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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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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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>> 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 class 12 maths chapter 8 question answer Exercise: 8.1
Question:1 Find the area of the region bounded by the curve
Answer:
Area of the region bounded by the curve
Area =
= 14/3 units
Question:2 Find the area of the region bounded by
Answer:
Area of the region bounded by the curve
Area =
Question:3 Find the area of the region bounded by
Answer:
The area bounded by the curves
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
Answer:
The area of the region bounded by
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
and Area of
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
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
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
Answer:
We can clearly see that given area is symmetrical about y-axis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of
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
Question: 10 Find the area bounded by the curve
Answer:
Points of intersections of
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
Question: 11 Find the area of the region bounded by the curve
Answer:
The combined figure of the curve
The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2
therefore the required area is
Question: 12 Choose the correct answer in the following
Area lying in the first quadrant and bounded by the circle
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
Answer:
The area bounded by the curve
the required area = OAB =
NCERT application of integrals class 12 solutions Exercise: 8.2
Question: 1 Find the area of the circle
Answer:
The area bounded by the circle
By solving the equation we get the intersecting point
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
Answer:
Given curves are
Point of intersection of these two curves are
We can clearly see that the required area is symmetrical about the x-axis
Therefore,
Area of OBCAO = 2
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
Therefore, the answer is
Question: 3 Find the area of the region bounded by the curves
Answer:
The area of the region bounded by the curves,
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
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:
Therefore we have Area of
So, the equation of line segment BC is
Therefore the area of BLMCB will be,
Equation of the line segment AC is,
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
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
Answer:
So, the smaller area enclosed by the circle,
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
Answer:
The area lying between the curve,
The points of intersection of these curves are
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.
NCERT application of integrals class 12 solutions Miscellaneous: Exercise
Question:1 Find the area under the given curves and given lines:
Answer:
The area bounded by the curve
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
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
Answer:
the area between the curves
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)
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
Answer:
the area of the region lying in the first quadrant and bounded by
The required area (ABCD) =
The area of the shaded region is 7/3 units
Question:4 Sketch the graph of
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
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
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
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
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
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
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
Question:11 Using the method of integration find the area bounded by the curve
[ Hint: The required region is bounded by lines
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=1-x
The area bounded by the curve
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+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:
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
Answer:
Hence the required area
Therefore the correct answer is B.
Question:17 Choose the correct answer.
T he area bounded by the curve
Answer:
The required area is
Question:18 Choose the correct answer.
The area of the circle
Answer:
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) =
For the region to the right of y-axis 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
Answer:
Given :
Area of shaded region = area of BCDB + are of ADCA
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.
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.
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
The application of integrals class 12 ncert solutions offers several key features, including
Expertly crafted by knowledgeable subject specialists.
Designed to enhance students' fundamental understanding of the applications of integrals.
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Supports students in completing their assignments and preparing for competitive exams.
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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.
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.
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.
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.
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.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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