NCERT Solutions for Class 12 Maths Application of Integrals
Hi Students, Welcome to Amans Maths Blogs (AMB). In this post, you will get the NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2.
As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the CBSE NCERT Syllabus. Thus, NCERT Solutions helps the students to solve the exercise questions as given in NCERT Books.
NCERT Solutions for class 12 is highly recommended by the experienced teacher for students who are going to appear in CBSE Class 12 and JEE Mains and Advanced and NEET level exams. Here You will get NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2 of all questions given in NCERT textbooks of class 12 in details with step by step process.
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NCERT Solutions for Class 12 Maths Application of Integrals
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 1.
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.
NCERT Solutions:
Required area is represented by the shaded area OBCDO.
On solving the given equation of circle 4x2 + 4y2 = 9 and the equation of parabola x2 = 4y, we get the point of intersection as
Putting x2 = 4y in 4x2 + 4y2 = 9, then
Putting y = -9/2 in x2 = 4y then.
Here this is NOT possible as square of any number is always positive.
Putting y = 1/2 in x2 = 4y then
Thus, the point of intersection as 
.
Since the required area is symmetrical about y-axis, then
Area OBCDO = 2(Area OBCO). Now, we draw BM perpendicular to OA. Thus, the coordinate of M is (√2, 0).
Area OBCO = Area OMBCO – Area OMBO
Therefore the required area is
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 2.
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1.
NCERT Solutions:
Required area bounded by the given curves (x – 1)2 + y2 = 1 and x2 + y2 = 1 is represented by the shaded area OBCAO as below.
Now, On solving the given equations
(x – 1)2 + y2 = 1 …(1)
and x2 + y2 = 1 …(2),
we get
The point of intersection as 
From the diagram, the required area is symmetrical about x-axis.
Thus, Area OBCAO = 2 x (Area OCAO). We join AB that intersect OC at M and hence AM is perpendicular to OC.
Thus, the coordinates of M is M(1/2, 0).
Therefore, the required area OBCAO is 
units.
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 3.
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3.
NCERT Solutions:
Required area bounded by the given curves y = x2 + 2, y = x, x = 0 and x = 3 is represented by the shaded area OCBAO as below.
Thus,
Area OCBAO = Area ODBAO – Area ODCO.
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 4.
Using integration find the area of region bounded by the triangle whose vertices are (– 1, 0), (1, 3) and (3, 2).
NCERT Solutions:
Required area is given by the shaded triangle ABC whose coordinates are A(-1, 0), B(1, 3) and C(1, 3).
Since the equation of the line AB is
Since the equation of the line BC is
Since the equation of the line AC is
Now,
Area (Triangle ABC)
= Area (Triangle ALB) + Area (Quadrilateral BLMC) – Area (Triangle AMC)
= (3 + 5 – 4)
= 4 units.
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 5.
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.
NCERT Solutions:
Required area is given by the shaded triangle ABC whose the equations of the sides are y = 2x + 1, y = 3x + 1 and x = 4 as below.
On solving the given equations y = 2x + 1, y = 3x + 1 and x = 4, we get the vertices of the triangle ABC as A(0, 1), B(4, 13) and C(4, 9).
From the figure, we get that
Area (Triangle ABC) = Area(OLBAO) – Area (OLCAO)
Choose the correct answer in the following exercises 6 and 7.
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 6.
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
(A) 2(π – 2)
(B) π – 2
(C) 2π – 1
(D) 2(π + 2)
NCERT Solutions:
(B)
Required area bounded by the circle x2 + y2 = 4 and the line x + y = 2 is the shaded area ACBA as below.
Now, from the figure, we get that
Area (ACBA) = Area (OACBO) – Area (OAB)
NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2: Ques No 7.
Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3
(B) 1/3
(C) 1/4
(D) 3/4
NCERT Solutions:
(B)
Required area bounded by the curves y2 = 4x and y = 2x is the shaded area OBAO as below.
Now, on solving the given equations, we get the point of intersection as O(0, 0) and A(1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C is C(1, 0).
Thus, Area (OBAO) = Area (OCA) – Area (OCABO)
