NCERT Solutions class 12 Maths Exercise 8.2 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
Download NCERT solutions for Applications of Integrals as PDF.

NCERT Solutions class 12 Maths Applications of Integrals
1. Find the area of the circle
which is interior to the parabola
Ans. Step I. Equation of the circle is 

……….(i)
Here centre is (0, 0) and radius is 
Equation of parabola is
…..(ii)
Step II. To find values of
and 
Putting
in eq. (i), 





or 
or 
Putting
in
,

Putting
in
,


Points of intersections of circle (i) and parabola (ii) are A
and B
Step III. Area OBM = Area between parabola (ii) and
axis
=

=

=
= 
=
……….(iii)
Step IV. Now area BDM = Area between circle (i) and
axis
=

=

= 
= 
= 
= 
=
…….(iv)
Step V. Required shaded area = Area AOBDA
= 2 (Area OBD) = 2 (Area OBM + Area MBD)
=
= 
= 
=
= 
=

NCERT Solutions class 12 Maths Exercise 8.2
2. Find the area bounded by the curves
and
Ans. Equations of two circles are

……….(i)
And
……….(ii)
From eq. (i), 
Putting this value in eq. (ii),




Putting
in
,

The two points of intersections of circles (i) and (ii) are
and
.
Now, from eq. (i),
in first quadrant and from eq. (ii),
in first quadrant.
Required area OACBO = 2 x Area OAC = 2 (Area OAD + Area DAC)
= 
= 
= 
= 
=
=
sq. units
NCERT Solutions class 12 Maths Exercise 8.2
3. Find the area of the region bounded by the curves
and
Ans. Equation of the given curve is

……….(i)

Here Vertex of the parabola is (0, 2).
Equation of the given line is
….….(ii)
Table of values for the line 
![]() |
0 | 1 | 2 |
![]() |
0 | 1 | 2 |
We know that it is a straight line passing through the origin and having slope 1 i.e., making an angle of
with
axis.
Here also, Limits of integration area given to be
to 
Area bounded by parabola (i) namely
the
axis and the ordinates
to
is the area OACD and 
=
= (9 + 6) – 0 = 15 ……….(iii)
Again Area bounded by parabola (ii) namely
the
axis and the ordinates
to
is the area OAB and 
=
=
……….(iii)
Required area = Area OBCD = Area OACD – Area OAB
= Area given by eq. (iii) – Area given by eq. (iv)
=
sq. units
NCERT Solutions class 12 Maths Exercise 8.2
4. Using integration, find the area of the region bounded by the triangle whose vertices are
(1, 3) and (3, 2).
Ans. Here, Vertices of triangle are A
B (1, 3) and

C (3, 2).
Equation of the line is



Area of
= Area bounded by line AB and
axis
=

= 
=
= 
= 
=
=
……….(i)
Again equation of line BC is

Area of trapezium BLMC = Area bounded by line BC and
axis
=
= 
= 
= 
=
= 
=
= 5 ……….(ii)
Again equation of line AC is

Area of triangle ACM = Area bounded by line AC and
axis
=
=
= 
= 
= 
= 
=
= 4 ……….(iii)
Required area = Area of
+ Area of Trapezium BLMC – Area of 
= 3 + 5 – 4 = 4 sq. units
NCERT Solutions class 12 Maths Exercise 8.2
5. Using integration, find the area of the triangular region whose sides have the equations
and
Ans. Equations of one side of triangle is

………..(i)
second line of triangle is
………..(ii)
third line of triangle is
……….(iii)
Solving eq. (i) and (ii), we get
and 
Point of intersection of lines (i) and (ii) is A (0, 1)
Putting
in eq. (i), we get 
Point of intersection of lines (i) and (iii) is B (4, 9)
Putting
in eq. (i), we get 
Point of intersection of lines (ii) and (iii) is C (4, 13)
Area between line (ii) i.e., AC and
axis
=
=
= 
= 24 + 4 = 28 sq. units ……….(iv)
Again Area between line (i) i.e., AB and
axis
=
=
= 
= 16 + 4 = 20 sq. units ……….(v)
Therefore, Required area of 
= Area given by (iv) – Area given by (v)
= 28 – 20 = 8 sq. units
NCERT Solutions class 12 Maths Exercise 8.2
6. Choose the correct answer:
Smaller area enclosed by the circle
and the line
is:
(A)
(B)
(C)
(D)
Ans. Step I. Equation of circle is
…….…(i)

………(ii)
Also, equation of the line is
……….(iii)
Table of values
![]() |
0 | 2 |
![]() |
2 | 0 |
Therefore graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).
Step II. From the graph of circle (i) and straight line (iii), it is clear that points of intersections of circle (i) and straight line (iii) are A (2, 0) and B (0, 2).
Step III. Area OACB, bounded by circle (i) and coordinate axes in first quadrant =
= 
= 
= 
=
sq. units ……….(iv)
Step IV. Area of triangle OAB, bounded by straight line (iii) and coordinate axes
=
= 
= 
=
= 2 sq. units ……….(v)
Step V. Required shaded area = Area OACB given by (iv) – Area of triangle OAB by (v)
=
sq. units
Therefore, option (B) is correct.
NCERT Solutions class 12 Maths Exercise 8.2
7. Choose the correct answer:
Area lying between the curves
and
is:
(A)
(B)
(C)
(D)
Ans. Equation of curve (parabola) is
…..(i)

=
…..(ii)
Equation of another curve (line) is
…..(iii)
Solving eq. (i) and (iii), we get
or
and
or 
Therefore, Points of intersections of circle (i) and line (ii) are O (0, 0) and A (1, 2).
Now Area OBAM = Area bounded by parabola (i) and
axis =
=
= 
=
………(iv)
Also, Area
OAM = Area bounded by parabola (iii) and
axis
=
=
= 
=
………(v)
Now Required shaded area OBA = Area OBAM – Area of
OAM
=
sq. units
Therefore, option (B) is correct.
NCERT Solutions class 12 Maths Exercise 8.2
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