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Miscellaneous Exercise 8

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

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)

Comments

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

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

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

Comments

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

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)

Comments

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

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 = 41 x dy                        =41 y2 dy    [as, y = 4x2]                        =1241y dy                         =12×23[y3/2]41                         =13[(4)3/2  (1)3/2]                         =13(8  1)                         =13×7                                  =73 square units

Comments

Sketch the graph of and evaluate

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, 

Comments

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

The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB

Comments

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

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)

Comments

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

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)

Comments

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

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)

Comments

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

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)

Comments

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]

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

Comments

Find the area bounded by curves 

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.

Comments

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)

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)

Comments

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

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)

Comments

Find the area of the region 

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

Comments

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

A. – 9

B. 

C. 

D.  


Required Area = ?? 02ydx??+10ydx∫-2 0ydx+∫01ydx

                       =?? 02x3dx??+10x3dx=???[x44]02???+[x44]10=??[0164]??+[140]=??4??+14=4+14=174 sq. units=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. units

Thus, the correct answer is D.

Comments

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.

Thus, the correct answer is C.

Comments

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

A. 

B. 

C. 

D. 

The given equations are

x2 + y2 = 16       … (1)

y2 = 6x               … (2)

Area bounded by the circle and parabola

=2[area(OADO)+area(ADBA)]=2[206x‾‾√ dx+4216x2‾‾‾‾‾‾‾√ dx]=2206x‾‾√ dx+24216x2‾‾‾‾‾‾‾√ dx=26‾√20x dx+24216x2‾‾‾‾‾‾‾√ dx=26‾√×23[x32]20+2[x216x2‾‾‾‾‾‾‾√+162sin1(x4)]42 =463(22‾√0)+2[{0+8sin1(1)}{23‾√+8sin1(12)}]=1633+2[8×π223‾√8×π6]=1633+2(4π23‾√4π3)=1633+8π43‾√8π3=163+24π438π3=16π+1233=43[4π+3‾√] square units=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23x3202+2x216-x2+162sin-1x424 =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=16π43(4π+3‾√)=16π16π3433=32π3433=43[8π3‾√] square units∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units

Thus, the correct answer is C.

Comments

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

A.

B.

C.

D.

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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