NCERT Solutions for Class 12 Maths Exercise 8.2 Chapter 8 Application of Integrals – FREE PDF Download
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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.2) Exercise 8.2
1. Find the area of the circle 
which is interior to the parabola 
……….(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 ,
x=±2–√x=±2
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
= 
= ∣∣∣∣∫1232(32)2−y2−−−−−−−−√dy∣∣∣∣|∫1232(32)2−y2dy| 
= 
= 
= 
= 
= 
…….(iv)
Step V. Required shaded area = Area AOBDA
= 2 (Area OBD) = 2 (Area OBM + Area MBD)
= 
= 
= 
= 
= 
= 
= 2√6+94sin−11−19−−−−−√26+94sin−11−19 [Since,cos−1x=sin−11−x2−−−−−√][Since,cos−1x=sin−11−x2]
= 2√6+94sin−189−−√26+94sin−189
= 2√6+94sin−122√326+94sin−1223 sq. units
2. Find the area bounded by the curves 
and
……….(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
3. Find the area of the region bounded by the curves and
and
……….(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
4. Using integration, find the area of the region bounded by the triangle whose vertices are (1, 3) and (3, 2).
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
5. Using integration, find the area of the triangular region whose sides have the equations and
and 
………..(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
6. Choose the correct answer:
Smaller area enclosed by the circle 
and the line 
is:
(A) 
(B) 
(C) 
(D) 
…….…(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.
7. Choose the correct answer:
Area lying between the curves 
and 
is:
(A) 
(B) 
(C) 
(D) 
…..(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.