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NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

Edited By Ramraj Saini | Updated on Sep 14, 2023 10:14 PM IST | #CBSE Class 12th
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NCERT Application Of Integrals Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals are discussed here. These NCERT solutions are created by expert team at Creers360 keeping in mind of latest syllabus of CBSE 2023-24. In geometry, you must have learned some formulas to calculate areas of simple geometrical figures like triangles, rectangles, trapeziums, circles, etc. But how do you calculate areas enclosed by curves? In this article, you will get NCERT solutions for class 12 maths chapter 8 application of integrals. This article also includes application of integrals class 12 solutions. Important topics that are going to be discussed in this chapter 8 class 12 maths are the area under simple curves, the area between lines and arcs of circles, parabolas, and ellipses. Interested students can find all NCERT Solutions for Class 12 Maths in one place

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  1. NCERT Application Of Integrals Class 12 Questions And Answers
  2. NCERT Application Of Integrals Class 12 Questions And Answers PDF Free Download
  3. Application Of Integrals Class 12 NCERT Solutions - Important Formulae
  4. NCERT Application Of Integrals Class 12 Questions And Answers (Intext Questions and Exercise)
  5. Application of integrals class 12 - Topics
  6. NCERT solutions for class 12 maths - Chapter wise
  7. Key Features of NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals
  8. NCERT solutions for class 12 subject wise
  9. NCERT Solutions Class Wise
  10. NCERT Books and NCERT Syllabus
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In the applications of integrals class 12 ncert solutions, questions from all these topics are covered. In this composition of Class 12 Maths Chapter 8 NCERT solutions application of integrals, you will learn some important applications of integrals class 12. If you are interested to check all NCERT solutions from classes 6 to 12 in a single place, which will help you to learn CBSE maths and science. Here you will get NCERT solutions for class 12.

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NCERT Application Of Integrals Class 12 Questions And Answers PDF Free Download

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Application Of Integrals Class 12 NCERT Solutions - Important Formulae

>> Area Enclosed by a Curve and Lines: The area enclosed by the curve y = f(x), the x-axis, and the lines x = a and x = b (where b > a) is given by the formula:

Area = ∫[a, b]y.dx = ∫[a, b]f(x).dx

>> Area Bounded by Curve and Horizontal Lines: The area of the region bounded by the curve x = φ(y) as its y-axis and the lines y = c and y = d is given by the formula: Area = ∫[c, d]x.dy = ∫[c, d]φ(y).dy

>> Area Between Two Curves and Vertical Lines: The area enclosed between two given curves y = f(x) and y = g(x), and the lines x = a and x = b is given by the formula:

Area = ∫[a, b][f(x) - g(x)].dx (Where f(x) ≥ g(x) in [a, b])

>> Area Between Curves with Different Intervals: If f(x) ≥ g(x) in [a, c] and f(x) ≤ g(x) in [c, b], where a < c < b, then the resultant area between the curves is given as:

Area = ∫[a, c][f(x) - g(x)].dx + ∫[c, b][g(x) - f(x)].dx

Free download Application Of Integrals Class 12 NCERT Solutions for CBSE Exam.

NCERT Application Of Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT class 12 maths chapter 8 question answer Exercise: 8.1

Question:1 Find the area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x -axis in the first quadrant.

Answer:

Area of the region bounded by the curve y2=x and the lines x=1,x=4 and the x -axis in the first quadrant

Area = 14ydy=14xdx

[x3232]14=23[(4)32(1)32]

=23[81]

= 14/3 units

Question:2 Find the area of the region bounded by y2=9x,x=2,x=4 and the x -axis in the first quadrant.

Answer:

Area of the region bounded by the curve y2=9x,x=2,x=4 and the x -axis in the first quadrant

Area = 24ydy=249xdx=324xdx

3[x3232]24=3.23[(4)32(2)32]

=2[822]

=[1642] units

Question:3 Find the area of the region bounded by x2=4y,y=2,y=4 and the y -axis in the first quadrant.

Answer:

The area bounded by the curves x2=4y,y=2,y=4 and the y -axis in the first quadrant is ABCD.


= \int^4_{2} x dy

=242ydy

=224ydy

=2{y3232}24

=43{(4)32(2)32}

=43{822}

={32823} units.

Question:4 Find the area of the region bounded by the ellipse x216+y29=1.

Answer:

The area bounded by the ellipse : x216+y29=1.


1646972152541

Area will be 4 times the area of EAB.

Therefore, Area of EAB=04ydx

=0431x216dx

=340416x2dx

=34[x216x2+162sin1x4]04

=34[21616+8sin1(1)08sin1(0)]

=34[8π2]

=34[4π]=3π

Therefore the area bounded by the ellipse will be =4×3π=12π units.

Question: 5 Find the area of the region bounded by the ellipse x24+y29=1

Answer:

The area bounded by the ellipse : x24+y29=1

1646972197883

The area will be 4 times the area of EAB.

Therefore, Area of EAB=02ydx

=0231x24dx

=32024x2dx

=32[x24x2+42sin1x2]02

=32[2π2]

=3π2

Therefore the area bounded by the ellipse will be =4×3π2=6π units.

Question: 6 Find the area of the region in the first quadrant enclosed by x -axis, line x=3y and the circle x2+y2=4
Answer:

The area of the region bounded by x=3y and x2+y2=4 is ABC shown:

1646972234725

The point B of the intersection of the line and the circle in the first quadrant is (3,1) .

Area ABC = Area ABM + Area BMC where, M is point in x-axis perpendicular drawn from the line.

Now,area of ABM=12×AM×BM=12×3×1=32 ............(1)

and Area of BMC=32ydx

=324x2dx

=[x24x2+42sin1x2]32

=[2×π232432sin1(32)]

=[π3π22π3]

=[π322π3]

=[π332] ..................................(2)

then adding the area (1) and (2), we have then

The Area under ABC =32+π332=π3 units.

Question: 7 Find the area of the smaller part of the circle x2+y2=a2 cut off by the line
x=a2

Answer:

1646972273424
we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC
Area of ABC = a2aydx=a2aa2x2dx=[x2a2x2+a22sin1xa]a2a
=[a2a2a2+a22sin1aaa22a2(a2)2a22sin1aa2]
=[a20+a22sin11a22a22a22sin112]
=[0+a22π2a24a22π4]
=a24(π21)
Area of ABCD = 2 X Area of ABC
=2×a24(π21)=a22(π21)
Therefore, the area of the smaller part of the circle is a22(π21)

Question:8 The area between x=y2 and x=4 is divided into two equal parts by the line x=a , find the value of a .

Answer:

1646972308309 we can clearly see that given area is symmetrical about x - axis
It is given that
Area of OED = Area of EFCD
Area of OED = 0aydx=0axdx=[x3232]0a=a3232=2a323
Area of EFCD = a4ydx=a4xdx=[x3232]a4=432a3232=2(8a32)3=2(8a32)3
Area of OED = Area of EFCD
2a323=2(8a32)32a32=8a32=4a=(4)23
Therefore, the value of a is a=(4)23

Question:9 Find the area of the region bounded by the parabola y=x2 and y=|x| .

Answer:

1646972345767 We can clearly see that given area is symmetrical about y-axis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of y=x2 and y=|x| is (1 , 1) and (-1 , 1)
Now,
Area od OCAO = Area OAM - Area of OCMO
Area of OAM = 12.OM.AM=12.1.1=12
Area of OCMO = 01ydx=01x2dx=[x33]01=13
Therefore,
Area od OCAO =1213=16
Now,
Area of the region bounded by the parabola y=x2 and y=|x| is = 2 X Area od OCAO =2×16=13 Units

Question: 10 Find the area bounded by the curve x2=4y and the line x=4y2 .

Answer:

1646972388198 Points of intersections of y=x2 and x=4y2 is
A(1,14) and B(2,1)
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO- Area of OMBO

Area of OMBCO = 02ydx=02x+24dx=[x28]02+[x2]02=48+22=32

Area of OMBO = 02ydx=02x24dx=[x312]02=812=23

Area of OBCO = Area of OMBCO- Area of OMBO
=3223=56
Similarly,
Area of OCAO = Area of OCALO - Area of OALO

Area of OCALO = 10ydx=10x+24dx=[x28]10+[x2]10=18(1)2=18+12=38

Area of OALO = 10ydx=10x24dx=[x312]10=(1)12=112

Area of OCAO = Area of OCALO - Area of OALO
=38112=9224=724
Now,
Area of OBAO = Area of OBCO + Area of OCAO
=56+724=20+724=2724=98

Therefore, area bounded by the curve x2=4y and the line x=4y2 is 98 units

Question: 11 Find the area of the region bounded by the curve y2=4x and the line x=3 .

Answer:
The combined figure of the curve y2=4x and x=3
15947265752881594726573011
The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2 × Area of OAB
=2[03ydx]=2032xdx=4[x3/23/2]03=83
therefore the required area is 83 units.

Question: 12 Choose the correct answer in the following

Area lying in the first quadrant and bounded by the circle x2+y2=4 and the lines x=0 and x=2 is

(A)π (B)π2 (C)π3 (D)π4

Answer:

The correct answer is A
The area bounded by circle C(0,0,4) and the line x=2 is
15947268860371594726883361
The required area = area of OAB
02ydx=024x2dx
=[x24x2+42sin1x2]02=2(π/2)=π

Question: 13 Choose the correct answer in the following.

Area of the region bounded by the curve y2=4x , y -axis and the line y=3 is

(A) 2 (B) 94 (C) 93 (D) 92

Answer:

The area bounded by the curve y2=4x and y =3

15947271505371594727147434
the required area = OAB =
03xdy=03y24dy=14.[y33]03=94


NCERT application of integrals class 12 solutions Exercise: 8.2

Question: 1 Find the area of the circle 4x2+4y2=9 which is interior to the parabola x2=4y .

Answer:

The area bounded by the circle 4x2+4y2=9 and the parabola x2=4y .
15947278183451594727815379
By solving the equation we get the intersecting point D(2,12) and B(2,12)
So, the required area (OBCDO)=2 times the area of (OBCO)
Draw a normal on the x-axis (M = 2,0 )


Thus the area of OBCO = Area of OMBCO - Area of OMBO

02(94x2)4dx02x24dx=120294x21402x2dx=14[x94x2+92sin12x3]0214[x33]02=14[2+98sin1223]112(2)3=212+98sin1223=12(26+94sin1223)
S0, total area =
=2×12(26+94sin1223)=26+94sin1223

Question:2 Find the area bounded by curves (x1)2+y2=1 and x2+y2=1 .

Answer:

1646972457138 Given curves are (x1)2+y2=1 and x2+y2=1

Point of intersection of these two curves are

A=(12,32) and B=(12,32)

We can clearly see that the required area is symmetrical about the x-axis

Therefore,

Area of OBCAO = 2 × Area of OCAO

Now, join AB such that it intersects the x-axis at M and AM is perpendicular to OC

Coordinates of M = (12,0)

Now,

Area OCAO = Area OMAO + Area CMAC

=[0121(x1)2dx+1211x2dx]
=[x121(x1)2+12sin1(x1)]012+[x2+12sin1x]121

=[141(12)2+12sin1(121)012sin1(1)]+[0+12sin1(1)141(12)212sin1(12)]
=[38+12(π6)12(π2)]+[12(π2)3812(π6)]
=[38+π6]+[π638]
=2[38+π6]
Now,
Area of OBCAO = 2 × Area of OCAO

=2×2[38+π6]
=2π332

Therefore, the answer is 2π332

Question: 3 Find the area of the region bounded by the curves y=x2+2,y=x,x=0 and x=3 .

Answer:

The area of the region bounded by the curves,

y=x2+2,y=x,x=0 and x=3 is represented by the shaded area OCBAO as

1646972497382

Then, Area OCBAO will be = Area of ODBAO - Area of ODCO

which is equal to

03(x2+2)dx03xdx

=(x33+2x)03(x32)03

=[9+6][92]=1592=212units.

Question: 4 Using integration find the area of region bounded by the triangle whose vertices are (1,0),(1,3) and (3,2) .

Answer:

So, we draw BL and CM perpendicular to x-axis.

Then it can be observed in the following figure that,

Area(ACB)=Area(ALBA)+Area(BLMCB)Area(AMCA)

We have the graph as follows:

1646972546438

Equation of the line segment AB is:

y0=301+1(x+1) or y=32(x+1)

Therefore we have Area of ALBA

=1132(x+1)dx=32[x22+x]11

=32[12+112+1]=3units.

So, the equation of line segment BC is

y3=2331(x1) or y=12(x+7)

Therefore the area of BLMCB will be,

=1312(x+7)dx=12[x22+7x]13

=12[92+21+127]=5units.

Equation of the line segment AC is,

y0=203+1(x+1) or y=12(x+1)

Therefore the area of AMCA will be,

=1213(x+1)dx=12[x22+x]13

=12(92+312+1)=4units.

Therefore, from equations (1), we get

The area of the triangle ABC=3+54=4units.

Question:5 Using integration find the area of the triangular region whose sides have the equations y=2x+1,y=3x+1 and x=4 .

Answer:

The equations of sides of the triangle are y=2x+1,y=3x+1, and x=4 .

ON solving these equations, we will get the vertices of the triangle as A(0,1),B(4,13), and C(4,9)

1646972586017

Thus it can be seen that,

Area(ACB)=Area(OLBAO)Area(OLCAO)

=04(3x+1)dx04(2x+1)dx

=[3x22+x]04[2x22+x]04

=(24+4)(16+4)=2820=8units.

Question:6 Choose the correct answer.

Smaller area enclosed by the circle x2+y2=4 and the line x+y=2 is

(A) 2(π2) (B) π2 (C) 2π1 (D) 2(π+2)

Answer:

So, the smaller area enclosed by the circle, x2+y2=4 , and the line, x+y=2 , is represented by the shaded area ACBA as

1646972621266

Thus it can be observed that,

Area of ACBA = Area OACBO - Area of (OAB)

=024x2dx02(2x)dx

=(x24x2+42sin1x2)02(2xx22)02

=[2.π2][42]

=(π2)units.

Thus, the correct answer is B.

Question:7 Choose the correct answer.

Area lying between the curves y2=4x and y=2x is

(A) 23 (B) 13 (C) 14 (D) 34

Answer:

The area lying between the curve, y2=4x and y=2x is represented by the shaded area OBAO as

1646972657802

The points of intersection of these curves are O(0,0) and A(1,2) .

So, we draw AC perpendicular to x-axis such that the coordinates of C are (1,0).

Therefore the Area OBAO = Area(OCA)Area(OCABO)

=2[x22]012[x3232]01

=|143|=|13|=13units.

Thus the correct answer is B.


NCERT application of integrals class 12 solutions Miscellaneous: Exercise

Question:1 Find the area under the given curves and given lines:

(i) y=x2,x=1,x=2 and x -axis

Answer:

The area bounded by the curve y=x2,x=1,x=2 and x -axis
15947281267411594728124007
The area of the required region = area of ABCD
=12ydx=12x2dx=[x33]12=73
Hence the area of shaded region is 7/3 units

Question:1 Find the area under the given curves and given lines:

(ii) y=x4,x=1,x=5 and x -axis

Answer:

The area bounded by the curev y=x4,x=1,x=5 and x -axis

15947282868341594728284206
The area of the required region = area of ABCD
=15ydx=12x4dx=[x55]12=62515=624.8
Hence the area of the shaded region is 624.8 units

Question:2 Find the area between the curves y=x and y=x2 .

Answer:

the area between the curves y=x and y=x2 .
AOI graph1594728523616
The curves intersect at A(1,1)
Draw a normal to AC to OC(x-axis)
therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO)
=01xdx01x2dx=[x22]01[x33]01=1/21/3=16
Thus the area of shaded region is 1/6 units

Question:3 Find the area of the region lying in the first quadrant and bounded by y=4x2,x=0,y=1 and y=4 .

Answer:

the area of the region lying in the first quadrant and bounded by y=4x2,x=0,y=1 and y=4 .

15947286788651594728676557
The required area (ABCD) =
=14xdy=14y2dy=12.23[y3/2]14=13[81]=73
The area of the shaded region is 7/3 units

Question:4 Sketch the graph of y=|x+3| and evaluate 60|x+3|dx.

Answer:

y=|x+3|

the given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

1646972694635

Integral to be evaluated is

60|x+3|dx=63(x3)dx+30(x+3)dx=[x223x]63+[x22+3x]30=(92+9)(18+18)+0(929)=9

Question:5 Find the area bounded by the curve y=sinx between x=0 and x=2π .

Answer:

The graph of y=sinx is as follows

1646972720677

We need to find the area of the shaded region

ar(OAB)+ar(BCD)

=2ar(OAB)

=2×0πsinxdx=2×[cosx]0π=2×[(1)(1)]=4

The bounded area is 4 units.

Question:6 Find the area enclosed between the parabola y2=4ax and the line y=mx .

Answer:

1646972749429

We have to find the area of the shaded region OBA

The curves y=mx and y 2 =4ax intersect at the following points

(0,0)and(4am2,4am)

y2=4axy=2ax

The required area is

04am2(2axmx)=2a[2x323]04am2m[x22]04am2=32a23m38a2m3=8a23m3units

Question:7 Find the area enclosed by the parabola 4y=3x2 and the line 2y=3x+12 .

Answer:

1646972775838

We have to find the area of the shaded region COB

2y=3x+12y=32x+64y=3x2y=3x24

The two curves intersect at points (2,3) and (4,12)

Required area is

24(32x+63x24)dx=[3x24+6xx34]24=[12+2416][312+2]=20(7)=27 units

Question:8 Find the area of the smaller region bounded by the ellipse x29+y24=1 and the line x3+y2=1 .

Answer:

1646972803949

We have to find the area of the shaded region

The given ellipse and the given line intersect at following points

(0,2)and(3,0)

x29+y24=1y=239x2

Since the shaded region lies above x axis we take y to be positive

x3+y2=1y=23(3x)

The required area is

2303(9x2(3x))dx=23[x2(9x2)+92sin1x33x+x22]03=23([92×π29+92]0)=23(9π492)=32(π2)units

Question:9 Find the area of the smaller region bounded by the ellipse x2a2+y2b2=1 and the line xa+yb=1 .

Answer:


1646972828912

The area of the shaded region ACB is to be found

The given ellipse and the line intersect at following points

(0,b)and(a,0)

x2a2+y2b2=1y=baa2x2

Y will always be positive since the shaded region lies above x axis

xa+yb=1y=ba(ax)

The required area is

ba0a(a2x2(ax))dx=ba[x2a2x2+a22sin1xaax+x22]0a=ba[(a22×π2a2+a22)]=ba(πa24a22)=ab4(π2)units


Question:10 Find the area of the region enclosed by the parabola x2=y, the line y=x+2 and the x -axis.

Answer:

1646972855917

We have to find the area of the shaded region BAOB

O is(0,0)

The line and the parabola intersect in the second quadrant at (-1,1)

The line y=x+2 intersects the x axis at (-2,0)

ar(BAOB)=ar(BAC)+ar(ACO)=21(x+2)dx+10(x2)dx=[x22+2x]21+[x33]10=(122)(24)+0(13)=56 units

The area of the region enclosed by the parabola x2=y, the line y=x+2 and the x -axis is 5/6 units.

Question:11 Using the method of integration find the area bounded by the curve |x|+|y|=1.

[ Hint: The required region is bounded by lines x+y=1,xy=1,x+y=1 and xy=1 ]

Answer:

1646972880742

We need to find the area of the shaded region ABCD

ar(ABCD)=4ar(AOB)

Coordinates of points A and B are (0,1) and (1,0)

Equation of line through A and B is y=1-x

ar(AOB)=01(1x)dx=[xx22]01=(112)0=12 unitsar(ABCD)=4ar(AOB)=4×12=2 units

The area bounded by the curve |x|+|y|=1 is 2 units.

Question:12 Find the area bounded by curves {(x,y);yx2andy=|x|} .

Answer:

1646972905611

We have to find the area of the shaded region

In the first quadrant

y=|x|=x

Area of the shaded region=2ar(OADO)

=201(xx2)dx=2[x22x33]01=2(1213)0=123=13 units

The area bounded by the curves is 1/3 units.

Question:13 Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2,0),B(4,5)andC(6,3).

Answer:

1646972934221

Equation of line joining A and B is

y0x2=5042y=5x25

Equation of line joining B and C is

y5x4=5346y=9x

Equation of line joining A and C is

y0x2=3062y=3x432

ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)

ar(ABL)=24(5x25)dx=[5x245x]24=(2020)(510)=5 units

ar(LBCM)=46(9x)dx=[9xx22]46=(5418)(368)=8 units

ar(ACM)=26(3x432)dx=[3x283x2]26=(2729)(323)=6 units

ar(ABC)=8+5-6=7

Therefore the area of the triangle ABC is 7 units.

Question:14 Using the method of integration find the area of the region bounded by lines:
2x+y=4,3x2y=6andx3y+5=0.

Answer:

1646972962041

We have to find the area of the shaded region ABC

ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC)

The lines intersect at points (1,2), (4,3) and (2,0)

x3y=5y=x3+53

ar(ACLM)=14(x3+53)dx=[x26+5x3]14=(426+5×43)(16+53)=152 units

2x+y=4y=42x

ar(ALB)=12(42x)dx=[4xx2]12=(84)(41)=1 unit

3x2y=6y=3x23

ar(BMC)=24(3x23)dx=[3x243x]24=(1212)(36)=3 units

ar(ABC)=15213=72 units

Area of the region bounded by the lines is 3.5 units

Question:15 Find the area of the region {(x,y);y24x,4x2+4y29} .

Answer:

1646972988754

We have to find the area of the shaded region OCBAO

Ar(OCBAO)=2ar(OCBO)

For the fist quadrant

4x2+4y2=9y=94x2

y2=4xy=2x

In the first quadrant, the curves intersect at a point (12,2)

Area of the unshaded region in the first quadrant is

012(94x22x)dx=[x294x2+98sin12x3]0124[x3/23]01/2=24+98sin11323

The total area of the shaded region is-
= Area of half circle - area of the shaded region in the first quadrant

π2×(32)22(2423+98sin113)=98(π2sin113)+26 units

Question:16 Choose the correct answer.

Area bounded by the curve y=x3 , the x -axis and the ordinates x=2 and x=1 is

(A) 9 (B) 154 (C) 154 (D) 174

Answer:

1646973022742

Hence the required area

=21ydx

=21x3dx=[x44]21

=[x44]20+[x44]01

=[0(2)44]+[140]

=4+14=154

Therefore the correct answer is B.

Question:18 Choose the correct answer.

The area of the circle x2+y2=16 exterior to the parabola y2=6x is

(A) 43(4π3) (B) 43(4π+3) (C) 43(8π3) (D) 43(8π+3)

Answer:

1646973081315

The area of the shaded region is to be found.

Required area =ar(DOC)+ar(DOA)

The region to the left of the y-axis is half of the circle with radius 4 units and centre origin.

Area of the shaded region to the left of y axis is ar(1) = π422=8π units

For the region to the right of y-axis and above x axis

x2+y2=16y=16x2

y2=6xy=6x

The parabola and the circle in the first quadrant intersect at point

(2,23)

Remaining area is 2ar(2) is

ar(2)=02(16x26x)dx=[x216x2+162sin1(x4)263x32]02=[12+162×π64123]=4π3233

Total area of shaded region is

ar(1)+2ar(2)=8π+8π3433=43(8π3) units

Question:19 Choose the correct answer The area bounded by the y -axis, y=cosx and y=sinx when 0xπ2 is

(A) 2(21) (B) 21 (C) 2+1 (D) 2

Answer:

Given : y=cosx and y=sinx

1646973113069


Area of shaded region = area of BCDB + are of ADCA

=121xdy+112xdy

=121cos1y.dy+112sin1xdy

=[y.cos1y1y2]121+[x.sin1x+1x2]112

=cos1(1)12cos1(12)+112+12sin1(12)+1121

=π42+12+π42+121

=221

=21

Hence, the correct answer is B.

If you are looking for exercises solutions for chapter application of integrals class 12 then these are listed below.

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More about NCERT Solutions for Class 12 Maths chapter 8

Generally, one question (5 marks) is asked from this ch 8 maths class 12 in the 12th board final exam and you can score these 5 marks very easily with help of these NCERT solutions for class 12 maths chapter 8 application of integrals. In this chapter 8 class 12 maths, there are 2 exercises with 20 questions. In the NCERT class 12 maths ch 8 question answer, these questions are prepared and explained in a detailed manner using diagrams.

There are a total of 14 solved examples are given in the NCERT textbook to give a better understanding of the concepts related to the application of integrals class 12. Applications of integrals class 12 are indispensable for the Board exam. At the end of the chapter, 19 questions are given in a miscellaneous exercise. In the NCERT solutions for class 12 maths chapter 8 application of integrals article, you will get solutions to miscellaneous exercises too.

Application of integrals class 12 - Topics

8.1 Introduction

8.2 Area under Simple Curves

8.2.1 The area of the region bounded by a curve and a line

8.3 Area between Two Curves

Also read,

Let's understand these topics with help of examples

  • How to find the area under simple curves- To find the area bounded by the curve y= f(x), \:x-axis and the ordinates x = a and x = b . Let assume that area under the curve as composed of large numbers of very thin vertical strips. Consider an arbitrary strip of width dx and height y ,then area of the elementary strip(dA) = ydx , where, y = f(x). This small area called the elementary area.

\\A=\int_{a}^{b}dA\\A=\int_{a}^{b}ydx\\A=\int_{a}^{b}f(x)dx

  • How to find the area of the region bounded by a curve and a line- In this subsection, we will find the area of the region bounded by a line and a circle, a line and an ellipse, a line and a parabola. Equations of the above-said curves will be in their standard forms only.

NCERT solutions for class 12 maths - Chapter wise

Key Features of NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

The application of integrals class 12 ncert solutions offers several key features, including

  1. Expertly crafted by knowledgeable subject specialists.

  2. Designed to enhance students' fundamental understanding of the applications of integrals.

  3. Clear and organized material.

  4. All questions are answered using informative diagrams.

  5. Supports students in completing their assignments and preparing for competitive exams.

NCERT solutions for class 12 subject wise

NCERT Solutions Class Wise

Benefits of NCERT solutions:

  • These NCERT maths chapter 8 class 12 solutions are very easy to understand as they are explained and are prepared in a detailed manner.
  • Scoring good marks in the 12th board exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 8 application of integrals.
  • NCERT solutions for Class 12 Maths Chapter 8 PDF download provides in-depth knowledge of the concepts as well as different ways to solve the problems.
  • NCERT maths chapter 8 class 12 solutions are prepared by experts who know how to answer to score well in the board exam.
  • You will also get some short tricks to check whether the answer is correct or not.

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Frequently Asked Questions (FAQs)

1. What is the weightage of the chapter Application of integrals for CBSE board exam ?

Generally, one question of 5 marks is asked from this chapter in the 12th board final exam but if you want to obtain the full 5 five marks that demands practice, good strategy, and command of concepts, and therefore NCERT textbooks, NCERT syllabus becomes very important. it is advised for students to precisely go through the basics of class 12 maths chapter 8 question answer to score good marks. 

2. How does the NCERT solutions are helpful in the CBSE board exam ?

Only knowing the answer is not enough to score good marks in the exam. One should know how to write the answer to board exam in order to get good marks. NCERT solutions are provided by experts who knows how best to write the answer in the board exam in order to get good marks. Practice class 12 maths chapter 8 question answer to command the concepts.

3. What are the strategies for achieving maximum marks in class 12 math chapter 8 application of integrals ncert solutions?

The NCERT Solutions for the application of integration class 12 at Careers360 are created by a team of experts through extensive research on each concept. The solutions are presented in a thorough and comprehensive manner, enabling students to perform well on class exams and board exams. Additionally, the solutions aid students in completing their assignments efficiently and on time. Interested students can download the application of integrals class 12 pdf.

4. Where can I find the complete solutions of NCERT for class 12 maths ?

Here you will get the detailed and comprehensive NCERT Solutions for class 12 maths by clicking on the link. if  you want to study other material like NCERT Syllabus, NCERT Notes, NCERT Exercise Solutions, etc then you can browse the careers360 official website. 

5. What are the important topics in chapter Application of integrals ?

Some applications of integrals like finding area under simple curves, area of the region bounded by a curve and a line, and area between two curves are the important topics covered in this chapter. practice class 12 maths chapter 8 ncert solutions to score well in the exam.

6. Which is the official website of NCERT ?

NCERT official is the official website of the "https://ncert.nic.in" where you can get NCERT textbooks and syllabus from class 1 to 12 that can be downloaded and studied offline. Careers360 created class 12 application of integrals ncert solutions. practice them to command the concepts.

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Hello there! Thanks for reaching out to us at Careers360.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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