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NCERT solution class 12 chapter 2 Application of Integrals exercise 2.3 mathematics part 2

EXERCISE 2.3


Page No 375:

Question 1:

Find the area under the given curves and given lines:

(i) y = x2x = 1, x = 2 and x-axis

(ii) y = x4x = 1, x = 5 and x –axis

Answer:

  1. The required area is represented by the shaded area ADCBA as

  1. The required area is represented by the shaded area ADCBA as

Question 2:

Find the area between the curves y = x and y = x2

Answer:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x2, is A (1, 1).

We draw AC perpendicular to x-axis.

∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)

Question 3:

Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and = 4

Answer:

The area in the first quadrant bounded by y = 4x2x = 0, y = 1, and = 4 is represented by the shaded area ABCDA as

 

Area of ABCDA = ∫14 x dy                        =∫14 y2 dy    as, y = 4×2                        =12∫14y dy                         =12×23y3/214                         =1343/2 – 13/2                         =138 – 1                         =13×7                                  =73 square units

Question 4:

Sketch the graph of and evaluate

Answer:

The given equation is 

The corresponding values of and y are given in the following table.

x

– 6

– 5

– 4

– 3

– 2

– 1

0

y

3

2

1

0

1

2

3

On plotting these points, we obtain the graph of  as follows.

It is known that, 

Question 5:

Find the area bounded by the curve y = sin between x = 0 and x = 2π

Answer:

The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB

Question 6:

Find the area enclosed between the parabola y2 = 4ax and the line y mx

Answer:

The area enclosed between the parabola, y2 = 4ax, and the line, y mx, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and .

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

Question 7:

Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12

Answer:

The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A (–2, 3) and (4, 12).

We draw AC and BD perpendicular to x-axis.

∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)

Question 8:

Find the area of the smaller region bounded by the ellipse and the line 

Answer:

The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

Question 9:

Find the area of the smaller region bounded by the ellipse  and the line 

Answer:

The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

Question 10:

Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis

Answer:

The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as

The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).

Area of OACO = ∫-12x + 2 dx  –  ∫-12 x2 dx⇒Area of OACO = x22 + 2x-12 – 13×3-12⇒Area of OACO = 222+22 – -122+2-1 – 1323 – -13⇒Area of OACO = 2 + 4 – 12-2 – 138 + 1⇒Area of OACO = 6 + 32 – 3⇒Area of OACO = 3 + 32 = 92 square units

 

Question 11:

Using the method of integration find the area bounded by the curve 

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – = 11]

Answer:

The area bounded by the curve, , is represented by the shaded region ADCB as

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

It can be observed that the given curve is symmetrical about x-axis and y-axis.

∴ Area ADCB = 4 × Area OBAO

Page No 376:

Question 12:

Find the area bounded by curves 

Answer:

The area bounded by the curves, , is represented by the shaded region as

It can be observed that the required area is symmetrical about y-axis.

Question 13:

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

Answer:

The vertices of ΔABC are A (2, 0), B (4, 5), and C (6, 3).

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment CA is

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

Question 14:

Using the method of integration find the area of the region bounded by lines:

2x + y = 4, 3x – 2y = 6 and x – 3+ 5 = 0

Answer:

The given equations of lines are

2x + y = 4 … (1)

3x – 2y = 6 … (2)

And, x – 3+ 5 = 0 … (3)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis.

Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)

Question 15:

Find the area of the region 

Answer:

The area bounded by the curves, , is represented as

The points of intersection of both the curves are.

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about x-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

Therefore, the required area is  units

Question 16:

Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is

A. – 9

B. 

C. 

D. 

Answer:

Required Area =

∫-2 0ydx+∫01ydx

=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. unitsThus, the correct answer is D.

Question 17:

The area bounded by the curvex-axis and the ordinates x = –1 and x = 1 is given by

[Hint: y = x2 if x > 0 and y = –x2 if x < 0]

A. 0

B. 

C. 

D. 

Answer:

Thus, the correct answer is C.

Question 18:

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is

A. 

B. 

C. 

D. 

Answer:

The given equations are

x2 + y2 = 16       … (1)

y2 = 6x               … (2)

Area bounded by the circle and parabola

=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23×3202+2×216-x2+162sin-1×424 =46322-0+20+8sin-11-23+8sin-112=1633+28×π2-23-8×π6=1633+24π-23-4π3=1633+8π-43-8π3=163+24π-43-8π3=16π+1233=434π+3 square units

Area of circle = π (r)2

= π (4)2

= 16π square units

∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units

Thus, the correct answer is C.

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when 

A. 

B. 

C. 

D. 

Answer:

The given equations are

y = cos x … (1)

And, y = sin x … (2)

Required area = Area (ABLA) + area (OBLO)

Integrating by parts, we obtain

Thus, the correct answer is B.


 

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