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# NCERT Solutions class 12 Maths Exercise 8.2

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## 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)

=  =

=

=  =

=

### 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

### 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

### 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

### 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

### 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.

### 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

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 20 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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