NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals: In geometry, you must have learnt some formulas to calculate areas of simple geometrical figures like triangles, rectangles, trapeziums, circles, etc. But how to calculate are areas enclosed by curves? In this article, you will get NCERT solutions for class 12 maths chapter 8 application of integrals in which such calculations are made easier. Important topics that are going to be discussed in this chapter are the area under simple curves, an area between lines and arcs of circles, parabolas, and ellipses. In CBSE NCERT solutions for class 12 maths chapter 8 applications of integrals article, questions from all these topics are covered. In chapter 7 Integrals, you have learnt integration by finding the area of the bounded curve in the definite integral. Integration has a lot of applications in the field of physics, chemistry, medical science, and real life. In solutions of NCERT for class 12 maths application of integrals, you will learn some important applications in mathematics. Check all NCERT solutions from class 6 to 12 at a single place, which will help you to learn CBSE maths and science. In this chapter, there are two exercises & a miscellaneous exercise.
Generally, one question( 5 marks) is asked from this chapter in the 12th board final exam and you can score these 5 marks very easily with h elp of these CBSE NCERT solutions for class 12 maths chapter 8 application of integrals. In this chapter , there are 2 exercises with 20 questions. In the solutions of NCERT for class 12 maths chapter 8 application of integrals, these questions are prepared and explained in a detailed manner using diagrams. Before going to these solutions, first you should try to solve every NCERT problems including examples by yourself. There are total of 14 solved examples are given in the NCERT textbook to give a better understanding of the concepts. At the end of the chapter, 19 questions are given in a miscellaneous exercise. In the CBSE NCERT solutions for class 12 maths chapter 8 application of integrals article, you will get solutions to miscellaneous exercise too.
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 and the ordinates and . 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.
 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 abovesaid curves will be in their standard forms only.
Topics and subtopics of NCERT Grade 12 Maths Chapter8 Application of integrals
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
NCERT solutions for class 12 maths chapter 8 Application of Integrals Solved Exercise Questions
CBSE NCERT solutions for class 12 math chapter 8 applications of integrals Exercise: 8.1
Question:1 Find the area of the region bounded by the curve and the lines and the axis in the first quadrant.
Answer:
Area of the region bounded by the curve and the lines and the axis in the first quadrant
Area =
= 14/3 units
Question:2 Find the area of the region bounded by and the axis in the first quadrant.
Answer:
Area of the region bounded by the curve and the axis in the first quadrant
Area =
units
Question:3 Find the area of the region bounded by and the axis in the first quadrant.
Answer:
The area bounded by the curves and the axis in the first quadrant is ABCD.
Question:4 Find the area of the region bounded by the ellipse
Answer:
The area bounded by the ellipse :
Area will be 4 times the area of EAB.
Therefore,
Therefore the area bounded by the ellipse will be
Question: 5 Find the area of the region bounded by the ellipse
Answer:
The area bounded by the ellipse :
The area will be 4 times the area of EAB.
Therefore,
Therefore the area bounded by the ellipse will be
Question: 6
Find the area of the region in the first quadrant enclosed by
axis, line
and the circle
Answer:
The area of the region bounded by and is ABC shown:
The point B of the intersection of the line and the circle in the first quadrant is .
Area ABC = Area ABM + Area BMC where, M is point in xaxis perpendicular drawn from the line.
Now,area of ............(1)
and Area of
..................................(2)
then adding the area (1) and (2), we have then
The Area under ABC
Question: 7
Find the area of the smaller part of the circle
cut off by the line
Answer:
we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC
Area of ABC =
Area of ABCD = 2 X Area of ABC
Therefore, the area of the smaller part of the circle is
Question:8 The area between and is divided into two equal parts by the line , find the value of .
Answer:
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 =
Area of EFCD =
Area of OED = Area of EFCD
Therefore, the value of a is
Question:9 Find the area of the region bounded by the parabola and .
Answer:
We can clearly see that given area is symmetrical about yaxis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of
is (1 , 1) and (1 , 1)
Now,
Area od OCAO = Area OAM  Area of OCMO
Area of OAM =
Area of OCMO =
Therefore,
Area od OCAO
Now,
Area of the region bounded by the parabola
and
is = 2 X Area od OCAO
Units
Question: 10 Find the area bounded by the curve and the line .
Answer:
Points of intersections of
is
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO Area of OMBO
Area of OMBCO =
Area of OMBO =
Area of OBCO = Area of OMBCO Area of OMBO
Similarly,
Area of OCAO = Area of OCALO  Area of OALO
Area of OCALO =
Area of OALO =
Area of OCAO = Area of OCALO  Area of OALO
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Therefore, area bounded by the curve
and the line
is
Question: 11 Find the area of the region bounded by the curve and the line .
Answer:
The combined figure of the curve
and
The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2
Area of OAB
therefore the required area is
units.
Question: 12 Choose the correct answer in the following
Area lying in the first quadrant and bounded by the circle and the lines and is
Answer:
The correct answer is A
The area bounded by circle C(0,0,4) and the line x=2 is
The required area = area of OAB
Question: 13 Choose the correct answer in the following.
Area of the region bounded by the curve , axis and the line is
Answer:
The area bounded by the curve and y =3
the required area = OAB =
Solutions of NCERT for class 12 maths chapter 8 application of integrals Exercise: 8.2
Question: 1 Find the area of the circle which is interior to the parabola .
Answer:
The area bounded by the circle
and the parabola
.
By solving the equation we get the intersecting point
and
So, the required area (OBCDO)=2 times the area of (OBCO)
Draw a normal on the xaxis (M =
)
Thus the area of OBCO = Area of OMBCO  Area of OMBO
S0, total area =
Question:2 Find the area bounded by curves and .
Answer:
Given curves are
and
Point of intersection of these two curves are
and
We can clearly see that the required area is symmetrical about the xaxis
Therefore,
Area of OBCAO = 2
Area of OCAO
Now, join AB such that it intersects the xaxis at M and AM is perpendicular to OC
Coordinates of M =
Now,
Area OCAO = Area OMAO + Area CMAC
Now,
Area of OBCAO = 2
Area of OCAO
Therefore, the answer is
Question: 3 Find the area of the region bounded by the curves and .
Answer:
The area of the region bounded by the curves,
and is represented by the shaded area OCBAO as
Then, Area OCBAO will be = Area of ODBAO  Area of ODCO
which is equal to
Question: 4 Using integration find the area of region bounded by the triangle whose vertices are and .
Answer:
So, we draw BL and CM perpendicular to xaxis.
Then it can be observed in the following figure that,
We have the graph as follows:
Equation of the line segment AB is:
or
Therefore we have Area of
So, the equation of line segment BC is
or
Therefore the area of BLMCB will be,
Equation of the line segment AC is,
or
Therefore the area of AMCA will be,
Therefore, from equations (1), we get
The area of the triangle
Question:5 Using integration find the area of the triangular region whose sides have the equations and .
Answer:
The equations of sides of the triangle are .
ON solving these equations, we will get the vertices of the triangle as
Thus it can be seen that,
Question:6 Choose the correct answer.
Smaller area enclosed by the circle and the line is
Answer:
So, the smaller area enclosed by the circle, , and the line, , is represented by the shaded area ACBA as
Thus it can be observed that,
Area of ACBA = Area OACBO  Area of
Thus, the correct answer is B.
Question:7 Choose the correct answer.
Area lying between the curves and is
Answer:
The area lying between the curve, and is represented by the shaded area OBAO as
The points of intersection of these curves are and .
So, we draw AC perpendicular to xaxis such that the coordinates of C are (1,0).
Therefore the Area OBAO =
Thus the correct answer is B.
NCERT solutions for class 12 maths chapter 8 application of integrals Miscellaneous: Exercise
Question:1 Find the area under the given curves and given lines:
Answer:
The area bounded by the curve
and
axis
The area of the required region = area of ABCD
Hence the area of shaded region is 7/3 units
Question:1 Find the area under the given curves and given lines:
Answer:
The area bounded by the curev and axis
The area of the required region = area of ABCD
Hence the area of the shaded region is 624.8 units
Question:2 Find the area between the curves and .
Answer:
the area between the curves
and
.
The curves intersect at A(1,1)
Draw a normal to AC to OC(xaxis)
therefore, the required area (OBAO)= area of (OCAO)  area of (OCABO)
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 and .
Answer:
the area of the region lying in the first quadrant and bounded by and .
The required area (ABCD) =
The area of the shaded region is 7/3 units
Question:4 Sketch the graph of and evaluate
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)
Integral to be evaluated is
Question:5 Find the area bounded by the curve between and .
Answer:
The graph of y=sinx is as follows
We need to find the area of the shaded region
ar(OAB)+ar(BCD)
=2ar(OAB)
The bounded area is 4 units.
Question:6 Find the area enclosed between the parabola and the line .
Answer:
We have to find the area of the shaded region OBA
The curves y=mx and y ^{ 2 } =4ax intersect at the following points
The required area is
Question:7 Find the area enclosed by the parabola and the line .
Answer:
We have to find the area of the shaded region COB
The two curves intersect at points (2,3) and (4,12)
Required area is
Question:8 Find the area of the smaller region bounded by the ellipse and the line .
Answer:
We have to find the area of the shaded region
The given ellipse and the given line intersect at following points
Since the shaded region lies above x axis we take y to be positive
The required area is
Question:9 Find the area of the smaller region bounded by the ellipse and the line .
Answer:
The area of the shaded region ACB is to be found
The given ellipse and the line intersect at following points
Y will always be positive since the shaded region lies above x axis
The required area is
Question:10
Find the area of the region enclosed by the parabola
the line
and the
axis.
Answer:
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)
The area of the region enclosed by the parabola the line and the axis is 5/6 units.
Question:11 Using the method of integration find the area bounded by the curve
[ Hint: The required region is bounded by lines and ]
Answer:
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=1x
The area bounded by the curve is 2 units.
Question:12 Find the area bounded by curves .
Answer:
We have to find the area of the shaded region
In the first quadrant
y=x=x
Area of the shaded region=2ar(OADO)
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
Answer:
Equation of line joining A and B is
Equation of line joining B and C is
Equation of line joining A and C is
ar(ABC)=ar(ABL)+ar(LBCM)ar(ACM)
ar(ABC)=8+56=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:
Answer:
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)
Area of the region bounded by the lines is 3.5 units
Question:15 Find the area of the region .
Answer:
We have to find the area of the shaded region OCBAO
Ar(OCBAO)=2ar(OCBO)
For the fist quadrant
In the first quadrant, the curves intersect at a point
Area of the unshaded region in the first quadrant is
The total area of the shaded region is
= Area of half circle  area of the shaded region in the first quadrant
Question:16 Choose the correct answer.
Area bounded by the curve , the axis and the ordinates and is
Answer:
Hence the required area
Therefore the correct answer is B.
Question:17 Choose the correct answer.
T he area bounded by the curve , axis and the ordinates and is given by
Answer:
The required area is
Question:18 Choose the correct answer.
The area of the circle exterior to the parabola is
Answer:
The area of the shaded region is to be found.
Required area =ar(DOC)+ar(DOA)
The region to the left of the yaxis 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) =
For the region to the right of yaxis and above x axis
The parabola and the circle in the first quadrant intersect at point
Remaining area is 2ar(2) is
Total area of shaded region is
Question:19 Choose the correct answer The area bounded by the axis, and when is
Answer:
Given : and
Area of shaded region = area of BCDB + are of ADCA
Hence, the correct answer is B.
NCERT solutions for class 12 maths chapter wise
chapter 1 
Solutions of NCERT for class 12 maths chapter 1 Relations and Functions 
chapter 2 
CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions 
chapter 3 

chapter 4 
Solutions of NCERT for class 12 maths chapter 4 Determinants 
chapter 5 
CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 
chapter 6 
NCERT solutions for class 12 maths chapter 6 Application of Derivatives 
chapter 7 

chapter 8 
CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals 
chapter 9 
NCERT solutions for class 12 maths chapter 9 Differential Equations 
chapter 10 
Solutions of NCERT for class 12 maths chapter 10 Vector Algebra 
chapter 11 
CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry 
chapter 12 
NCERT solutions for class 12 maths chapter 12 Linear Programming 
chapter 13 
Solutions of NCERT for class 12 maths chapter 13 Probability 
NCERT solutions for class 12 subject wise
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