Careers360 Logo
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

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.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise
  2. Areas of Bounded Regions Excercise:20.3
  3. RD Sharma Chapter-wise Solutions

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

Areas of Bounded Regions Excercise:20.3


Areas of Bounded Region Exercise 20.3 Question 1

Answer:
12 sq.units.
Given:
y2=6x
x2=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
y2=6x
y=6x ...…(i)
And
x=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
1799432
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)
A=066xdx06x26dxA=06(6xx26)dx
On integration the above definite integration
A=06(6xx26)dx=[6x3/23/2x318]06=[6(6)3/23/2(6)318]
=12sq units.

Areas of Bounded Region Exercise 20.3 Question 2

Answer: 4 sq.units
Given:
4y2=9x
3x2=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,
4y2=9x
y=32x .........(1)
And,
3x2=16y
y=3x216
Equating (1)and (2)
32x=3x216x22=432x=4
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
https://www.sarthaks.com/?qa=blob&qa_blobid=11099588031339998552
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]
A=0432xdx043x216dxA=04(32x3x216)dx
On integrating the above definate integration ,
The required area
A=04[3x23x216]dx=[x3/2x316]04=[(4)3/2(4)316]=[86416]=[84]=4

Areas of Bounded Region Exercise 20.3 Question 3

Answer:
16 sq.units
Given:
y=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=x (i)and y=x (ii)
Solving eqation (1) and( 2)
y2=x=y
y2=y
y(y1)=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
https://www.sarthaks.com/?qa=blob&qa_blobid=14342760982036744916
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]
A=01xdx01xdxA=01(xx)dx
On integrating the above definate integration,
=01(y1y2)dx=01(xx)dx=[23xxx22]01=[2311(1)22][0]=[2312]
=16 sq.units

Areas of Bounded Region Exercise 20.3 Question 4

Answer:283 sq.units
Given:y=4x2 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=4x2 …(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
https://www.sarthaks.com/?qa=blob&qa_blobid=6555794839998903474
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=2034ydy
On intregrating the above definate integration,
=2[(4y)3/23/2]03=223[43/213/2] , as area in not negative.
=283 sq units
The area bounded by the curved y=4x2 and the lined y=0, y=3 is 283 sq units

Areas of Bounded Region Exercise 20.3 Question 5

Answer:ab4[π2] Square units.
Hint:
Given:
{(x,y):x2a2+y2b21xa+yb}
Solution:

 Let R={(x,y):x2a2+y2b21xa+yb}R1={(x,y):x2a2+y2b21} And R2={(x,y):1xa+yb} Then R=R1R2
considered x2a2+y2b2=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)
R1=(x2a2+y2b21) Represents the area inside the ellipse
xa+xb=1 Represent a straight line cutting x-axis at A(a,0) and y- axis at B(0,b)
R2=(xa+xb1) Represent the area above the straight line
R=R1R2 Represent the smaller shaded arera bounded by the line and the ellips in the shaded region, consider a vertical strip with length=|y2y1| and width=dx such that P(x,y2) lies on ellipse and (x,y1) lies on the straight line area of approximating rectangle=|y2y1|dx
The approximating rectangle move from x = 0 to x = a
Area of the shaded region=0a|y2y1|dx=0a(y2y1)dx.[As,y1|y2y1|=y2y1]
A=0a(baa2x2ba(ax))dxA=a(baa2x2)dx0aba(ax)dxA=ba[{xaa2x2+12a2sin1(xa)}]0aba[axx22]0aA=ba[12a2×π2a22]A=ab2[π21]A=ab4[π2] Square units 

Areas of Bounded Region Exercise 20.3 Question 6

Answer:
4 square units
Hint:
Given:
A(2,1),B(3,4)and C(5,2)
Solution:
https://www.sarthaks.com/?qa=blob&qa_blobid=16818331701441575165
The equation of AB,
yy1=(y2y1x2x1)(xx1)y1=(4132)(x2)y1=31(x2)y=3x5 (1) 
The equation of BC,
y4=(2453)(x3)=22(x3)y=x+7(2)
The equation of AC,
y1=(2152)(x2)y1=13(x2)y=13x23+1y=13x+13..(3)
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]
A=23(y1y3)dx+35(y2y1)dx=23[(3x5)(13x+13)]dx+35[(x+7)(13x+13)]dx=23[3x513x13]dx+35[x+713x+13]dx
=23(8x3163)dx+35(43x+203)dx=83(x222x)23(4x26203)35=83[(926)(24)][(5031003)(620)]=83[32+2][503+14]=43[83]
=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

Answer:
152 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
https://www.sarthaks.com/?qa=blob&qa_blobid=4507152498576551315
The equation of AB
y=y1=(y2y1x2x1)(xx1)yy1=(510+1)(x+1)y1=41(x+1)y1=4x+4Y=4x+5..(1)
The equation of BC,
y5=(2530)(x0)=33(x0)y5=xy=5x(2)
The equation oa AC,
y1=(213+1)(x+1)y1=14(x+1)y1=14x+14y=14(x+5)(3)
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 ,AreaA=A1+A2
A1=10[(4x+5)14(x+5)]dx=10[4x+5x454]dx=10(154x+154)dx=154(x22+x)10=154[(0)(121)]=158
 And, A2=03(y2y3)dx=03[(5x)(14x+54)]dx=03[5x14x54]dx=03(54x+154)dx=54(3xx22)03=54[992]=458
So the enclosed area of the triangle is 158+458=152 square units

Areas of Bounded Region Exercise 20.3 Question 8

Answer:
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)
https://www.sarthaks.com/?qa=blob&qa_blobid=1472378851162222032
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
04(3x+1)dx04(2x+1)dx=[3x22+x]04[2x22+x]04=[(24+4)0][(16+4)0]=2820
=8 sq units

Areas of Bounded Region Exercise 20.3 Question 9

Answer:
2[23+9π492sin1(13)] square units
Hint:
Given:
{(x,y):y28x,x2+y29}
Solution:
To find area {(x,y):y28x,x2+y29}
y2=8x.....(1)
x2+y2=9.....(2)
On solving the equation (1) and (2)
x2+8x=9x2+8x9=0(x+9)(x1)=0x=9 or x=1

And when x=1 then y= ±22
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 (±3,0),(0,±3)
Point of intersection of parabola and circle is (1,22) and (122)
The sketch of the curves is as below
https://www.sarthaks.com/?qa=blob&qa_blobid=13041479697829642655
Or, required area=(region ODCO+REGION DBCD)
=2[018xdx+139x2dx]=2[(2223xx)01+(x29x2+92sin1x3)13]=2[(42311)+{(3299+92sin1(1))(1291+92sin113)}]=2[423+{(92π2)(22292sin1(13))}]=2[423+9π4292sin1(13)]
Hence, the required area is 2[23+9π492sin1(13)] square unit

Areas of Bounded Region Exercise 20.3 Question 10

Answer:
(433+163π) Square unit
Hint:
Given:
x2+y2=16
y2=6x
Solution:
https://www.sarthaks.com/?qa=blob&qa_blobid=3302140514589017426
x2+y2=16 …(1)
Equation of circle with centre (0,0) and radius 4
Equation of parabola:
y2=6x ….(2)
Intersecting piont of (1) and (2)
x2+6x16=0x2+8x2x16=0x(x+8)2(x+8)=0(x+8)(x2)=0x=8,2 but ,x8,x=2y2=12y=±12y=±23
So points are (2,23) and (2,23)
Area of shaded region
=2[036xdx+2416x2dx]=26[x3/23/2]02+22416x2dx=436(2)3/2+2[x216x2+162sin1x4]24=436×22+16sin1121216sin112=83×23+16π24316π6
433+8π8π3
=(433+163π) square unit

Areas of Bounded Region Exercise 20.3 question 11.

Answer:
(8π323)sq. units
Hint:
a2x2dx=x2a2x2+a22sinxa+c .
Given: The equation of the given curves is
x2+y2=4 ........(I)
(x2)2+y2=4 .........(II)
Solution: The equation of the given curves is
x2+y2=4 ........(I)
(x2)2+y2=4 .........(II)
Clearly x2+y2=4 represents a circle with center (0, 0) and radius 2. Also, (x2)2+y2=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(1,3) and D(1,3).
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 = Δx and area = y1Δx. As it can move from x = 0 to x = 1
 Area OACO=01y1dx AreaO ACO=014(x2)2dx
Similarly, approximating rectangle in the region CABC has length = y2 , width = Δx and area = y2Δx
As it can move from x =1 to x =2
 Area CABC=12y2dx=124x2dx
Hence,required area A is given by
A=2[014(x2)2dx+124x2dx]A=2[[(x2)24(x2)2+42sin1(x2)2]01+[x24x2+42sin1x2]12]A=2{32+2sin1(12)2sin1(1)+2sin1(1)322sin112}
=2[32(π6)+2(π2)+2(π2)2(π6)]=2(32π3+2π)=2(4π33)=(8π323) sq.units. 

Areas of Bounded Region Exercise 20.3 question 12.

Answer:
92sq. units .
Hint: Using the formula
Given: The equations of the given curves are
y2=x ..........(I)
x+y=2 ........(II)
Solution: Plot the two curves
y2=x ..........(I)
x+y=2 ........(II)

Solving (I) and (II) ,we have
y2+y=2(y+2)(y1)=0y=2,1
We have to determine the area of shaded region.
Required Area
=21(2y)dy21y2dy=2yy22y33]21=(21213)(442+83)=92 square units. 

Areas of Bounded Region Exercise 20.3 question 13.

Answer:
[43a3/2+[8π3a163a2163sin1(3a4)]] square unit.  where a=9+2739
Hint: Using the identity formula a2x2dx=x2a2x2+a22sin1xa+c. .
Given: Here we know that R=(x,y):y23x,3x2+3y216
Solution:

Here we know that R=(x,y):y23x,3x2+3y216
We can write it as
R=(x,y):y23x(x,y):3x2+3y216=R1R2
Here R1=(x,y):y23x which represents the region inside the parabola, y2 = 3x with vertex (0, 0) and x-axis as it axis
R=(x,y):3x2+3y216 represents the interior of 3x2+3y2=16 circle having (0, 0) as centre and 43 as radius
So the region R which is intersection of points R1 and R2 is shaded in the figure
3x2+3y2=16 …… (1)
y2=3x…… (2)
Solving both the equations
3x2 + 9x – 16 = 0
So we get
x=(9±273)6
Here x=(9±273)6 is the rejecting negative value
Substituting y = 0 in equation (1)
x=43
We know that the circle (1) cuts x-axis at P (43,0) and P’ ((43,0)
So the required area can be written as
Required area = 2 [area of ODPAO] = 2 [area of ODAO + area of ADPA] .
 Re quired area =2[0a3xdx+a4,β3x2dx]
Intergrating w.r.t
=2(3[x3/23/2]0a+[x2163x2+163sin1(x4/3)]a4/s)
Substituting the value of x
[43a3/2+[8π3a163a2163sin1(3a4)]] square unit.  where a=9+2739

Areas of Bounded Region Exercise 20.3 question 15.

Answer:
272
Hint:
First we need to find the point of intersection of the given curve
Given: Equation of parabola y=3x2 and equation of line 3xy+6=0
Solution:

Equation of parabola y=3x2 ---(i)
Equation of line 3xy+6=0 ---(ii)
Putting the value of y in equation (ii)
3x3x2+6=03x2+3x+6=03x23x6=0x2x2=0x22x+x2=0x(x2)+1(x2)=0
(x+1)(x2)=0x+1=0,x2=0x=1,x=2
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)
Required area=12(YLineYParabola)dx
=12(3x+63x2)dx=[3x22+6x3x23]12=(32×4+128)(326+1)=101+72=20+72=272
Hence , area of the region is 272.

Areas of Bounded Region Exercise 20.3 question 16

Answer:
163ab sq. units .
Hint:
Use basic concepts.
Given:
y2=4ax and x2=4by .
Solution: To find area enclosed by
y2=4ax ..........(1)
x2=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 (4a13b23,4a23b13)

A rough sketch is given as:

The shaded region is required area and it is sliced into rectangle of width=Δx and length(y2y1)
Area of rectangle=(y1y2)Δx
This approximation rectangle slides from x=0 to, x=4a13b23 , so
Required area = Region OQAPO
=04a14b23(y1y2)dx=04a1ab2(2axx24b)dx=[2a23xxx312b]04a13b2=32a3a13b23a16b13=64ab212b=163ab sq.units 
Hence the required area is 163ab sq. units .

Areas of Bounded Region Exercise 20.3 question 17.

Answer: π3
Hint:
use basic concepts.
Given: The equations of the curves are
x2+y2=4
x=3y
Solution: Obviously x2+y2=4 is a circle having centre at (0, 0) and radius 2 units. Forgraph of line
x=3y.
x
0
1
y
0
0.58


For intersecting point of given circle and line
Putting x=3 in x2+y2=4 we get
(3y)2+y2=43y2+y2=44y2=4y=±1x=±3
Intersecting points are (3,1),(3,1)
Shaded region is required region.

Now required area =03x3dx+034x2
=13[x22]03+[x4x22+42sinx2]32=123(30)+[2sin1(32+2sin132)]
=32+[2π2322π3]=32+π322π3=π2π3=π3( proved )

Areas of Bounded Region Exercise 20.3 question 18

Answer:
9 sq. units.
Hint:use basic concepts.
Given:
he given equations of the curves are
y=x, x=2y+3
Solution: We have,
y=x, x=2y+3
Solving we get,
y=2y+3,y0y2=2y+3,y0y22y3=0,y0(y3)(y+1)=0,y0y=3

The graph of function y=x is part of parabolay=x2 lying above x –axis.
The graph is as shown in the figure.

From the figure, area of shaded region,
A=03(2y+3y2)dy=[2y22+3yy33]03=[189+990]=9 sq units 

Areas of Bounded Region Exercise 20.3 question 19

Answer:
4a239(3+4π)
Hint:
a2x2dx=x2a2x2+a22sin1a+c .
Given:
Given curves x2+y2=16a2 and y2=6ax .
Solution:
The point of intersection of circle
x2+y2=16a2

And parabola, y2=6ax is
x2+6ax16a2=0(x+8a)(x2a)=0x=2a,y=±23a
The required common area ,A=2[APOA]
=202aydx+22a4aydy=202a6axdx+22a4a(4a)2x2dx=26a[23x3/2]02a+2[x2(4a)2x2+12(4a)2sin1x4a]2a4a=26a23(2a)3/2+2[(02a3a)+8a2(sin1sin112)]=22343a2+2[23a2+8a2(π2π6)]=1633a243a2+16a2π3=433a2+16πa23=4a23(3+4π) sq.units. 

Areas of Bounded Region Exercise 20.3 question 20.

Answer:
(4π323)sq. units
Hint::
a2x2dx=x2a2x2+a22sin1a+c .
Given:
Given curves are x2+y2=8x and y2=4x .
Solution:
We have given equations
x2+y2=8x ...............(1)
y2=4x ........(2)
Equation (1) can be written as
(xy)2+y2=42
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.
A=04(8xx24x)dx=04((42)(x4)22x)dx=[(x4)2(4)2(x4)2+162sin1(x4)42(2x3)3/2]04=[8sin1043×8][8×(π2)]=323+4π=(4π323) sq.units 

Areas of Bounded Regions exercise 20.3 question 21

Answer:
123sq. unnits .
Hint:
Given:
Given curve y2=5x2 and y=2x2+9 .
Solution:


y2=5x2 Represents a parabola with vetex(0,0) and opening upward , symmetrical about +ve axis
y=2x2+9 Represents the wider parabola, with vertex at C (0,9)
To find point of intersection, solve the two equations
5x2=2x2+93x2=±9x=±3y=15
Thus A(3,15) and ’A(3,15) are point of intersection of two parabolas.
Shaded area A’OA=2 x area (OCAO)
Consider a vertical stip of length |y2y1| and width dx
Area of approximating rectangle |y2y1|dx
The approximating rectangle moves from x=0 to x=3
 Area (OCAO)=03|y2y1|dx=(y2y1)dx{|y2y1|=y2y1asy2>y1}=03(2x2+95x2)dx=03(93x2)dx=[(9x3x33)]03=9333=63
Shaded area B’A’B=2 area OCAO= 2×63=123sq.units

Areas of Bounded Regions exercise 20.3 question 22.

Answer:
323 sq. Units
Hint:
Use concept.
Given:
The given equations are y=2x2 and y=x2+4
Solution:
To find the area enclosed by,
y=2x2 ........(i)
y=x2+4 ........(ii)
On solving the equation (i) and (ii),
2x2=x2+4
or x2=4
or x=±2
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)
=202(x2+42x2)dx=202(4x2)dx=2[4xx33]02=2[(883)0]=323 sq.units 

Areas of Bounded Regions exercise 20.3 question 23.

Answer:
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=12(3x+7)
BC is: y=12(11x)
AC is: y=12(x+5)
Required area
=1211(3x+7)dx+1213(11x)dx1213(3x+5)dx=[112(3x+7)2]1114[(11x)2]1314[(x+5)2]13=7+912=4 sq.units 

(ii)

Equation of,
AB is: y=32x+4
BC is: y=4x2
AC is: y=12x+2
Required area
=20(32x+4)dx+02(4x2)dx22(12x+2)dx=[3x24+4x]20+[4xx24]02[x24+2x]22=5+78=4sq. units

(III)


Equation of ,
AB is: y=x+3
BC is : y=345x2
AC is: y=263x4
Required area
=24(x+3)dx+46(345x2)dx26(263x4)dx=[x22+3x]24+[34x25x24]46[26x4+3x28]26
=[(8+12)(2+6)+(10245)(6820)(39272)(1332)]=208+574826+12=7 sq.units 

Areas of Bounded Regions exercise 20.3 question 24

Answer:
16 sq.units 
Hint:
Given: The given equations are y=x and y=x .
Solution:

 Area of the bounded region =01xxdx=[x3/23/2x22]01=[2312]=16 squnits 


Areas of Bounded Regions exercise 20.3 question 25

Answer:
8π3sq. units
Hint: :
a2x2dx=x2a2x2+a22sin1a+c .
Given:
The given equations are x2+y2=16 and y=3x .
Solution:

x2+y2=16 Represents a circle with centreo(0,0) and cutting the x axis at A(4,0)
y=3x Represents straight passing through o(0,0) .
Point of intersection is obtained by solving the two equations.
x2+y2=16 and y=3x
x2+(3x)2=16
4x2=16
x=±2
y=±23
B(2,23) and B(2,23) are points of intersection of circle and straight line
Shaded area (OBQAO) =area(OBPO)+area(PBQAP)

=023xdx+2416x2dx=3[x22]02+[12x16x2+162sin1(x4)24]=23+8π2238π6=23+4π234π3=8π3 sq.units 

Areas of Bounded Regions exercise 20.3 question 26

Answer:
163sq. units .
Hint:
Use concept.
Given:
The given equations are y2=2x+1 and xy1=0
Solution:
To find area bounded by
y2=2x+1 ...........(1)
xy1=0 ........(2)
On solving the equation (i) and (ii)
xy=1 or y2=2(y1)+1 or y2=2y1 or (y+1)(y3)=0 or y=3 or 1x=4,0
Equation (i) is a parabola with vertex (12,0) (−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
=13(1+yy212)dy=1213(2+2yy2+1)dy=1213(3+2yy2)dy=12[3y+y2y33]13=12[(9+99)(3+1+13)]=12[9+53]=163 sq.units 

Areas of Bounded Regions exercise 20.3 question 27

Answer:
643sq. units
Hint:
use concept.
Given:
The given equations are y=x1 and (y1)2=4(x+1)
Solution:

We have.y=x1 and (y1)2=4(x+1)
(x11)2=4(x+1)
(x2)2=4(x+1)x2+44x=4x+4x2+44x4x4=0x28x=0x=0 or x=8y=1 or 7
Consider a horizontal strip of length |x2x1| and width dy where P(x2,y) lies on straightline and Q(x1,y) lies on the parabola.
Area of approximating rectangle=|x2x1| and it moves from y =1 to y =- 7
 Required area =area(OADO)=17|x2x1|dy
=17|x2x1|dy{|x2x1|=x2x1 as x2>x1}=17[(1+y)14{(y1)24}]dy=17{1+y14(y1)2+1}dy=17{2+y14(y1)2}dy
=[2y+y22112(y1)3]17=[14+49212×6×6×6][2+12+112×2×2×2]=[14+49218][2+12+23]=[412+56]=643 sq.units 

Areas of Bounded Regions exercise 20.3 question 28

Answer:
92unit2
Hint:
Use concept.
Given:
The given equations are y=x2 and x+y+2=0 .
Solution:


 Area of the bounded region =12x2(2x)dx=[x23+2x+x22]12=[83+6][13+122]=92 unit 2

Areas of Bounded Regions exercise 20.3 question 29

Answer:
92sq. units
Hint:
Use concept.
Given:
The given equations are y=2x2 and y+x=0 .
Solution:
To find area bounded by
y=2x2 ......(1)
y+x=0 .......(2)
Equation (1) represents a parabola with vertex (0,2) and downward, meets axes at (±0,2) .
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 =(y2y1)Δx. It slides from x =-1 to x = 2 so,
Required area =Region ABPCOA
A=12(y1y2)dx=12(2x2+x)dx=[2xx33+x22]12=[(483+2)(2+13+12)]=[103+76]=276=92 sq.units 

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

Answer:
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
3xy3=02x+y1=0x2y1=0

Area of bounded region =
=033x3(x12)dx+35122x(x12)dx=[3x223xx24+12x]03+[12x2x22x24+12x]35=[272994+32]=[6025254+5236+9+9432]=11 sq.units 

Areas of Bounded Regions exercise 20.3 question 30 sub question 2

Answer:
6.5 sq. units
Hint:
Given:
3x2y+1=0,2x+3y21=0,x5y+9=0
Solution:
3x2y+1=0,y1=(3x+1)22x+3y21=0,y2=(212x3)x5y+9=0,y3=(x+9)5
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

13y1dx+36y2dx16y3dx=31(3x+1)2dx+36(212x)3dx16(x+9)5dx=12(3x22+x)13+13(21xx2)36+15(x22+9x)16=12[14]+13[36]13[1252]=7+1212.5=6.5 sq.units 

Areas of Bounded Regions exercise 20.3 question 31

Answer:
116sq. units
Hint:
Use concept.
Given:
The given equation are y=x and y=x2+2 .
Solution: To find area bounded by x =0, x =1
and
y=x ........(1)
y=x2+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
A=01(y1y2)dx=01(x2+2x)dx=[x33+2xx22]01=[(13+212)0]=(2+1236)=116 sq.units 

Areas of Bounded Regions exercise 20.3 question 32

Answer:
4 sq. Units
Hint:
Use concept.
Given:
Equation of the given curve x=y2 and x=3y2
Solution: To find area bounded by
x=y2 …(i)
And x3=y2 …(ii)
On solving the equation (i) and (ii),
y2=32y2
Or 3y2=3
Or y=±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]
=2[01y1dx+13y2dx]=2[01xdx+133x2dx]=2[(23xx)01+(23(3x2)3x2(2))2]1]=2[(230)+{(0)(23).1.1.(2)}]=2[23+43]
= 4 sq.units

Areas of Bounded Regions exercise 20.3 question 33

Answer:
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=52x9
Equation of BC is given by
y=x+12
Equation of AC is given by
y=34x2

Area of ABC
=46(yAByAC)dx+68(yBCyAC)dx=46(52x934x+2)dx+08(x+1234x+2)dx=46(74x7)dx+68(7x4+14)dx=[7x287x]46+[7x28+14x]68=[(63242)(1428)]+[(56+112)(632)+84]=[6324214+2856+112+63284]=6356=7 sq.units 

Areas of Bounded Regions exercise 20.3 question 34

Answer:
(5π412) sq. Units.
Hint:
Given: {(x,y):|x1|y(5x2)}
Solution: Equation of the curve is y=(5x)2 or y2+x2=5 ,which is a circle with
centre at (0, 0) and? radius? 52.

Equation of the line is y=|x1|
Consider, y =x -1 and y=5x2
Eliminating y ,we get
x1=5x2x2+12x=5x22x22x4=0x2x2=0(x2)(x+1)=0x=2,1
The required area is
A=125x2dx11(x+1)dx12(x1)dx=[x25x2+52sin1(x5)]12+[x22x]11+[x22+x]12=[2254+52sin1(25)+1251+52sin1(15)]+[121121]+[42+2+121]
=[1+52sin1(25)+22+52sin1(15)]212=2+[52sin1(25)+52sin1(15)]212=52×π212=5π412
=(5π412) sq.units
Therefore, the area of the region is (5π412) sq.units.

Areas of Bounded Regions exercise 20.3 question 35

Answer:
1sq.unit.
Hints:
Use concept.
Given:
The curves y=|x1| and y=1.
Solution:
To find area bounded by y=|x1| and y=1
y={x1, if x01x, if x<0
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
=01(y1y2)dx+12(y1y3)dx=01(11+x)dx+12(1x+1)dx=01xdx+12(2x)dx=(x22)01+(2xx22)12=(120)+[(42)(212)]=12+(22+12)
A = 1 sq. units.

Areas of Bounded Regions exercise 20.3 question 36

Answer:
4π sq. Units
Hints:
Use concept.
Given:
The first quadrant enclosed by x-axis, the line y = x and the circle x2+y2=32
Solution:

Point of intersection, x =4
Area of a shaded region = 04xdx+44232x2dx
=[x22]04+[x232x2+16sin1x42]442=8+16π284π=4π=4π Sq.units. 

Areas of Bounded Regions exercise 20.3 question 37

Answer:
32π343sq.units
Hints:
Given: Equation of circle x2+y2=32 exterior to parabola y2=6x
Solution: First finding intersection point by solving the equation of two curves
x2+y2=16 (i)  and y2=6x (ii) x2+6x=16x2+6x16=0
x2+8x2x16=0x(x+8)2(x+8)=0(x+8)(x2)=0x=8
(not possible y2 cannot be -ve )

Or x=2 (only allowed value)
y=±23
A1=026xdx=[6x3232]02=263×232=23×23×22=83 sq.units 
A2=24(16x2)dx=[x216x2+162sin1x4]24=0+8sin18sin112164=8×π28×π612=4π4π312=8π312 sq.units 
 Area =2(A1+A2)=2(83+8π312) Required area =π×1616x3+43163=32π34332π343sq.units
Therefore, the area of the circle is 32π343sq.units

Areas of Bounded Regions exercise 20.3 question 38

Answer:
92 sq. Units
Hints:
Use concept .
Given:
x2=y is a parabola line y=x+2
Solution:x2=y is a upward parabola line y=x+2
They meet at pts (2,4) and (-1,1)
By solving the equation,
x2=y,y=x+2x2x2=0x=2 or 1x=4 or 1
Required area
= 12 Area of line - 12 Area of parabola


=12(x+2)dx12x2dx=[x22]12+[2x]12[x33]12=42(1)22+22(21)(233(1)33)=212+4+2(83+13)=8123=512=92 sq.units 


Areas of Bounded Regions exercise 20.3 question 39

Answer: The Area of the region {(x,y):0yx2+3;0y2x+3;0x3} is 383 sq.units.
Hints:
Given: {(x,y):0yx2+3;0y2x+3;0x3}
Solution: To find area given equation are
y=x3+3...... (i)
y=2x+3....... (ii)
And x=3 .......(iii)
Solving the above three equations to get the intersectons points,
x2+3=2x+3 Or x22x=0 Or x(x2)=0 And x=0 or x=2y=3 or y=7
Equation (i) represents a parabola with vertices (3,0) and axis as y - axis
Equation (ii) represents a line passing through (0,3) and (32,0)
The points of intersection are A(0,3) and B(2,7)
These are shown in the graph below:

Required area =
=23y1dx+02y2dx=23(2x+3)dx+02(x2+3)dx=(x3+3x)23+(x23+x)02=[(9+9)(4+6)]+[(83+2)(0)]=[1810]+[143]=8+143
=383 sq.units
Therefore, the area of the region is 383 sq.units

Areas of Bounded Regions exercise 20.3 question 40

Answer:
π8 sq. Units
Hints:
Use concept of definite integrals.
Given:
The curve y=1x2 , line y =x and the positive x-axis
Solution:

y=1x2y2=1x2x2+y2=1
Hence,
y=1x2 represents the upper half of the circle x2+y2=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
y=xy=1x2x=1x2x2=1x22x2=1x=±12y=±12
D(12,12) And D(12,12) are two points of intersection between the circles and the straight line.
And D(12,12) is the intersection point of y=1x2 and y =x .
Required area = Shaded area (ODAEO) = Area (ODEO) + Area (EDAE)…..(1)
Now, area (ODEO) = 012xdx
=[x22]012=12(12)=14 sq.units ..(2)
Area (EDAE) = 1211x2dx
=[12x1x2+×12×sin1(x1)]121=0+12sin1(1)12×12×1(12)212sin1(12)=12×π21412×π4.{ using, sin1(1)=π2 and sin1(12)=π4}=π4π814=π814 sq.units  (3) 
Area (ODAEO) = 14+π414=π8 sq.units

Areas of Bounded Region Exercise 20.3 Question 41

Answer:
152 sq. units
Hints:
Use concept.
Given:
Lines y=4x+5,y=5x and 4y=x+5
Solution:
To find area bounded by lines
y=4x+5 {Say AB}
y=x5 {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) =01(y1y2)dx

=10(4x+5x454)dx=10(15x4+154)dx=154(x22+x)10=154[(0)(121)]=154×12

AR ( ABD) = 158 sq. units
AR ( BDC) = 03(y2y3)dx
=03[(5x)(x4+54)]dx=03[5xx454]dx=03(5x4+154)dx=54(3xx22)=54(992)
AR ( BDC) = 458 sq. units
Using equation (1), (2) and (3),
AR ( ABC) = 158+458
=608
ar( ABC) = =152 sq. units

Areas of Bounded Region Exercise 20.3 Question 42

Answer:
(6π932)sq. units
Hints:
Use concept.
Given:
The two curves x2+y2=9 and (x3)2+y2=9
Solution:
To find area enclosed by
x2+y2=9 ……. (1)
(x3)2+y2=9 …….. (2)
Equation (1) represents a circle with centre (0,0) and meets axis at (±3,0) , (0,±3)
Equation (2) is a circle with center (3,0) and meets axis at (0,0) , (6,0)
they intersect each other at (32,332) and (32,332) . 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[0329(x3)2dx+3239x2dx]
=2[{(x3)29(x3)2+92sin1(x3)3[]}032+{x29x2+92sin1(x3)}32]23]=2[{(34994+92sin1(36))(0+92sin1(1))}+{(0+92sin1(1))(34994+92sin1(12)).
=2[{93892π6+92π2}+{92π293892π6}]=2[9383π4+9π4+9π49383π4]=2[12π41838]=(6π932) sq. units 

Areas of Bounded Regions Exercise 20.3 Question 43

Answer:
π2 sq. units
Hints:
Use concept.
Given:
{(x,y):x2+y24,x+y2}
Solution: The equation of the given curves are
x2+y2=4 ……(1)
x+y=2 …….(2)
Clearly x2+y2=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 = y2y1
Width = x and
Area = (y2y1)x
Since the approximating rectangle can move from x=0 to x=2, the required area is given by
A=02(y2y1)dx We have y1=2x and y2=4x2 Thus, A=02(4x22+x)dxA=02(4x2)dx202dx+02xdx
A=[x4x22+a22sin1(x2)]022(x)02+(x22)02A=42sin1(22)4+2A=2sin1(1)2A=2×π22A=π2
Therefore, the area of the region is π2 sq. units

Areas of Bounded Regions Exercise 20.3 Question 44

Answer: (3π23) sq. units 
Hints:
Use concept.
Given:
{(x,y):x29+y241x3+y2}
Solution:
To find the area of a region
{(x,y):x29+y241x3+y2}
Here,
x29+y24=1 …..(1)
x3+y2=1 …..(2)
Equation (1) represents an ellipse with centre at origin and meets axis at (±3,0) ,(0,±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 (y2y2)x which slides from x=0 to x=3, so
Required area = Region APBQA
A=03(y1y2)dx=03[239x2dx23(3x)dx]=23[x29x2+92sin1(x3)3x+x22]03=23[{0+92π29+92}{0}]
=23[9π492]A=(3π23) sq. units 
Therefore, the area of the region is (3π23) sq. units 

Areas of Bounded Regions Exercise 20.3 Question 45

Answer:
(4π33) sq. units 
Hints:
Use concept.
Given:
The curve y=4x2 ,x2+y24x=0 and the x-axis.
Solution:
Give, equation of a curves are
y=4x2 ……(1)
x2+y24x=0 ……(2)
Consider the curve
y=4x2=y24x=0
y2+x2=0 which represents a circle with centre (0, 0) and radius 2 units.
Now, consider the curve x2+y24x=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
x2(4x)24x=0 or 44x=0 or x=1
On substituting x =1 in Eq. (1), we get y=3
Thus, the point of intersection is (1.3) .
Clearly, required area = Area of shaded region OABO
=01y(sacoricicic k)dx+12y(f trs cic cie)dx=014xx2dx+124x2dx=01(x24x)dx+0222x2dx
=01[x22(2)(x)+44]dx+[x24x2+42sin1(x2)]12=014(x2)2dx+[2sin1(1){123+2sin1(12)}]
=[(x2)24xx2+2sin1(x22)]01+[2π2322π6]=[{32+2sin1(12)}{2sin1(1)}]+(ππ332)=322sin1(12)+2sin1(1)+2π332[sin1(x)=sin1x]
=2π22π6+2π33=ππ3+2π33=π+π33=(4π33) sq. units 

Areas of Bounded Regions Exercise 20.3 Question 46

Answer:
12 .sq. Units.
Hints:
Use concept.
Given:
The curves y=|x1| and y=|x1|+1
Solution:
To find area enclosed by
y=|x1|y={(x1), if x1<0(x1), if x10
y={1x, if x<1x1, if x1 And y=|x1|+1y={+(x1)+1, if x1<0(x1)+1, if x10y={x, if x<1x+2, if x1
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 =(y2y1)x and it slides from x=12 to x =1
Region ABDA is sliced into rectangle of area =(y3y4)x and it slides from x =1 to x=32 .
So, using equation (1),
Required area = Region BDCB + Region ABDA
=121(y1y2)dx+132(y3y4)dx=121(x1+x)dx+132(x+2x+1)dx=121(2x1)dx+132(32x)dx=[x2x]121+[3xx2]132
=[(11)(1412)]+[(9294)(31)]=14+942A=12 sq. units 

Areas of Bounded Regions Exercise 20.3 Question 47

Answer:
A=332sq. units
Hints:
Use concept.
Given:
The curves 3x2+5y=32 and y=|x2| .
Solution:
To find area enclosed by
3x2+5y=32
3x2=5(y325) ……. (1)
And
y=|x2|y={(x2), if x2<1(x2), if x21y={2x, if x<2x2, if x2........(2)
Equation (1) represents a downward parabola with vertex (0,325) and equation (2) represents lines. A rough sketch of curves is given as:

Required area = Region ABECDA
A = Region ABEA + Region AECDA
=23(y3y4)dx+22(yy2)dx
=23(323x25x+2)dx+22(323x252+x)dx=1523(323x25x+105)dx+22(323x210+5x5)dx=15[23(423x25x)dx+22(223x2+5x)dx]
=15[(42xx35x22)23+(22xx3+5x22)22]=15[{(12627452)(84810)}+{(448+10)(44+8+10)}]=15[{153266}+{46+26}]=15[212+72]
A=332sq. units


Areas of Bounded Regions Exercise 20.3 Question 48

Answer:
12524 sq. units
Hints:
Use concept.
Given:
The parabola y=4xx2 and y=x2
Solution:
To area enclosed by
y=4xx2y=x24x+44y+4=(x2)2(y4)=(x2)2 And y=x2x(y+14)=(x12)2
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 (12,14) and meets axis at (1,0), (0,0). Points of intersection of parabola are (0,0) and (52,154) .
A rough sketch of the curves is as under:

Shaded region is required area it is sliced into rectangles with area = (y2y1)x . It slides from x =0 to x=52 , so
Required area = Region OQAP
A=052(y1y2)dx=052[4xx2x2+x]dx=052[5x2x2]dx=[5x2223x2]052=[(125825024)(0)]A=12524 sq. units 
Therefore, the area enclosed by the parabola y=4xx2 and y=x2 is 12524 sq. units.

Areas of Bounded Regions Exercise 20.3 Question 49

Answer:
121:4
Hints:
Use basic concepts.
Given:
The parabola y=4xx2 and y=x2x .
Solution:
The curves are
y=4xx2(y4)=(x2)2 And y=x2x(y+14)2=(x12)2
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 (12,14) and meets axis at (1,0), (0,0) and (52,154) . A rough sketch of the curves is as under:

Area of the region above x-axis
A1 = Area of region OBACO
= Region OBCO + Region BACB
=01y1dx+152(y1y2)dx
=01(4xx2)dx+152(4xx2x2+x)dx=(4x22x33)01+[5x222x33]152=(213)+[(125825024)(5223)]=53+12524116=12124 sq. units 
Area of the region below x-axis
A2 = Area of region OPBO
= Region OBCO + Region BACB
=|01y2dx|=|01(x2x)dx|=|(x33x22)01|=|(1312)(0)|=|16|A2=16 sq. units 
A1A2=12124:424A1A2=121:24

Areas of Bounded Regions Exercise 20.3 Question 50

Answer:
4 sq. units.
Hints:
Use concept of definite integrals.
Given:
The curves y=|x1| and y=3|x|
Solution:
To find area bounded by the curve

y=|x1|y={1x, if x<1.......(1)x1, if x1........(2))
And y=3|x|
y={3+x, if x<0......(3)3x, if x0......(4)

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
=12(y1y2)dx+01(y1y3)dx+10(y4y3)dx=12(3xx+1)dx+01(3x1+x)dx+10(3+x1+x)dx=12(42x)dx+012dx+10(2+2x)dx=[4x2]12+[2x]01+[2x+x2]10=[(84)(41)]+[20]+[(0)(2+1)]=(43)+2+1
A = 4 sq. units.
Therefore, the area bounded by the curve is 4 sq. units.

Areas of Bounded Region Exercise 20.3 question 51.

Answer:
M=2.
Hints:
Use concept of definite integrals
Given:
The parabola y2=4ax and the line y = mx is a212 .
Solution:

Area of the bounded region = a212
a212=0a4axmxdxa212=[2ax1232mx22]0a
a212=4a23ma22m=2
Therefore, the value of m is 2.

Areas of Bounded Region Exercise 20.3 question 52.

Answer: The value of a is 2.
Hints:
Given: The parabola y2=16ax and x2=16ay cgfy
Solution:

Area of the bounded region = 10243
x=y216ay=0;16a
 Area =016a16axdx016ax216adx=[4ax4232]016a[x348a]016a=23×4a×(16a)320116a×(16a)330
A=1024310243=8a3×(16a)×4a256a331024=256a2a2=4a=2


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.

  • Teachers are very fond of the solutions of the RD Sharma and most of the questions which they assign for homework are taken in from the RD Sharma class 12th exercise 20.3.

  • The RD Sharma class 12 chapter 20 exercise 20.3 Is important to practice thoroughly as it has been observed that most of the questions that are asked in the board exams are the similar pattern that has been provided in the RD Sharma solutions.

  • The RD Sharma solutions are of the similar pattern of the NCERT, therefore students who are appearing for public examinations can also prepare using these solutions.

  • The RD Sharma class 12th exercise 20.3 Is recommended because it can be easily accessed by downloading the study material from the careers360 website and that is also free of cost.

NEET Highest Scoring Chapters & Topics
This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.
Download E-book

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

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.

Articles

Back to top