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## NCERT Solutions class 12 Maths Applications of Integrals

1. Find the area under the given curves and given lines:

(i) and axis.

(ii) and axis.

Ans. (i)Equation of the curve (parabola) is

……..(i)

Required area bounded by curve (i), vertical line  and axis

=

=

=

=  sq. units

(ii)Equation of the curve

…..(i)

It is clear that curve (i) passes through the origin because  from (i)

Table of values for curve for  and  (given)

 1 2 3 4 5 1

Required shaded area between the curve , vertical lines  and axis

=  =

=  =

=

= 624.8 sq. units

### 2. Find the area between the curves  and

Ans. Equation of one curve (straight line) is  …..(i)

Equation of second curve (parabola) is  …..(ii)

Solving eq. (i) and (ii), we get  or  and  or

Points of intersection of line (i) and parabola (ii) are O (0, 0) and A (1, 1).

Now Area of triangle OAM

= Area bounded by line (i) and axis

=  =  =

=  sq. units

Also Area OBAM = Area bounded by parabola (ii) and axis

=  =  =

=  sq. units

Required area OBA between line (i) and parabola (ii)

= Area of triangle OAM – Area of OBAM

=  =  sq. units

### 3. Find the area of the region lying in the first quadrant and bounded by  and

Ans. Equation of the curve (parabola) is

……….(i)

……….(ii)

Here required shaded area of the region lying in first quadrant bounded by parabola (i),

and the horizontal lines  and  is

=  =

=

=

=

=  =  sq. units

### 4. Sketch the graph of  and evaluate

Ans. Equation of the given curve is  ……….(i)

0 for all real

Graph of curve is only above the axis i.e., in first and second quadrant only.

=

If  0

….(ii)

And

=

If  0

…….(iii)

Table of values for  for

 0 0 1 2 3

Table of values for  for

 0 1 2 3

Now,

=  +

=  +

=

=

=

=  sq. units

### 5. Find the area bounded by the curve  between  and

Ans. Equation of the curve is  ……….(i)

for  i.e., graph is in first and second quadrant.

And  for  i.e., graph is in third and fourth quadrant.

If tangent is parallel to axis, then

Table of values for curve  between  and

 0 0 1 0 0

Now Required shaded area = Area OAB + Area BCD

=

=

=

=

=

= 2 + 2 = 4 sq. units

### 6. Find the area enclosed by the parabola  and the line

Ans. Equation of parabola is  ……..(i)

The area enclosed between the parabola  and line  is represented by shaded area OADO.

Here Points of intersection of curve (i) and line  are O (0, 0) and A.

Now Area ODAM = Area of parabola and axis

=

=

=

=

=  ……….(ii)

Again Area of  = Area between line  and axis

=  =

=

=  =  ……….(ii)

Requires shaded area = Area ODAM – Area of

=

=

=

### 7. Find the area enclosed by the parabola  and the line

Ans. Equation of the parabola is

……….(i)

Equation of the line is  ……(ii)

In the graph, points of intersection are B (4, 12) and C.

Now, Area ABCD =

=

=

= 45 sq. units

Again, Area CDO + Area OAB =

=

=  = 18 sq. units

Required area = Area ABCD – (Area CDO + Area OAB)

= 45 – 18 = 27 sq. units

### 8. Find the area of the smaller region bounded by the ellipse  and the line

Ans. Equation of the ellipse is

………(i)

Here intersection of ellipse (i) with axis are

A (3, 0) and A’ and intersection of ellipse (i) with  axis are B (0, 2) and B’.

Also, the points of intersections of ellipse (i) and line  are A (3, 0) and B (0, 2).

Area OADB = Area between ellipse (i) (arc AB of it) and axis

=

=

=

=

=

=  =  sq. units……….(ii)

Again Area of triangle OAB = Area bounded by line AB and axis

=

=

=

=  sq. units ….(iii)

=

=  sq. units

### 9. Find the area of the smaller region bounded by the ellipse  and the line

Ans. Equation of ellipse is  ……..(i)

Area between arc AB of the ellipse and axis

=

=

=

=

=  ………..(ii)

Also Area between chord AB and axis

=

=

=

=

=

Now Required area

= Area between arc AB of the ellipse and axis – Area between chord AB and axis

=  sq. units

10.Find the area of the region enclosed by the parabola  the line  and  axis.

Ans. Equation of parabola is  …………(i)

Equation of line is  …………(ii)

Here the two points of intersections of parabola (i) and line (ii) are A and B (2, 4).

Area ALODBM = Area bounded by parabola (i) and axis

=  =

=

=  =  = 3 sq. units

AlsoArea of trapezium ALMB = Area bounded by line (ii) and axis

=  =

=

=

=  sq. units

Now Required area = Area of trapezium ALMB – Area ALODBM

=  sq. units

### 11.Using the method of integration, find the area enclosed by the curve

Ans. Equation of the curve (graph) is

……….(i)

The area bounded by the curve (i) is represented by the shaded region ABCD.

The curve intersects the axes at points A (1, 0), B (0, 1), C and D

It is observed clearly that given curve is symmetrical about axis and axis.

Area bounded by the curve

= Area of square ABCD

= 4 x OAB

=

=

=

=  = 2 sq. units

### 12.Find the area bounded by the curves .

Ans. The area bounded by the curves  is represented by the shaded region.

It is clearly observed that the required area is symmetrical about axis.

Required area

= Area between parabola  and axis between limits  and

=  =

=  ………..(i)

And Area of ray  and axis,

=  =  =  ………..(ii)

= Area between ray  for  and axis – Area between parabola  and axis in first quadrant

= Area given by eq. (ii) – Area given by eq. (i) =  sq. units

### 13.Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).

Ans. Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).

Equation of side AB is

=

Equation of side BC is

=

Equation of side AC is

=

Now, Required shaded area = Area  + Area of trapezium BLMC – Area

=

=

=

=

=

= 5 + 8 – 6 = 7 sq. units

### 14.Using the method of integration, find the area of the region bounded by the lines:  and

Ans. Equation of one line  is ,

Equation of second line  is

And Equation of third line  is

Here, vertices of triangle ABC are A (2, 0), B (4, 3) and C (1, 2).

Now, Required area of triangle

= Area of trapezium CLMB – Area   Area

=

=

=

=

=

=  sq. units

### 15.Find the area of the region .

Ans. Equation of parabola is  ……(i)

And equation of circle is  ……(ii)

Here, the two points of intersection of parabola (i) and circle (ii) are A and B

Required shaded area OADBO (Area of the circle which is interior to the parabola)

= 2 x Area OADO = 2 [Area OAC + Area CAD]

=

=

=

=

=  sq. units

Area bounded by the curve  the axis and the ordinate  and  is:

(A)

(B)

(C)

(D)

Ans. Equation of the curve is       Y

To find: Area OBN ( for ) and Area OAM ( for )

Required area = Area OBN + Area OAM

=

=

=

=  =  sq. units

Therefore, option (D) is correct,

The area bounded by the curve axis and the ordinates  and  is given by:

(A) 0

(B)

(C)

(D)

Ans. Equation of the curve is

if  ……….(i)

And  if  ………..(ii)

Required area = Area ONBO + Area OAMO

=

=

=  sq. units

Therefore, option (C) is correct.

The area of the circle  exterior to the parabola

(A)

(B)

(C)

(D)

Ans. Equation of the circle is  …..(i)

This circle is symmetrical about axis and  axis.

Here two points of intersection are B and B’

Required area = Area of circle – Area of circle interior to the parabola

=  Area OBAB’O

=  2 x Area OBACO

= 2[Area OBCO + Area BACB]

=

=

=

=

=

=

=

=

=

=

=  sq. units

The area bounded by the axis,  and  when  is:

(A)

(B)

(C)

(D)

Ans. Here both graphs intersect at the point B.

Required shaded area = Area OABC – Area OBC

= Area OABC – (Area OBM + Area BCM)

=

=

=

=

=

=  =  sq. units

Therefore, option (B) is correct.

## NCERT Solutions class 12 Maths Miscellaneous

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