# NCERT Solutions class 12 Maths Exercise 8.1 (Ex 8.1) Chapter 8 Application of Integrals

## NCERT Solutions for Class 12 Maths Exercise 8.1 Chapter 8 Application of Integrals – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 (Ex 8.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. You can download them in PDF format for free.

# NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.1) Exercise 8.1

1. Find the area of the region bounded by the curve  and the lines  and the  axis.

Ans. Equation of the curve (rightward parabola) is

……….(i)

Required area (shaded region)

[From eq. (i)]

=  =

=   =  sq. units

### 2. Find the area of the region bounded by  and the axis in the first quadrant.

Ans. Equation of the curve (rightward parabola) is

……….(i)

Required area (shaded region) bounded by curve  (vertical lines ) and axis in first quadrant.

∣∣∣24ydx∣∣∣=∣∣∣243x−−√dx∣∣∣|∫24ydx|=|∫243xdx|       [From eq. (i)]

=

=

=   =  sq. units

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

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

Required (shaded) area bounded by curve  (Horizontal lines ) and axis in first quadrant.

=  =  sq. units

### 4. Find the area of the region bounded by the ellipse

Ans. Equation of ellipse is  ……….(i)

Here

From eq. (i),

y=±3416x2−−−−−−√y=±3416−x2……….(ii)

For arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Intersections of ellipse (i) with axis

Putting  in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 4) and .

Intersections of ellipse (i) with axis

Putting  in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

[ At end B of arc AB of ellipse;  and at end A of arc AB ; ]

=

=

sq. units

### 5. Find the area of the region bounded by the ellipse

Ans. Equation of ellipse is

Here

From eq. (i),

y=±324x2−−−−−√y=±324−x2……….(ii)

for arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Intersections of ellipse (i) with axis

Putting  in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 2) and .

Intersections of ellipse (i) with axis

Putting  in eq. (i),

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

[ At end B of arc AB of ellipse;  and at end A of arc AB ; ]

=

sq. units

### 6. Find the area of the region in the first quadrant enclosed by axis, line  and the circle

Ans. Step I. To draw the graphs and shade the region whose area we are to find.

Equation of the circle is  ……….(i)

We know that eq. (i) represents a circle whose centre is (0, 0) and radius is 2.

Equation of the given line is

……….(ii)

We know that eq. (ii) being of the form  where   represents a straight line passing through the origin and making angle of  with axis.

Step II. To find values of  and  Putting  from eq. (ii) in eq. (i),

Putting  in , and

Therefore, the two points of intersections of circle (i) and line (ii) are A and D.

Step III. Now shaded area OAM between segment OA of line (ii) and axis

=  =  =  =  sq. units……….(iii)

Step IV. Now shaded area AMB between are AB of circle and axis.

= =

From eq. (ii),

=

=  =   sq. units……….(iv)

Step V. Required shaded area OAB = Area of OAM + Area of AMB

sq. units

### 7. Find the area of the smaller part of the circle  cut off by the line

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

……….(ii)

Here Area of smaller part of the circle  cut off by the line  = Area of ABMC = 2 x Area of ABM

= [From eq. (ii)]

2[0+a22.π2a2×a2a22sin112]2[0+a22.π2−a2×a2−a22sin−112]

2[πa24a2×a2a22.π4]2[πa24−a2×a2−a22.π4]

2[πa24a22πa28]2[πa24−a22−πa28]

2a2[π412π8]2a2[π4−12−π8]

2a2[2π2π8]2a2[2π−2−π8]

a24(π2)=πa24a22a24(π−2)=πa24−a22

=>a22(π21)=>a22(π2−1)sq. units

### 8. The area between  and  is divided into two equal parts by the line  find the value of

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

Now area bounded by parabola (i) and vertical line  is divided into two equal parts by the vertical line

Area OAMB = Area AMBDNC

=

=

=

units

### 9. Find the area of the region bounded by the parabola  and

Ans. The required area is the area included between the parabola  and the modulus function

To find: Area between the parabola  and the ray  for

Here, Limits of integration

Now, for  , table of values,

if   if

 0 1 2 0 1 2

 0 0 1 2

Now, Area between parabola  and axis between limits  and

=  =  ………..(i)

And Area of ray  and axis,

=  =  ………..(ii)

Required shaded area in first quadrant = 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

Similarly, shaded area in second quadrant = 1616sq. units

Therefore, total area of shaded region in the above figure = 161616161313sq. units

### 10. Find the area bounded by the curve x2=4yx2=4y and the line

Ans. Step I. Graphs and region of Integration

Equation of the given curve is

……..(i)

Equation of the given line is

……….(ii)

 0 1 0 0

Step II. Putting  from eq. (i) in eq. (ii),

x(x2)+1(x2)=0x(x−2)+1(x−2)=0

or

For ,from eq. (i),  (2, 1)

For  from eq. (i),

Therefore, the two points of intersection of parabola (i) and line (ii) are C and D (2, 1).

Step III. Area CMOEDN between parabola (i) and axis =  =

=  =  sq. units……..(iii)

Step IV. Area of trapezium CMND between line (ii) and axis =  =

=  =  =

=  sq. units……….(iv)

Required shaded area = Area given by eq. (iv) – Area given by eq. (iii)

sq. units

### 11. Find the area of the region bounded by the curve  and the line

Ans. Equation of the (parabola) curve is
………(i)
y=2x−−√y=2x ………(ii)
Here required shaded area OAMB = 2 x Area OAM

=  =
=  =  sq. units

### 12. Choose the correct answer:

Area lying in the first quadrant and bounded by the circle  and the lines  and  is

(A)

(B)

(C)

(D)

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

……….(ii)

Required area =  =

=  sq. units

Therefore, option (A) is correct.

### 13. Choose the correct answer:

Area of the region bounded by the curve axis and the line  is:

(A) 2

(B)

(C)

(D)

Ans. Equation of the curve (parabola) is

Required area = Area OAM =  = ∣∣∣03y24dy∣∣∣|∫03y24dy|
14∣∣∣(y33)30∣∣∣14|(y33)03| =  sq. units
Therefore, option (B) is correct.