} &y=\sqrt{2 y+3}, y \geq 0 \\ &y^{2}=2 y+3, y \geq 0 \\ &y^{2}-2 y-3=0, y \geq 0 \\ &(y-3)(y+1)=0, y \geq 0 \\ &y=3 \end{aligned}

The graph of function $y=\sqrt{x}$ is part of parabola$y=x^2$ lying above x –axis.
The graph is as shown in the figure.

From the figure, area of shaded region,
\begin{aligned} A &=\int_{0}^{3}\left(2 y+3-y^{2}\right) d y \\ &=\left[\frac{2 y^{2}}{2}+3 y-\frac{y^{3}}{3}\right]_{0}^{3} \\ &=\left[\frac{18}{9}+9-9-0\right] \\ &=9 \text { sq units } \end{aligned}

Areas of Bounded Region Exercise 20.3 question 19

$\frac{4a^2}{3}9(\sqrt{3}+4\pi)$
Hint:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{1}{a}+c \text {. }$ .
Given:
Given curves $x^2+y^2=16a^2$ and $y^2=6ax$ .
Solution:
The point of intersection of circle
$x^2+y^2=16a^2$

And parabola, $y^2=6ax$ is
\begin{aligned} &x^{2}+6 a x-16 a^{2}=0 \\ &(x+8 a)(x-2 a)=0 \\ &x=2 a, y=\pm 2 \sqrt{3} a \end{aligned}
The required common area ,$A=2[APOA]$
\begin{aligned} &=2 \int_{0}^{2 a} y d x+2 \int_{2 a}^{4 a} y d y \\ &=2 \int_{0}^{2 a} \sqrt{6 a} \sqrt{x} d x+2 \int_{2 a}^{4 a} \sqrt{(4 a)^{2}-x^{2}} d x \\ &=2 \cdot \sqrt{6 a}\left[\frac{2}{3} x 3 / 2\right]_{0}^{2 a}+2\left[\frac{x}{2} \sqrt{(4 a)^{2}-x^{2}}+\frac{1}{2}(4 a)^{2} \sin -1 \frac{x}{4 a}\right]_{2 a}^{4 a} \\ &=2 \cdot \sqrt{6 a} \frac{2}{3}(2 a)^{3 / 2}+2\left[(0-2 a \sqrt{3 a})+8 a^{2}\left(\sin -^{1}-\sin -1 \frac{1}{2}\right)\right] \\ &=2 \cdot 2 \sqrt{3} \cdot \frac{4}{3} a^{2}+2\left[-2 \sqrt{3} a^{2}+8 a^{2}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\right] \\ &=\frac{16}{3} \sqrt{3} a^{2}-4 \sqrt{3} a^{2}+16 a^{2} \frac{\pi}{3} \\ &=\frac{4 \sqrt{3}}{3} a^{2}+\frac{16 \pi a^{2}}{3} \\ &=\frac{4 a^{2}}{3}(\sqrt{3}+4 \pi) \text { sq.units. } \end{aligned}

Areas of Bounded Region Exercise 20.3 question 20.

$\left (4 \pi-\frac{32}{3} \right ) \text{sq. units}$
Hint::
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{1}{a}+c$ .
Given:
Given curves are $x^2+y^2=8x$ and $y^2=4x$ .
Solution:
We have given equations
$x^2+y^2=8x$ ...............(1)
$y^2=4x$ ........(2)
Equation (1) can be written as
$(x-y)^2+y^2=4^2$
So equation (i) represents a circle with centre (4, 0) and radius 4.
Again, clearly equation (ii) represents parabola with vertex (0, 0) and axis as x -axis. The curve (i) and (ii) are shown infigure and the required region is shaded. On solving equation (i) and (ii) we have points of intersection 0(0, 0) and A(4, 4), C(4, - 4)
Now, we have to find the area of region bounded.

By(i) and (ii) & above x - axis.
So requires region is OBAO.
\begin{aligned} A &=\int_{0}^{4}\left(\sqrt{8 x-x^{2}}-\sqrt{4 x}\right) d x \\ &=\int_{0}^{4}\left(\sqrt{\left(4^{2}\right)-(x-4)^{2}}-2 \sqrt{x}\right) d x \\ &=\left[\frac{(x-4)}{2} \sqrt{(4)^{2}-(x-4)^{2}}+\frac{16}{2} \sin -1 \frac{(x-4)}{4}-2\left(\frac{2 x}{3}\right)^{3 / 2}\right]_{0}^{4} \\ &=\left[8 \sin -{ }^{1} 0-\frac{4}{3} \times 8\right]-\left[8 \times\left(\frac{-\pi}{2}\right)\right] \\ &=\frac{-32}{3}+4 \pi \\ &=\left(4 \pi-\frac{32}{3}\right) \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 21

$12\sqrt{3}\; \text{sq. unnits}$ .
Hint:
Given:
Given curve $y^2=5x^2$ and $y=2x^2+9$ .
Solution:

$y^2=5x^2$ Represents a parabola with vetex(0,0) and opening upward , symmetrical about +ve axis
$y=2x^2+9$ Represents the wider parabola, with vertex at C (0,9)
To find point of intersection, solve the two equations
\begin{aligned} 5 x^{2} &=2 x^{2}+9 \\ 3 x^{2} &=\pm 9 \\ x &=\pm \sqrt{3} \\ y &=15 \end{aligned}
Thus $A(\sqrt{3},15)$ and ’$A'(-\sqrt{3},15)$ are point of intersection of two parabolas.
Shaded area A’OA=2 x area (OCAO)
Consider a vertical stip of length $\left |y_2-y_1 \right |$ and width dx
Area of approximating rectangle $\left |y_2-y_1 \right |dx$
The approximating rectangle moves from $x=0$ to $x=\sqrt{3}$
\begin{aligned} &\therefore \text { Area }(O C A O)=\int_{0}^{\sqrt{3}}\left|y_{2}-y_{1}\right| d x \\ &=\left(y_{2}-y_{1}\right) d x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \cdot\left\{\cdot\left|y_{2}-y_{1}\right|=y_{2}-y_{1} a s y_{2}>y_{1}\right\} \\ &=\int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x \\ &=\int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x \\ &=\left[\left(9 x-3 \frac{x^{3}}{3}\right)\right]_{0}^{\sqrt{3}} \\ &=9 \sqrt{3}-3 \sqrt{3} \\ &=6 \sqrt{3} \end{aligned}
$\therefore$ Shaded area B’A’B=2 area OCAO= $2\times 6\sqrt{3}=12\sqrt{3}\; \text{sq.units}$

Areas of Bounded Regions exercise 20.3 question 22.

$\frac{32}{3}$ sq. Units
Hint:
Use concept.
Given:
The given equations are $y=2x^2$ and $y=x^2+4$
Solution:
To find the area enclosed by,
$y=2x^2$ ........(i)
$y=x^2+4$ ........(ii)
On solving the equation (i) and (ii),
$2x^2=x^2+4$
or $x^2=4$
or $x=\pm 2$
$\therefore y=8$
Equation (1) represents a parabola with vertex (0, 0) and axis as y - axis.
Equation (2) represents a parabola with vertex (0,4) and axis as the y - axis.
Points of intersection of parabolas are A(2, 8) and B(– 2, 8).
These are shown in the graph below:

Required area=Region AOBCA
=2(RegionAOCA)
\begin{aligned} &=2 \int_{0}^{2}\left(x^{2}+4-2 x^{2}\right) d x \\ &=2 \int_{0}^{2}\left(4-x^{2}\right) d x \\ &=2\left[4 x-\frac{x^{3}}{3}\right]_{0}^{2} \\ &=2\left[\left(8-\frac{8}{3}\right)-0\right] \\ &=\frac{32}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 23.

4 sq. units
Hint: use concept.
Given:
The vertices of triangle are (-1,2), (1,5) and (3,4).
Solution:

Equation of,
AB is:$y=\frac{1}{2}(3x+7)$
BC is: $y=\frac{1}{2}(11-x)$
AC is: $y=\frac{1}{2}(x+5)$
Required area
\begin{aligned} &=\frac{1}{2} \int_{-1}^{1}(3 x+7) d x+\frac{1}{2} \int_{1}^{3}(11-x) d x-\frac{1}{2} \int_{-1}^{3}(3 x+5) d x \\ &=\left[\frac{1}{12}(3 x+7)^{2}\right]_{-1}^{1}-\frac{1}{4}\left[(11-x)^{2}\right]_{1}^{3}-\frac{1}{4}\left[(x+5)^{2}\right]_{-1}^{3} \\ &=7+9-12 \\ &=4 \text { sq.units } \end{aligned}

(ii)

Equation of,
AB is: $y =\frac{3}{2}x+4$
BC is: $y =4-\frac{x}{2}$
AC is: $y =\frac{1}{2}x+2$
Required area
\begin{aligned} &=\int_{-2}^{0}\left(\frac{3}{2} x+4\right) d x+\int_{0}^{2}\left(4-\frac{x}{2}\right) d x-\int_{-2}^{2}\left(\frac{1}{2} x+2\right) d x \\ &=\left[\frac{3 x^{2}}{4}+4 x\right]_{-2}^{0}+\left[4 x-\frac{x^{2}}{4}\right]_{0}^{2}-\left[\frac{x^{2}}{4}+2 x\right]_{-2}^{2} \\ &=5+7-8 \\ &=4 \; \text{sq. units} \end{aligned}

(III)

Equation of ,
AB is: $y =x+3$
BC is : $y =\frac{34-5x}{2}$
AC is: $y =\frac{26-3x}{4}$
Required area
\begin{aligned} &=\int_{2}^{4}(x+3) d x+\int_{4}^{6}\left(\frac{34-5 x}{2}\right) d x-\int_{2}^{6}\left(\frac{26-3 x}{4}\right) d x \\ &=\left[\frac{x^{2}}{2}+3 x\right]_{2}^{4}+\left[\frac{34 x}{2}-\frac{5 x^{2}}{4}\right]_{4}^{6}-\left[\frac{26 x}{4}+\frac{3 x^{2}}{8}\right]_{2}^{6} \end{aligned}
\begin{aligned} &=\left[(8+12)-(2+6)+(102-45)-(68-20)-\left(39-\frac{27}{2}\right)-\left(13-\frac{3}{2}\right)\right] \\ &=20-8+57-48-26+12 \\ &=7 \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 24

$\frac{1}{6} \text { sq.units }$
Hint:
Given: The given equations are $y=\sqrt{x}$ and $y=x$ .
Solution:

\begin{aligned} &\text { Area of the bounded region }\\ &=\int_{0}^{1} \sqrt{x}-x d x\\ &=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}\\ &=\left[\frac{2}{3}-\frac{1}{2}\right]\\ &=\frac{1}{6} \text { squnits } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 25

$\frac{8 \pi}{3}\; \text{sq. units}$
Hint: :
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{1}{a}+c \text {. }$ .
Given:
The given equations are $x^2+y^2=16$ and $y=\sqrt{3}x$ .
Solution:

$x^2+y^2=16$ Represents a circle with centreo(0,0) and cutting the x axis at A(4,0)
$y=\sqrt{3}x$ Represents straight passing through o(0,0) .
Point of intersection is obtained by solving the two equations.
$x^2+y^2=16$ and $y=\sqrt{3}x$
$x^{2}+(\sqrt{3} x)^{2} =16$
$4x^2=16$
$x=\pm 2$
$y=\pm 2\sqrt{3}$
$B(2,2\sqrt{3})$ and $B'(-2,-2\sqrt{3})$ are points of intersection of circle and straight line

\begin{aligned} &=\int_{0}^{2} \sqrt{3} x d x+\int_{2}^{4} \sqrt{16-x^{2}} d x \\ &=\sqrt{3}\left[\frac{x^{2}}{2}\right]_{0}^{2}+\left[\frac{1}{2} x \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1}\left(\frac{x}{4}\right)_{2}^{4}\right] \\ &=2 \sqrt{3}+8 \frac{\pi}{2}-2 \sqrt{3}-8 \frac{\pi}{6} \\ &=2 \sqrt{3}+4 \pi-2 \sqrt{3}-\frac{4 \pi}{3} \\ &=\frac{8 \pi}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 26

$\frac{16}{3 }\; \text{sq. units}$ .
Hint:
Use concept.
Given:
The given equations are $y^2=2x+1$ and $x-y-1=0$
Solution:
To find area bounded by
$y^2=2x+1$ ...........(1)
$x-y-1=0$ ........(2)
On solving the equation (i) and (ii)
\begin{aligned} &x-y=1 \\ &\text { or } y^{2}=2(y-1)+1 \\ &\text { or } y^{2}=2 y-1 \\ &\text { or }(y+1)(y-3)=0 \\ &\text { or } y=3 \text { or }-1 \\ &\therefore x=4,0 \end{aligned}
Equation (i) is a parabola with vertex $\left (\frac{-1}{2},0 \right )$ (−12,0) and passes through (0, 1), A (0, – 1)
Equation (ii) is a line passing through (1, 0) and (0, – 1).
Points of intersection of parabola and line are B (4, 3) and A (0, – 1)
These are shown in the graph below:

Required area=RegionABCDA
\begin{aligned} &=\int_{-1}^{3}\left(1+y-\frac{y^{2}-1}{2}\right) d y \\ &=\frac{1}{2} \int_{-1}^{3}\left(2+2 y-y^{2}+1\right) d y \\ &=\frac{1}{2} \int_{-1}^{3}\left(3+2 y-y^{2}\right) d y \\ &=\frac{1}{2}\left[3 y+y^{2}-\frac{y^{3}}{3}\right]_{-1}^{3} \\ &=\frac{1}{2}\left[(9+9-9)-\left(-3+1+\frac{1}{3}\right)\right] \\ &=\frac{1}{2}\left[9+\frac{5}{3}\right] \\ &=\frac{16}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 27

$\frac{64}{3}\; \text{sq. units}$
Hint:
use concept.
Given:
The given equations are $y=x-1$ and $(y-1)^2=4(x+1)$
Solution:

We have.$y=x-1$ and $(y-1)^2=4(x+1)$
$\therefore (x-1-1)^2=4(x+1)$
\begin{aligned} &(x-2)^{2}=4(x+1) \\ &x^{2}+4-4 x=4 x+4 \\ &x^{2}+4-4 x-4 x-4=0 \\ &x^{2}-8 x=0 \\ &x=0 \text { or } x=8 \\ &\therefore y=-1 \text { or } 7 \end{aligned}
Consider a horizontal strip of length $\left |x_2-x_1 \right |$ and width dy where $P(x_2,y)$ lies on straightline and $Q(x_1,y)$ lies on the parabola.
Area of approximating rectangle=$\left |x_2-x_1 \right |$ and it moves from y =1 to y =- 7
$\text { Required area }=\operatorname{area}(O A D O)=\int_{1}^{7}\left|x_{2}-x_{1}\right| d y$
\begin{aligned} &=\int_{-1}^{7}\left|x_{2}-x_{1}\right| d y \ldots \ldots \ldots\left\{\therefore\left|x_{2}-x_{1}\right|=x_{2}-x_{1} \text { as } x_{2}>x_{1}\right\} \\ &=\int_{-1}^{7}\left[(1+y)-\frac{1}{4}\left\{(y-1)^{2}-4\right\}\right] d y \\ &=\int_{-1}^{7}\left\{1+y-\frac{1}{4}(y-1)^{2}+1\right\} d y \\ &=\int_{-1}^{7}\left\{2+y-\frac{1}{4}(y-1)^{2}\right\} d y \end{aligned}
\begin{aligned} &=\left[2 y+\frac{y^{2}}{2}-\frac{1}{12}(y-1)^{3}\right]_{-1}^{7} \\ &=\left[14+\frac{49}{2}-\frac{1}{2} \times 6 \times 6 \times 6\right]-\left[-2+\frac{1}{2}+\frac{1}{12} \times 2 \times 2 \times 2\right] \\ &=\left[14+\frac{49}{2}-18\right]-\left[-2+\frac{1}{2}+\frac{2}{3}\right] \\ &=\left[\frac{41}{2}+\frac{5}{6}\right] \\ &=\frac{64}{3} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 28

$\frac{9}{2} \: \: \text{unit}^2$
Hint:
Use concept.
Given:
The given equations are $y=-x^2$ and $x+y+2=0$ .
Solution:

\begin{aligned} &\text { Area of the bounded region }\\ &=\int_{-1}^{2}-x^{2}-(-2-x) d x\\ &=\left[-\frac{x^{2}}{3}+2 x+\frac{x^{2}}{2}\right]_{-1}^{2}\\ &=\left[-\frac{8}{3}+6\right]-\left[\frac{1}{3}+\frac{1}{2}-2\right]\\ &=\frac{9}{2} \text { unit }^{2} \end{aligned}

Areas of Bounded Regions exercise 20.3 question 29

$\frac{9}{2}\text{sq. units}$
Hint:
Use concept.
Given:
The given equations are $y=2-x^2$ and $y+x=0$ .
Solution:
To find area bounded by
$y=2-x^2$ ......(1)
$y+x=0$ .......(2)
Equation (1) represents a parabola with vertex (0,2) and downward, meets axes at $(\pm 0, \sqrt{2})$ .
Equation (2) represents a line passing through (0,0) and (2,-2) .The points of intersection of line and parabola are (2,-2) and (1,-1) .
A rough sketch of curve is as follows:

Shaded region is sliced into rectangles with area $=(y_2-y_1)_{\Delta }x$. It slides from x =-1 to x = 2 so,
Required area =Region ABPCOA
\begin{aligned} &A=\int_{-1}^{2}\left(y_{1}-y_{2}\right) d x \\ &=\int_{-1}^{2}\left(2-x^{2}+x\right) d x \\ &=\left[2 x-\frac{x^{3}}{3}+\frac{x^{2}}{2}\right]_{-1}^{2} \\ &=\left[\left(4-\frac{8}{3}+2\right)-\left(-2+\frac{1}{3}+\frac{1}{2}\right)\right] \\ &=\left[\frac{10}{3}+\frac{7}{6}\right] \\ &=\frac{27}{6} \\ &=\frac{9}{2} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 30 sub question 1.

11 sq.units.
Hint:
Use concept of definite integrals.
Given:
The given equations are 3x - y -3 =0 , 2x + y - 1 =0 and x -2y -1 =0.
Solution:
We have
\begin{aligned} &3 x-y-3=0 \\ &2 x+y-1=0 \\ &x-2 y-1=0 \end{aligned}

Area of bounded region =
\begin{aligned} &=\int_{0}^{3} 3 x-3-\left(\frac{x-1}{2}\right) d x+\int_{3}^{5} 12-2 x-\left(\frac{x-1}{2}\right) d x \\ &=\left[\frac{3 x^{2}}{2}-3 x-\frac{x^{2}}{4}+\frac{1}{2} x\right]_{0}^{3}+\left[12 x-2 \frac{x^{2}}{2}-\frac{x^{2}}{4}+\frac{1}{2} x\right]_{3}^{5} \\ &=\left[\frac{27}{2}-9-\frac{9}{4}+\frac{3}{2}\right] \\ &=\left[60-25-\frac{25}{4}+\frac{5}{2}-36+9+\frac{9}{4}-\frac{3}{2}\right] \\ &=11 \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 30 sub question 2

6.5 sq. units
Hint:
Given:
$3x-2y+1=0 , 2x+3y-21=0 , x-5y+9=0$
Solution:
\begin{aligned} &\\ &3 x-2 y+1=0, y_{1}=\frac{(3 x+1)}{2}\\ &2 x+3 y-21=0, y_{2}=\left(\frac{21-2 x}{3}\right)\\ &x-5 y+9=0, y_{3}=\frac{(x+9)}{5} \end{aligned}
Point off intersection of (i) and (ii) is A(3,5)
Point off intersection of (ii) and (iii) is B(6,3)
Point off intersection of (iii) and (i) is C(1,2)

Area of the region bonded

\begin{aligned} &\int_{1}^{3} y_{1} \cdot d x+\int_{3}^{6} y_{2} \cdot d x-\int_{1}^{6} y_{3} \cdot d x \\ &=\int_{3}^{1} \frac{(3 x+1)}{2} \cdot d x+\int_{3}^{6} \frac{(21-2 x)}{3} \cdot d x-\int_{1}^{6} \frac{(x+9)}{5} \cdot d x \\ &=\frac{1}{2}\left(\frac{3 x^{2}}{2}+x\right)_{1}^{3}+\frac{1}{3}\left(21 x-x^{2}\right)_{3}^{6}+\frac{1}{5}\left(\frac{x^{2}}{2}+9 x\right)_{1}^{6} \\ &=\frac{1}{2}[14]+\frac{1}{3}[36]-\frac{1}{3}\left[\frac{125}{2}\right] \\ &=7+12-12.5 \\ &=6.5 \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 31

$\frac{11}{6}\text{sq. units}$
Hint:
Use concept.
Given:
The given equation are $y =x$ and $y =x^2+2$ .
Solution: To find area bounded by x =0, x =1
and
$y =x$ ........(1)
$y =x^2+2$........(2)
Equation(1) is a line passing through (2,2) and (0,0).Equation (2) is a parabola upward with vertex at (0,2).
A rough sketch of curve is as under:

Shaded region is sliced into rectangle of area=. .It slides from to ,so
Required Area=Region OABCO
\begin{aligned} A &=\int_{0}^{1}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{1}\left(x^{2}+2-x\right) d x \\ &=\left[\frac{x^{3}}{3}+2 x-\frac{x^{2}}{2}\right]_{0}^{1} \\ &=\left[\left(\frac{1}{3}+2-\frac{1}{2}\right)-0\right] \\ &=\left(\frac{2+12-3}{6}\right) \\ &=\frac{11}{6} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 32

4 sq. Units
Hint:
Use concept.
Given:
Equation of the given curve $x=y^2$ and $x=3-y^2$
Solution: To find area bounded by
$x=y^2$ …(i)
And $x-3=y^2$ …(ii)
On solving the equation (i) and (ii),
$y^2=3-2y^2$
Or $3y^2=3$
Or $y=\pm 1$
when y =1 then x =1 and when y = -1 then x =1
Equation (i) represents an upward parabola with vertex (0, 0) and axis -y
Equation (ii) represents a parabola with vertex (3, 0) and axis as x-axis
They intersect at A (1, – 1) and C (1, 1)
These are shown in the graph below:

Required area = Region OABCO = 2
= 2 Region OBCO
= 2[Region ODCO + Region BDCB]
\begin{aligned} &=2\left[\int_{0}^{1} y_{1} d x+\int_{1}^{3} y_{2} d x\right] \\ &=2\left[\int_{0}^{1} \sqrt{x} d x+\int_{1}^{3} \sqrt{\frac{3-x}{2} d x}\right] \\ &\left.=2\left[\left(\frac{2}{3} x \sqrt{x}\right)_{0}^{1}+\left(\frac{2}{3} \cdot\left(\frac{3-x}{2}\right) \sqrt{\frac{3-x}{2}} \cdot(-2)\right)^{2}\right]_{1}\right] \\ &=2\left[\left(\frac{2}{3}-0\right)+\left\{(0)-\left(\frac{2}{3}\right) .1 .1 .(-2)\right\}\right] \\ &=2\left[\frac{2}{3}+\frac{4}{3}\right] \end{aligned}
= 4 sq.units

Areas of Bounded Regions exercise 20.3 question 33

7sq. Units.
Hint:
Use conceptual part.
Given:
Triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C (8,4).
Solution:
Equation of AB is given by
$y=-\frac{5}{2}x-9$
Equation of BC is given by
$y=-x+12$
Equation of AC is given by
$y=\frac{3}{4}x-2$

Area of ABC
\begin{aligned} &=\int_{4}^{6}\left(y_{A B}-y_{A C}\right) d x+\int_{6}^{8}\left(y_{B C}-y_{A C}\right) d x \\ &=\int_{4}^{6}\left(\frac{5}{2} x-9-\frac{3}{4} x+2\right) d x+\int_{0}^{8}\left(-x+12-\frac{3}{4} x+2\right) d x \\ &=\int_{4}^{6}\left(\frac{7}{4} x-7\right) d x+\int_{6}^{8}\left(-\frac{7 x}{4}+14\right) d x \\ &=\left[\frac{7 x^{2}}{8}-7 x\right]_{4}^{6}+\left[-\frac{7 x^{2}}{8}+14 x\right]_{6}^{8} \\ &=\left[\left(\frac{63}{2}-42\right)-(14-28)\right]+\left[(-56+112)-\left(\frac{-63}{2}\right)+84\right] \\ &=\left[\frac{63}{2}-42-14+28-56+112+\frac{63}{2}-84\right] \\ &=63-56 \\ &=7 \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 34

$\left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right )$ sq. Units.
Hint:
Given: $\left\{(x, y):|x-1| \leq y \leq \sqrt{\left.\left(5-x^{2}\right)\right\}}\right.$
Solution: Equation of the curve is $y=\sqrt{(5-x)^2}$ or $y^2+x^2=5$ ,which is a circle with
centre at (0, 0) and? radius? $\frac{5}{2}$.

Equation of the line is $y=\left | x-1 \right |$
Consider, y =x -1 and $y=\sqrt{5-x^2}$
Eliminating y ,we get
\begin{aligned} &x-1=\sqrt{5-x^{2}} \\ &x^{2}+1-2 x=5-x^{2} \\ &2 x^{2}-2 x-4=0 \\ &x^{2}-x-2=0 \\ &(x-2)(x+1)=0 \\ &x=2,-1 \end{aligned}
The required area is
\begin{aligned} &A=\int_{-1}^{2} \sqrt{5-x^{2}} d x-\int_{-1}^{1}(-x+1) d x-\int_{1}^{2}(x-1) d x \\ &=\left[\frac{x}{2} \sqrt{5-x^{2}}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{x}{\sqrt{5}}\right)\right]_{-1}^{2}+\left[\frac{x^{2}}{2}-x\right]_{-1}^{1}+\left[\frac{-x^{2}}{2}+x\right]_{1}^{2} \\ &=\left[\frac{2}{2} \sqrt{5-4}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{2} \sqrt{5-1}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]+\left[\frac{1}{2}-1-\frac{1}{2}-1\right]+\left[\frac{-4}{2}+2+\frac{1}{2}-1\right] \end{aligned}
\begin{aligned} &=\left[1+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{2}{2}+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]-2-\frac{1}{2} \\ &=2+\left[\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)+\frac{\sqrt{5}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]-2-\frac{1}{2} \\ &=\frac{\sqrt{5}}{2} \times \frac{\pi}{2}-\frac{1}{2} \\ &=\frac{\sqrt{5} \pi}{4}-\frac{1}{2} \end{aligned}
=$\left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right )$ sq.units
Therefore, the area of the region is $\left (\frac{\sqrt{5}\pi}{4}-\frac{1}{2} \right )$ sq.units.

Areas of Bounded Regions exercise 20.3 question 35

1sq.unit.
Hints:
Use concept.
Given:
The curves $y=\left | x-1 \right |$ and $y=1$.
Solution:
To find area bounded by $y=\left | x-1 \right |$ and $y=1$
$y=\left\{\begin{array}{l} x-1, \text { if } x \geq 0 \\ 1-x, \text { if } x<0 \end{array}\right.$
A rough sketch of the curve is as under:

Shaded region is the required area. So
Required area = Region ABCA
A = Region ABDA + Region BCDB
\begin{aligned} &=\int_{0}^{1}\left(y_{1}-y_{2}\right) d x+\int_{1}^{2}\left(y_{1}-y_{3}\right) d x \\ &=\int_{0}^{1}(1-1+x) d x+\int_{1}^{2}(1-x+1) d x \\ &=\int_{0}^{1} x d x+\int_{1}^{2}(2-x) d x \\ &=\left(\frac{x^{2}}{2}\right)_{0}^{1}+\left(2 x-\frac{x^{2}}{2}\right)_{1}^{2} \\ &=\left(\frac{1}{2}-0\right)+\left[(4-2)-\left(2-\frac{1}{2}\right)\right] \\ &=\frac{1}{2}+\left(2-2+\frac{1}{2}\right) \end{aligned}
A = 1 sq. units.

Areas of Bounded Regions exercise 20.3 question 36

$4 \pi$ sq. Units
Hints:
Use concept.
Given:
The first quadrant enclosed by x-axis, the line y = x and the circle $x^2+y^2=32$
Solution:

Point of intersection, x =4
Area of a shaded region = $\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2}} d x$
\begin{aligned} &=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x}{2} \sqrt{32-x^{2}}+16 \sin ^{-1} \frac{x}{4 \sqrt{2}}\right]_{4}^{4 \sqrt{2}} \\ &=8+16 \frac{\pi}{2}-8-4 \pi=4 \pi=4 \pi \text { Sq.units. } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 37

$\frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units}$
Hints:
Given: Equation of circle $x^2+y^2=32$ exterior to parabola $y^2=6x$
Solution: First finding intersection point by solving the equation of two curves
\begin{aligned} &x^{2}+y^{2}=16 \ldots \text { (i) } \\ &\text { and } y^{2}=6 x \ldots \text { (ii) } \\ &\Rightarrow x^{2}+6 x=16 \\ &\Rightarrow x^{2}+6 x-16=0 \end{aligned}
\begin{aligned} &\Rightarrow x^{2}+8 x-2 x-16=0 \\ &\Rightarrow x(x+8)-2(x+8)=0 \\ &\Rightarrow(x+8)(x-2)=0 \\ & x=-8 \end{aligned}
(not possible $y^2$ cannot be -ve )

Or x=2 (only allowed value)
$y=\pm 2\sqrt{3}$
\begin{aligned} &A_{1}=\int_{0}^{2} \sqrt{6 x} d x=\left[\frac{\sqrt{6} x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\ &=\frac{2 \sqrt{6}}{3} \times 2^{\frac{3}{2}}=\frac{2 \sqrt{3} \times \sqrt{2}}{3} \times 2 \sqrt{2} \\ &=\frac{8}{\sqrt{3}} \text { sq.units } \end{aligned}
\begin{aligned} &A_{2}=\int_{2}^{4}\left(\sqrt{16-x^{2}}\right) d x=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2}^{4} \\ &=0+8 \sin ^{-1}-8 \sin ^{-1} \frac{1}{2}-\sqrt{16-4} \\ &=8 \times \frac{\pi}{2}-8 \times \frac{\pi}{6}-\sqrt{12} \\ &=4 \pi-\frac{4 \pi}{3}-\sqrt{12} \\ &=\frac{8 \pi}{3}-\sqrt{12} \text { sq.units } \end{aligned}
\begin{aligned} &\text { Area }=2\left(A_{1}+A_{2}\right) \\ &=2\left(\frac{8}{\sqrt{3}}+\frac{8 \pi}{3}-\sqrt{12}\right) \\ &\text { Required area }=\pi \times 16-\frac{16 x}{3}+4 \sqrt{3}-\frac{16}{\sqrt{3}} \\ &=\frac{32 \pi}{3}-\frac{4}{\sqrt{3}} \end{aligned}\frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units}
Therefore, the area of the circle is $\frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units}$

Areas of Bounded Regions exercise 20.3 question 38

$\frac{9}{2}$ sq. Units
Hints:
Use concept .
Given:
$x^2=y$ is a parabola line $y=x+2$
Solution:$x^2=y$ is a upward parabola line $y=x+2$
They meet at pts (2,4) and (-1,1)
By solving the equation,
\begin{aligned} &x^{2}=y, y=x+2 \\ &\Rightarrow x^{2}-x-2=0 \\ &\Rightarrow x=2 \text { or }-1 \\ &\Rightarrow x=4 \text { or } 1 \end{aligned}
Required area
= $\int_{-1}^{2}$ Area of line - $\int_{-1}^{2}$ Area of parabola

\begin{aligned} &=\int_{-1}^{2}(x+2) d x-\int_{-1}^{2} x^{2} d x \\ &=\left[\frac{x^{2}}{2}\right]_{-1}^{2}+[2 x]_{-1}^{2}-\left[\frac{x^{3}}{3}\right]_{-1}^{2} \\ &=\frac{4}{2}-\frac{(-1)^{2}}{2}+2 * 2-(2 *-1)-\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right) \\ &=2-\frac{1}{2}+4+2-\left(\frac{8}{3}+\frac{1}{3}\right) \\ &=8-\frac{1}{2}-3=5-\frac{1}{2}=\frac{9}{2} \text { sq.units } \end{aligned}

Areas of Bounded Regions exercise 20.3 question 39

Answer: The Area of the region $\left\{(x, y): 0 \leq y \leq x^{2}+3 ; 0 \leq y \leq 2 x+3 ; 0 \leq x \leq 3\right\}$ is $\frac{38}{3}$ sq.units.
Hints:
Given: $\left\{(x, y): 0 \leq y \leq x^{2}+3 ; 0 \leq y \leq 2 x+3 ; 0 \leq x \leq 3\right\}$
Solution: To find area given equation are
$y= x^3+3$...... (i)
$y= 2x+3$....... (ii)
And $x=3$ .......(iii)
Solving the above three equations to get the intersectons points,
\begin{aligned} &x^{2}+3=2 x+3 \\ &\text { Or } x^{2}-2 x=0 \\ &\text { Or } x(x-2)=0 \\ &\text { And } x=0 \text { or } x=2 \\ &y=3 \text { or } y=7 \end{aligned}
Equation (i) represents a parabola with vertices (3,0) and axis as y - axis
Equation (ii) represents a line passing through (0,3) and $\left (-\frac{3}{2},0 \right )$
The points of intersection are A(0,3) and B(2,7)
These are shown in the graph below:

Required area =
\begin{aligned} =& \int_{2}^{3} y_{1} d x+\int_{0}^{2} y_{2} d x \\ =& \int_{2}^{3}(2 x+3) d x+\int_{0}^{2}\left(x^{2}+3\right) d x \\ &=\left(x^{3}+3 x\right)_{2}^{3}+\left(\frac{x^{2}}{3}+x\right)_{0}^{2} \\ &=[(9+9)-(4+6)]+\left[\left(\frac{8}{3}+2\right)-(0)\right] \\ &=[18-10]+\left[\frac{14}{3}\right] \\ &=8+\frac{14}{3} \end{aligned}
=$\frac{38}{3}$ sq.units
Therefore, the area of the region is $\frac{38}{3}$ sq.units

Areas of Bounded Regions exercise 20.3 question 40

$\frac{\pi}{8}$ sq. Units
Hints:
Use concept of definite integrals.
Given:
The curve $y=\sqrt{1-x^2}$ , line y =x and the positive x-axis
Solution:

\begin{aligned} &y=\sqrt{1-x^{2}} \\ &\Rightarrow y^{2}=1-x^{2} \\ &\Rightarrow x^{2}+y^{2}=1 \end{aligned}
Hence,
$y=\sqrt{1-x^2}$ represents the upper half of the circle $x^2+y^2=1$ a circle with centre O(0,0) and radius 1 unit.y = x represents equation of a straight line passing through O(0,0)
Point of intersection is obtained by solving two equations
\begin{aligned} &y=x \\ &y=\sqrt{1-x^{2}} \\ &\Rightarrow x=\sqrt{1-x^{2}} \\ &\Rightarrow x^{2}=1-x^{2} \\ &\Rightarrow 2 x^{2}=1 \\ &\Rightarrow x=\pm \frac{1}{\sqrt{2}} \\ &\Rightarrow y=\pm \frac{1}{\sqrt{2}} \end{aligned}
$D\left (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right )$ And $D\left (\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right )$ are two points of intersection between the circles and the straight line.
And $D\left (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right )$ is the intersection point of $y=\sqrt{1-x^2}$ and y =x .
Required area = Shaded area (ODAEO) = Area (ODEO) + Area (EDAE)…..(1)
Now, area (ODEO) = $\int_{0}^{\frac{1}{\sqrt{2}}} x d x$
\begin{aligned} &=\left[\frac{x^{2}}{2}\right]_{0}^{\frac{1}{\sqrt{2}}} \\ &=\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right) \\ &=\frac{1}{4} \text { sq.units } \ldots . .(2) \end{aligned}
Area (EDAE) = $\int_{\frac{1}{\sqrt{2}}}^{1} \sqrt{1-x^{2}} d x$
\begin{aligned} &=\left[\frac{1}{2} x \sqrt{1-x^{2}}+\times \frac{1}{2} \times \sin ^{-1}\left(\frac{x}{1}\right)\right]_{\frac{1}{\sqrt{2}}}^{1}\\ &=0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ &=\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{4}-\frac{1}{2} \times \frac{\pi}{4} \ldots \ldots \ldots .\left\{\text { using, } \sin ^{-1}(1)=\frac{\pi}{2} \text { and } \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\right\}\\ &=\frac{\pi}{4}-\frac{\pi}{8}-\frac{1}{4}\\ &=\frac{\pi}{8}-\frac{1}{4} \text { sq.units } \ldots &\text { (3) } \end{aligned}
Area (ODAEO) = $\frac{1}{4}+\frac{\pi}{4}-\frac{1}{4}=\frac{\pi}{8}$ sq.units

Areas of Bounded Region Exercise 20.3 Question 41

$\frac{15}{2}$ sq. units
Hints:
Use concept.
Given:
Lines $y=4x+5,y=5-x$ and $4y=x+5$
Solution:
To find area bounded by lines
$y=4x+5$ {Say AB}
$y=x-5$ {Say BC}
$4y=x+5$ {Say AC}
By solving equation (1) and (2) , we get B(0,5)
By solving equation (2) and (3) , we get C(3,2)
By solving equation (1) and (3) , we get A(-1,1)
A rough sketch of the curve is as under:-

Shaded area ABC is the required area.
Required area = ar( ABD) + ( BDC)
ar( ABD) $=\int_{0}^{-1}(y_1-y_2)dx$

\begin{aligned} &=\int_{-1}^{0}\left(4 x+5-\frac{x}{4}-\frac{5}{4}\right) d x \\ &=\int_{-1}^{0}\left(\frac{15 x}{4}+\frac{15}{4}\right) d x \\ &=\frac{15}{4}\left(\frac{x^{2}}{2}+x\right)_{-1}^{0} \\ &=\frac{15}{4}\left[(0)-\left(\frac{1}{2}-1\right)\right] \\ &=\frac{15}{4} \times \frac{1}{2} \end{aligned}

AR ( ABD) = $\frac{15}{8}$ sq. units
AR ( BDC) = $\int_{0}^{3}(y_2-y_3)dx$
\begin{aligned} &=\int_{0}^{3}\left[(5-x)-\left(\frac{x}{4}+\frac{5}{4}\right)\right] d x \\ &=\int_{0}^{3}\left[5-x-\frac{x}{4}-\frac{5}{4}\right] d x \\ &=\int_{0}^{3}\left(\frac{-5 x}{4}+\frac{15}{4}\right) d x \\ &=\frac{5}{4}\left(3 x-\frac{x^{2}}{2}\right) \\ &=\frac{5}{4}\left(9-\frac{9}{2}\right) \end{aligned}
AR ( BDC) = $\frac{45}{8}$ sq. units
Using equation (1), (2) and (3),
AR ( ABC) = $\frac{15}{8}+\frac{45}{8}$
$=\frac{60}{8}$
ar( ABC) = $=\frac{15}{2}$ sq. units

Areas of Bounded Region Exercise 20.3 Question 42

$\left(6 \pi-\frac{9 \sqrt{3}}{2}\right)$sq. units
Hints:
Use concept.
Given:
The two curves $x^2+y^2=9$ and $(x-3)^2+y^2=9$
Solution:
To find area enclosed by
$x^2+y^2=9$ ……. (1)
$(x-3)^2+y^2=9$ …….. (2)
Equation (1) represents a circle with centre (0,0) and meets axis at $\left ( \pm 3,0 \right )$ , $\left ( 0,\pm 3 \right )$
Equation (2) is a circle with center (3,0) and meets axis at (0,0) , (6,0)
they intersect each other at $\left ( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right )$ and $\left ( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right )$ . A rough sketch of the curves is as under:

Shaded region is the required area.
Required area = Region OABCO
A = 2(Region OBCO)
= 2(Region ODCO + Region DBCD)
$=2\left[\int_{0}^{\frac{3}{2}} \sqrt{9-(x-3)^{2}} d x+\int_{\frac{3}{2}}^{3} \sqrt{9-x^{2}} d x\right]$
\begin{aligned} &\left.=2\left[\left\{\frac{(x-3)}{2} \sqrt{9-(x-3)^{2}}+\frac{9}{2} \sin ^{-1} \frac{(x-3)}{3[]}\right\}_{0}^{\frac{3}{2}}+\left\{\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right\}_{\frac{3}{2}}\right]_{2}^{3}\right] \\ &=2\left[\left\{\left(-\frac{3}{4} \sqrt{9-\frac{9}{4}}+\frac{9}{2} \sin ^{-1}\left(-\frac{3}{6}\right)\right)-\left(0+\frac{9}{2} \sin ^{-1}(-1)\right)\right\}+\left\{\left(0+\frac{9}{2} \sin ^{-1}(1)\right)-\left(\frac{3}{4} \sqrt{9-\frac{9}{4}}+\frac{9}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right) .\right.\right. \end{aligned}
\begin{aligned} &=2\left[\left\{-\frac{9 \sqrt{3}}{8}-\frac{9}{2}-\frac{\pi}{6}+\frac{9}{2}-\frac{\pi}{2}\right\}+\left\{\frac{9}{2}-\frac{\pi}{2}-\frac{9 \sqrt{3}}{8}-\frac{9}{2}-\frac{\pi}{6}\right\}\right] \\ &=2\left[-\frac{9 \sqrt{3}}{8}-\frac{3 \pi}{4}+\frac{9 \pi}{4}+\frac{9 \pi}{4}-\frac{9 \sqrt{3}}{8}-\frac{3 \pi}{4}\right] \\ &=2\left[\frac{12 \pi}{4}-\frac{18 \sqrt{3}}{8}\right] \\ &=\left(6 \pi-\frac{9 \sqrt{3}}{2}\right) \text { sq. units } \end{aligned}

Areas of Bounded Regions Exercise 20.3 Question 43

$\pi -2$ sq. units
Hints:
Use concept.
Given:
$\left\{(x, y): x^{2}+y^{2} \leq 4, x+y \geq 2\right\}$
Solution: The equation of the given curves are
$x^2+y^2=4$ ……(1)
$x+y=2$ …….(2)
Clearly $x^2+y^2=4$ represents a circle and $x+y=2$ is the equation of a straight line cutting x and y axis at (0,2) and (2,0) respectively.
The smaller region bounded by thesetwo curves is shaded in the following figure.

Length = $y_2-y_1$
Width = x and
Area = $(y_2-y_1)x$
Since the approximating rectangle can move from x=0 to x=2, the required area is given by
\begin{aligned} &\mathrm{A}=\int_{0}^{2}\left(y_{2}-y_{1}\right) d x\\ &\text { We have } y_{1}=2-x \text { and } y_{2}=\sqrt{4-x^{2}}\\ &\text { Thus, } A=\int_{0}^{2}\left(\sqrt{4-x^{2}}-2+x\right) d x\\ &\Rightarrow \mathrm{A}=\int_{0}^{2}\left(\sqrt{4-x^{2}}\right) d x-2 \int_{0}^{2} d x+\int_{0}^{2} x d x \end{aligned}
\begin{aligned} &\Rightarrow A=\left[\frac{x \sqrt{4-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{2}-2(x)_{0}^{2}+\left(\frac{x^{2}}{2}\right)_{0}^{2} \\ &\Rightarrow A=\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)-4+2 \\ &\Rightarrow A=2 \sin ^{-1}(1)-2 \\ &\Rightarrow A=2 \times \frac{\pi}{2}-2 \\ &\Rightarrow A=\pi-2 \end{aligned}
Therefore, the area of the region is $\pi -2$ sq. units

Areas of Bounded Regions Exercise 20.3 Question 44

Answer: $\left(\frac{3 \pi}{2}-3\right) \text { sq. units }$
Hints:
Use concept.
Given:
$\left\{(x, y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\right\}$
Solution:
To find the area of a region
$\left\{(x, y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\right\}$
Here,
$\frac{x^2}{9}+\frac{y^2}{4}=1$ …..(1)
$\frac{x}{3}+\frac{y}{2}=1$ …..(2)
Equation (1) represents an ellipse with centre at origin and meets axis at $(\pm 3,0)$ ,$(0,\pm 2)$ .
Equation (2) is a line that meets axis at (3,0) ,(0,2)
A rough sketch is as under:

Shaded region represents required area. This is sliced into rectangles with area $(y_2-y_2)x$ which slides from x=0 to x=3, so
Required area = Region APBQA
\begin{aligned} &A=\int_{0}^{3}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{3}\left[\frac{2}{3} \sqrt{9-x^{2}} d x-\frac{2}{3}(3-x) d x\right] \\ &=\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)-3 x+\frac{x^{2}}{2}\right]_{0}^{3} \\ &=\frac{2}{3}\left[\left\{0+\frac{9}{2} \cdot \frac{\pi}{2}-9+\frac{9}{2}\right\}-\{0\}\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right] \\ &A=\left(\frac{3 \pi}{2}-3\right) \text { sq. units } \end{aligned}
Therefore, the area of the region is $\left(\frac{3 \pi}{2}-3\right) \text { sq. units }$

Areas of Bounded Regions Exercise 20.3 Question 45

$\left(\frac{4 \pi}{3}-\sqrt{3}\right) \text { sq. units }$
Hints:
Use concept.
Given:
The curve $y=\sqrt{4-x^2}$ ,$x^2+y^2-4x=0$ and the x-axis.
Solution:
Give, equation of a curves are
$y=\sqrt{4-x^2}$ ……(1)
$x^2+y^2-4x=0$ ……(2)
Consider the curve
$y=\sqrt{4-x^2}=y^2-4x=0$
$y^2+x^2=0$ which represents a circle with centre (0, 0) and radius 2 units.
Now, consider the curve $x^2+y^2-4x=0$ which also represents a circle with centre (2,0) and radius 2 units.
Now Let us sketch the graph of given curves and find their points of intersection.

On substituting the value of y from Eq. (1) in Eq. (2), we get
$x^2(4-x)^2-4x=0$ or $4-4x=0$ or $x=1$
On substituting x =1 in Eq. (1), we get $y=\sqrt{3}$
Thus, the point of intersection is $\left ( 1.\sqrt{3} \right )$ .
Clearly, required area = Area of shaded region OABO
\begin{aligned} &=\int_{0}^{1} y_{(\text {sacoricicic } k)} d x+\int_{1}^{2} y_{(f \text { trs cic } c i e)} d x \\ &=\int_{0}^{1} \sqrt{4 x-x^{2}} d x+\int_{1}^{2} \sqrt{4-x^{2}} d x \\ &=\int_{0}^{1} \sqrt{-\left(x^{2}-4 x\right)} d x+\int_{0}^{2} \sqrt{2^{2}-x^{2}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{1} \sqrt{-\left[x^{2}-2(2)(x)+4-4\right]} d x+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{1}^{2} \\ &=\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x+\left[2 \sin ^{-1}(1)-\left\{\frac{1}{2} \sqrt{3}+2 \sin ^{-1}\left(\frac{1}{2}\right)\right\}\right] \end{aligned}
\begin{aligned} &=\left[\frac{(x-2)}{2} \sqrt{4 x-x^{2}}+2 \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{1}+\left[2 \cdot \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2 \cdot \frac{\pi}{6}\right] \\ &=\left[\left\{\frac{-\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{-1}{2}\right)\right\}-\left\{2 \sin ^{-1}(-1)\right\}\right]+\left(\pi-\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) \\ &=-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}(1)+\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\left[\therefore \sin ^{-1}(-x)=-\sin ^{-1} x\right] \end{aligned}
\begin{aligned} &=2 \cdot \frac{\pi}{2}-2 \cdot \frac{\pi}{6}+\frac{2 \pi}{3}-\sqrt{3} \\ &=\pi-\frac{\pi}{3}+\frac{2 \pi}{3}-\sqrt{3} \\ &=\pi+\frac{\pi}{3}-\sqrt{3}=\left(\frac{4 \pi}{3}-\sqrt{3}\right) \text { sq. units } \end{aligned}

Areas of Bounded Regions Exercise 20.3 Question 46

$\frac{1}{2}$ .sq. Units.
Hints:
Use concept.
Given:
The curves $y=\left | x-1 \right |$ and $y=-\left | x-1 \right |+1$
Solution:
To find area enclosed by
\begin{aligned} &y=|x-1| \\ &\Rightarrow y=\left\{\begin{array}{l} -(x-1), \text { if } x-1<0 \\ (x-1), \text { if } x-1 \geq 0 \end{array}\right. \end{aligned}
\begin{aligned} &\Rightarrow y=\left\{\begin{array}{l} 1-x, \text { if } x<1 \\ x-1, \text { if } x \geq 1 \end{array}\right. \\ &\text { And } y=-|x-1|+1 \\ &\Rightarrow y=\left\{\begin{array}{l} +(x-1)+1, \text { if } x-1<0 \\ -(x-1)+1, \text { if } x-1 \geq 0 \end{array}\right. \\ &y=\left\{\begin{array}{lr} x, \quad \text { if } x<1 \\ -x+2, \text { if } x \geq 1 \end{array}\right. \end{aligned}
A rough sketch of equation of lines (1),(2),(3),(4) is given as:

Shaded region is the required area.
Required area = Region ABCDA
Required area = Region BDCB + Region ABDA
Region BDCB is sliced into rectangles of area $=(y_2-y_1)x$ and it slides from $x=\frac{1}{2}$ to x =1
Region ABDA is sliced into rectangle of area $=(y_3-y_4)x$ and it slides from x =1 to $x=\frac{3}{2}$ .
So, using equation (1),
Required area = Region BDCB + Region ABDA
\begin{aligned} &=\int_{\frac{1}{2}}^{1}\left(y_{1}-y_{2}\right) d x+\int_{1}^{\frac{3}{2}}\left(y_{3}-y_{4}\right) d x \\ &=\int_{\frac{1}{2}}^{1}(x-1+x) d x+\int_{1}^{\frac{3}{2}}(-x+2-x+1) d x \\ &=\int_{\frac{1}{2}}^{1}(2 x-1) d x+\int_{1}^{\frac{3}{2}}(3-2 x) d x \\ &=\left[x^{2}-x\right]_{\frac{1}{2}}^{1}+\left[3 x-x^{2}\right]_{1}^{\frac{3}{2}} \end{aligned}
\begin{aligned} &=\left[(1-1)-\left(\frac{1}{4}-\frac{1}{2}\right)\right]+\left[\left(\frac{9}{2}-\frac{9}{4}\right)-(3-1)\right] \\ &=\frac{1}{4}+\frac{9}{4}-2 \\ &A=\frac{1}{2} \text { sq. units } \end{aligned}

Areas of Bounded Regions Exercise 20.3 Question 47

$A= \frac{33}{2}\; \text{sq. units}$
Hints:
Use concept.
Given:
The curves $3x^2+5y =32$ and $y=\left | x-2 \right |$ .
Solution:
To find area enclosed by
$3x^2+5y=32$
$3x^2=-5\left ( y-\frac{32}{5} \right )$ ……. (1)
And
\begin{aligned} &y=|x-2|\\ &\Rightarrow y= \begin{cases}-(x-2), & \text { if } x-2<1 \\ (x-2), & \text { if } x-2 \geq 1\end{cases}\\ &\Rightarrow y=\left\{\begin{array}{l} 2-x, \text { if } x<2 \\ x-2, \text { if } x \geq 2\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ........(2) \end{array}\right. \end{aligned}
Equation (1) represents a downward parabola with vertex $\left (0,\frac{32}{5} \right )$ and equation (2) represents lines. A rough sketch of curves is given as:

Required area = Region ABECDA
A = Region ABEA + Region AECDA
$=\int_{2}^{3}\left(y_{3}-y_{4}\right) d x+\int_{-2}^{2}\left(y-y_{2}\right) d x$
\begin{aligned} &=\int_{2}^{3}\left(\frac{32-3 x^{2}}{5}-x+2\right) d x+\int_{-2}^{2}\left(\frac{32-3 x^{2}}{5}-2+x\right) d x \\ &=\frac{1}{5} \int_{2}^{3}\left(\frac{32-3 x^{2}-5 x+10}{5}\right) d x+\int_{-2}^{2}\left(\frac{32-3 x^{2}-10+5 x}{5}\right) d x \\ &=\frac{1}{5}\left[\int_{2}^{3}\left(42-3 x^{2}-5 x\right) d x+\int_{-2}^{2}\left(22-3 x^{2}+5 x\right) d x\right] \end{aligned}
\begin{aligned} &=\frac{1}{5}\left[\left(42 x-x^{3}-\frac{5 x^{2}}{2}\right)_{2}^{3}+\left(22 x-x^{3}+\frac{5 x^{2}}{2}\right)_{-2}^{2}\right] \\ &=\frac{1}{5}\left[\left\{\left(126-27-\frac{45}{2}\right)-(84-8-10)\right\}+\{(44-8+10)-(-44+8+10)\}\right] \\ &=\frac{1}{5}\left[\left\{\frac{153}{2}-66\right\}+\{46+26\}\right] \\ &=\frac{1}{5}\left[\frac{21}{2}+72\right] \end{aligned}
$A= \frac{33}{2}\; \text{sq. units}$

Areas of Bounded Regions Exercise 20.3 Question 48

$\frac{125}{24}$ sq. units
Hints:
Use concept.
Given:
The parabola $y=4x-x^2$ and $y=x^2$
Solution:
To area enclosed by
\begin{aligned} &y=4 x-x^{2} \\ &\Rightarrow-y=x^{2}-4 x+4-4 \\ &\Rightarrow-y+4=(x-2)^{2} \\ &\Rightarrow-(y-4)=(x-2)^{2} \\ &\text { And } y=x^{2}-x \\ &\left(y+\frac{1}{4}\right)=\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). Points of intersection of parabola are (0,0) and $\left (\frac{5}{2},\frac{15}{4} \right )$ .
A rough sketch of the curves is as under:

Shaded region is required area it is sliced into rectangles with area = $(y_2-y_1)x$ . It slides from x =0 to $x=\frac{5}{2}$ , so
Required area = Region OQAP
\begin{aligned} \mathrm{A} &=\int_{0}^{\frac{{5}}{2}}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{\frac{5}{2}}\left[4 x-x^{2}-x^{2}+x\right] d x \\ &=\int_{0}^{\frac{5}{2}}\left[5 x-2 x^{2}\right] d x \\ &=\left[\frac{5 x^{2}}{2}-\frac{2}{3} x^{2}\right]_{0}^{\frac{5}{2}} \\ &=\left[\left(\frac{125}{8}-\frac{250}{24}\right)-(0)\right] \\ & A=\frac{125}{24} \text { sq. units } \end{aligned}
Therefore, the area enclosed by the parabola $y=4x-x^2$ and $y=x^2$ is $\frac{125}{24}$ sq. units.

Areas of Bounded Regions Exercise 20.3 Question 49

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}

Areas of Bounded Regions Exercise 20.3 Question 50

4 sq. units.
Hints:
Use concept of definite integrals.
Given:
The curves $y=\left | x-1 \right |$ and $y=3-\left | x \right |$
Solution:
To find area bounded by the curve

\begin{aligned} &y=|x-1|\\ &\Rightarrow y=\left\{\begin{array}{l} 1-x, \text { if } x<1 \; \; \; \; \; \; \; .......(1)\\ x-1, \text { if } x \geq 1\; \; \; \; \; \; \; ........(2)) \end{array}\right.\\ & \end{aligned}
And $y=3-\left | x \right |$
$\Rightarrow \quad y= \begin{cases}3+x, \text { if } x<0\; \; \; \; \; \; \; \; \; ......(3) \\ 3-x, \text { if } x \geq 0\; \; \; \; \; \; \; \; \; ......(4)\end{cases}$

Drawing the rough sketch of lines (1), (2), (3) and (4) as under:

Shaded region is the required area
Required area = Region ABCDA
A = Region ABFA + Region AFCEA + Region CDEC
\begin{aligned} &=\int_{1}^{2}\left(y_{1}-y_{2}\right) d x+\int_{0}^{1}\left(y_{1}-y_{3}\right) d x+\int_{-1}^{0}\left(y_{4}-y_{3}\right) d x \\ &=\int_{1}^{2}(3-x-x+1) d x+\int_{0}^{1}(3-x-1+x) d x+\int_{-1}^{0}(3+x-1+x) d x \\ &=\int_{1}^{2}(4-2 x) d x+\int_{0}^{1} 2 d x+\int_{-1}^{0}(2+2 x) d x \\ &=\left[4-x^{2}\right]_{1}^{2}+[2 x]_{0}^{1}+\left[2 x+x^{2}\right]_{-1}^{0} \\ &=[(8-4)-(4-1)]+[2-0]+[(0)-(-2+1)] \\ &=(4-3)+2+1 \end{aligned}
A = 4 sq. units.
Therefore, the area bounded by the curve is 4 sq. units.

Areas of Bounded Region Exercise 20.3 question 51.

M=2.
Hints:
Use concept of definite integrals
Given:
The parabola $y^2=4ax$ and the line y = mx is $\frac{a^2}{12}$ .
Solution:

Area of the bounded region = $\frac{a^2}{12}$
\begin{aligned} &\frac{a^{2}}{12}=\int_{0}^{a} \sqrt{4 a x}-m x d x \\ &\frac{a^{2}}{12}=\left[2 \sqrt{a} \frac{x^{\frac{1}{2}}}{\frac{3}{2}}-m \frac{x^{2}}{2}\right]_{0}^{a} \end{aligned}
\begin{aligned} &\frac{a^{2}}{12}=\frac{4 a^{2}}{3}-m \frac{a^{2}}{2} \\ &m=2 \end{aligned}
Therefore, the value of m is 2.
Answer: The value of a is 2.
Hints:
Given: The parabola $y^2=16ax$ and $x^2=16ay$ cgfy
Solution:

Area of the bounded region = $\frac{1024}{3}$
\begin{aligned} &x=\frac{y^{2}}{16 a} \\ &y=0 ; 16 a \end{aligned}
\begin{aligned} &\text { Area }=\int_{0}^{16 a} \sqrt{16 a x} d x-\int_{0}^{16 a} \frac{x^{2}}{16 a} d x \\ &=\left[4 \sqrt{a} \frac{x^{\frac{4}{2}}}{\frac{3}{2}}\right]_{0}^{16 a}-\left[\frac{x^{3}}{48 a}\right]_{0}^{16 a} \\ &=\frac{2}{3} \times 4 \sqrt{a} \times(16 a)^{\frac{3}{2}}-0-\frac{1}{16 a} \times \frac{(16 a)^{3}}{3}-0 \end{aligned}
\begin{aligned} &A=\frac{1024}{3} \\ &\frac{1024}{3}=\frac{8 \sqrt{a}}{3} \times(16 a) \times 4 \sqrt{a}-\frac{256 a^{3}}{3} \\ &1024=256 a^{2} \\ &a^{2}=4 \\ &a=2 \end{aligned}

The RD Sharma class 12th exercise 20.3 consist a total of 55 questions that covers up all the important factors of the chapter that is mentioned below:

• Finding Area using vertical stripes and horizontal stripes

• Finding the Area between two curves

• Finding the area between two curves using vertical stripes and horizontal stripes

• Area of the bounded region

The benefits of practicing from the RD Sharma class 12th exercise 20.3 Is mentioned below:-

• The RD Sharma class 12 solution chapter 20 exercise 20.3 Is Designed by experts to provide the best solutions to the questions and helpful tips to make it easy for them to solve mathematics.

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## RD Sharma Chapter-wise Solutions

1. Is a solution useful for class 12 and 11 students?

Yes,  the RD Sharma solutions are completely beneficial for any student who is appearing for board exams  for will be appearing in the coming year.

2. How do you find the area of a bounded region?

The area under a cover between two points can be found by doing a definite integral between the two points.

3. Can teachers also use the RD Sharma solutions?

Teachers make the most use of R D Sharma solutions by preparing question papers and providing assignments to the students through the usage of the solutions.

4. Does the careers360 website contain the updated version of R D Sharma solutions?

Yes, The careers360 website contains all the updated version RD Sharma solutions for every class in every exercise.

5. Does the RD Sharma class 12th exercise 20.3 contain all important questions?

Yes, RD Sharma class 12 exercise 20.3 contains all essential questions that cover the concepts important for the exams.

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