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.
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 help 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
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
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 x-axis 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 y-axis
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 =
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 x-axis (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 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 =
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 x-axis.
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 x-axis such that the coordinates of C are (1,0).
Therefore the Area OBAO =
Thus the correct answer is B.
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