Class XII Chapter 8 – Application of Integrals Maths...

Class XII Chapter 8 – Application of Integrals Maths

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Exercise 8.1

Question 1:

Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and

the x-axis.

Answer

The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the

x-axis is the area ABCD.


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Question 2:

Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first

quadrant.

Answer

The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis

is the area ABCD.


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Question 3:

Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first

quadrant.

Answer

The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis

is the area ABCD.


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Question 4:

Find the area of the region bounded by the ellipse

Answer

The given equation of the ellipse, , can be represented as

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

∴ Area bounded by ellipse = 4 × Area of OAB


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Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Question 5:

Find the area of the region bounded by the ellipse

Answer

The given equation of the ellipse can be represented as

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

∴ Area bounded by ellipse = 4 × Area OAB


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Therefore, area bounded by the ellipse =

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line and the

circle

Answer

The area of the region bounded by the circle, , and the x-axis is the

area OAB.


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The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ∆OCA + Area ACB

Area of OAC

Area of ABC

Therefore, area enclosed by x-axis, the line , and the circle in the first

quadrant =

Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line

Answer


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The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the

area ABCDA.

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

∴ Area ABCD = 2 × Area ABC


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Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, ,

is units.

Question 8:

The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find

the value of a.

Answer

The line, x = a, divides the area bounded by the parabola and x = 4 into two equal

parts.


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∴ Area OAD = Area ABCD

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

⇒ Area OED = Area EFCD


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From (1) and (2), we obtain

Therefore, the value of a is .

Question 9:

Find the area of the region bounded by the parabola y = x2 and

Answer

The area bounded by the parabola, x2 = y,and the line, , can be represented as


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The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).

Area of OACO = Area ∆OAB – Area OBACO

⇒ Area of OACO = Area of ∆OAB – Area of OBACO

Therefore, required area = units


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Question 10:

Find the area bounded by the curve x2 = 4y and the line x = 4y – 2

Answer

The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the

shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO


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Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =

Question 11:

Find the area of the region bounded by the curve y2 = 4x and the line x = 3

Answer

The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.


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The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

Question 12:

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0

and x = 2 is


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A. π

B.

C.

D.

Answer

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is

represented as

Thus, the correct answer is A.


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Question 13:

Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is

A. 2

B.

C.

D.

Answer

The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as

Thus, the correct answer is B.


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Exercise 8.2

Question 1:

Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y

Answer

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the

point of intersection as .

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

∴ Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are .

Therefore, Area OBCO = Area OMBCO – Area OMBO


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Therefore, the required area OBCDO is

units

Question 2:

Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1

Answer

The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by

the shaded area as


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On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of

intersection as A and B .

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

∴ Area OBCAO = 2 × Area OCAO

We join AB, which intersects OC at M, such that AM is perpendicular to OC.

The coordinates of M are .


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Therefore, required area OBCAO = units

Question 3:

Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3

Answer

The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented by

the shaded area OCBAO as


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Then, Area OCBAO = Area ODBAO – Area ODCO

Question 4:

Using integration finds the area of the region bounded by the triangle whose vertices are

(–1, 0), (1, 3) and (3, 2).

Answer

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

Area (∆ACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)


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Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain


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Area (∆ABC) = (3 + 5 – 4) = 4 units

Question 5:

Using integration find the area of the triangular region whose sides have the equations y

= 2x +1, y = 3x + 1 and x = 4.

Answer

The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.

On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C

(4, 9).


Area (∆ACB) = Area (OLBAO) –Area (OLCAO)

Question 6:

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is

A. 2 (π – 2)


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B. π – 2

C. 2π – 1

D. 2 (π + 2)

Answer

The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is

represented by the shaded area ACBA as


Area ACBA = Area OACBO – Area (∆OAB)


Question 7:

Area lying between the curve y2 = 4x and y = 2x is

A.


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B.

C.

D.

Answer

The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded

area OBAO as

The points of intersection of these curves are O (0, 0) and A (1, 2).

We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).

∴ Area OBAO = Area (∆OCA) – Area (OCABO)


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Miscellaneous Solutions

Question 1:

Find the area under the given curves and given lines:

(i) y = x2, x = 1, x = 2 and x-axis

(ii) y = x4, x = 1, x = 5 and x –axis

Answer

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

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


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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


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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 (∆OCA) – Area (OCABO) … (1)

Question 3:

Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y

= 1 and y = 4


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Answer

The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is

represented by the shaded area ABCDA as

Question 4:

Sketch the graph of and evaluate

Answer


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The given equation is

The corresponding values of x 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,


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Question 5:

Find the area bounded by the curve y = sin x 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


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

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (∆OCA)


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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)


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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


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


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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)


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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 OABCO as


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

∴ Area OABCO = Area (BCA) + Area COAC


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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

– y = 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


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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)


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Answer

The vertices of ∆ABC 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


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Area (∆ABC) = 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 – 3y + 5 = 0

Answer

The given equations of lines are

2x + y = 4 … (1)

3x – 2y = 6 … (2)

And, x – 3y + 5 = 0 … (3)


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The area of the region bounded by the lines is the area of ∆ABC. AL and CM are the

perpendiculars on x-axis.

Area (∆ABC) = Area (ALMCA) – Area (ALB) – Area (CMB)

Question 15:

Find the area of the region

Answer

The area bounded by the curves, , is represented as


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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

Question 16:

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

A. – 9

B.


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C.

D.

Answer


Question 17:

The area bounded by the curve , x-axis and the ordinates x = –1 and x = 1 is

given by

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


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A. 0

B.

C.

D.

Answer

Thus, the correct answer is C.


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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


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Area of circle = π (r)2

= π (4)2

= 16π units

Thus, the correct answer is C.


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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


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Therefore, the required area is units

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APPLICATION OF INTEGRATION

INTRODUCTION

Area under Simple Curves

(i)

Area bounded by the curve y = f(x), the x-axis and between the ordinates at x = a and x = b

is given by

Area =∫ 𝒚 𝒅𝒙 𝒃

𝒂 = ∫ 𝒇(𝒙) 𝒅𝒙

𝒃

𝒂

(ii)

Area bounded by the curve y = f(x), the y axis and between abscissas at y = c and y =

d is given by

Area = ∫ 𝒙 𝒅𝒙𝒅

𝒄 = ∫ 𝒈(𝒚)𝒅𝒚

𝒅

𝒄

Where y = 𝒇(𝒙) => 𝑥 = 𝑔(𝑦)

Note: If area lies below x-axis or to left side of y-axis, then it is negative and in such a

case we like its absolute value. (numerical value)

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4.

Area bounded by two curves 𝒚 = 𝒇(𝒙) and 𝒚 = 𝒈(𝒙), such that 𝟎 ≤ 𝒈(𝒙) ≤ 𝒇(𝒙) for

all 𝒙 ∈ [𝒂, 𝒃] and between the ordinates at 𝒙 = 𝒂, 𝒙 = 𝒃 is given by

Area = ∫ {𝒇(𝒙) − 𝒈(𝒙)} 𝒅𝒙𝒃

𝒂

Finding the area enclosed between a curve, X- axis and two ordinates or a curve , Y- axis

and two abscissa

WORKING RULE

1. Draw the rough sketch of the given curve

2. Find whether the required area is included between two ordinate or two abscissa

3. (a) If the required area is included between two ordinates x = a and x= b then use

the formula ∫ 𝒚 𝒅𝒙 𝒃

𝒂

(b) If the required area is included between two abscissa y = c and = d then use the

Formula ∫ 𝒙 𝒅𝒚𝒅

𝒄

Finding the area included between two curves

WORKING RULE

1. Draw the graph of the given curves.

2. Mark the region whose area is to be determined.

3. Find whether the area is bounded between two given curves and two ordinates or

between the two given curves and two abscissa.

(a) If the required area is bounded between 2 ordinates , use the formula

∫ [𝒇 (𝒙) − 𝒈 (𝒙)𝒃

𝒂

]𝒅𝒙

(b) If the required area is included between two abscissa use the formula

∫ [𝒇(𝒚) − 𝒈(𝒚)] 𝒅𝒚𝒃

𝒂

SOME IMPORTANT POINTS TO BE KEPT IN MIND FOR SKETCHING THE GRAPH

1. 𝒚𝟐 = 4ax is a parabola with vertex at origin,

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symmetric to X axis and right of origin

2. 𝒚𝟐 = - 4ax is a parabola with vertex at origin,

symmetric to X axis and left of origin

3. 𝒙𝟐 = 4ay is a parabola with vertex at origin,

symmetric to y axis and above origin

4. 𝒙𝟐 = - 4ay is a parabola with vertex at origin,

symmetric to y axis and below origin

5. 𝒙𝟐

𝒂𝟐 + 𝒚𝟐

𝒃𝟐 = 1 is an ellipse symmetric to both axis,

Cut x axis at (±a,0) and y axis at (0,±𝒃)

6. 𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐 is a circle symmetric to both the axes

with centre at origin and radius r

7. (𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 = 𝒓𝟐 is a circle with centre at (h,k) and radius r.

8. ax +by + c = 0 representing a straight line

9. Graph of 𝒚 = |𝒙|

10. Graph of 𝒚 = |𝒙 − 𝟑|

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IMPORTANT SOLVED PROBLEMS

1. Calculate the area of the region bounded by the parabola y =x2 and x = y2

Solution

Parabola y 2 = x is symmetrical to X- axis and x2 = y is symmetrical to y axis

Solving both the equations we get the point of intersection of the two curves as (0,0) and

(1,1)

Required area = ∫ (√𝒙 𝟏

𝟎− 𝒙𝟐) 𝒅𝒙

= [𝒙

𝟑𝟐

𝟑

𝟐

− 𝒙𝟑

𝟑]𝟏𝟎

= 𝟐

𝟑 −

𝟏

𝟑 =

𝟏

𝟑 sq. units

2. Find the area of the region ; { (x, y ) :𝒚𝟐 ≤ 𝟒𝒙 , 𝟒𝒙𝟐 + 𝟒𝒚𝟐 ≤ 𝟗}

Solution:

Curves y2 =4x, parabola symmetric to x axis and the circle 𝒙𝟐 + 𝒚𝟐 = 𝟗

𝟒 circle with centre

at (0,0 ) and radius 𝟑

𝟐

Sketch both the curves and shaded the area

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The point of intersection of y2 =4x and 𝒙𝟐 + 𝒚𝟐 = 𝟗

𝟒 are the points (0, 0 ) (

𝟏

𝟐 , √𝟐)

Required area = 2 × Area of OBALO

= 2×[ area of OBLO + area of BLAB]

= 2 (∫ 𝟐√𝒙𝟏

𝟐𝟎

+ ∫ √𝟗

𝟒− 𝒙𝟐

𝟑

𝟐𝟏

𝟐

) dx

= 2 {(𝟐𝒙

𝟑𝟐

𝟑

𝟐

)𝟏

𝟐

𝟎 + [

𝒙

𝟐√

𝟗

𝟒− 𝒙𝟐 +

𝟗

𝟖𝐬𝐢𝐧−𝟏 𝟐𝒙

𝟑]

𝟑

𝟐𝟏

𝟐

}

= 𝟗𝝅

𝟖 −

𝟗

𝟒𝐬𝐢𝐧−𝟏 (

𝟏

𝟑) +

√𝟐

𝟔 sq. units

3.Find the area bounded between the lines y = 2x + 1 , y = 3x + 1 , x= 4 using integration

Solution:

Draw the rough sketch and shaded the area

Area enclosed = ∫ (𝒇 (𝒙) − 𝒈(𝒙))𝒅𝒙𝟒

𝟎

= ∫ (𝟑𝒙 + 𝟏 − 𝟐𝒙 − 𝟏𝟒

𝟎) dx

= ∫ 𝒙 𝒅𝒙𝟒

𝟎

= [𝒙𝟐

𝟐]𝟒𝟎

= 8 sq.units

4. Find the area of the region { (x,y): 𝟎 ≤ 𝒚 ≤ 𝒙𝟐 + 𝟏, 𝟎 ≤ 𝒚 ≤ 𝒙 + 𝟏, 𝟎 ≤ 𝒙 ≤ 𝟐}

Solution

Sketch the region whose area is to be found out.

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The point of intersection of y = x2 +1 and y = x+1 are he points (0,1 ) and (1,2 )

The required area = area of the region OPQRSTO

= area of the region OTQPO + area of the region TSRQT

= ∫ (𝒙𝟐 + 𝟏 )𝒅𝒙 + ∫ (𝒙 + 𝟏 )𝟐

𝟏

𝟏

𝟎𝒅𝒙

= [𝒙𝟑

𝟑 + 𝒙]

𝟏𝟎

+ [𝒙𝟐

𝟐+ 𝒙]

𝟐𝟏

= 𝟐𝟑

𝟔 sq.units

5. Find the area cut off from the parabola 4y = 3 x2 by the line 2y = 3x + 12

Solution

Given 4y = 3x2 and 3x – 2y +12 = 0

Solve both the equation we get the point of

intersecction of both the curves

(-2,3) and (4,12)

Required area = area of AOBA

= ∫ [𝟑𝒙+𝟏𝟐

𝟐

𝟒

−𝟐−

𝟑𝒙𝟐

𝟒 ] dx

= 27 sq.unts

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PRACTICE PROBLEMS

LEVEL I

1. Find the areaof the region bounded by the parabola y2 = 4ax , its axis and two ordinates x= 4

and x = 9

2. Find the area bounded by the parabola x2 = y , y axis and the line y =1

3. Find the area bounded by the curve y = 4x – x2 , x axis and the ordinates x = 1 and x = 3

4. Find the area of the region bounded by the curve y2 = 4x and the line x = 3

LEVEL II

1. Find the area of the region { (x,y) : x2 + y2≤ 4, 𝑥 + 𝑦 ≥ 2}

2. Find the of the circle x2 + y2 = a2

3. Find the area of the region { (x , y) : 𝑦2 ≤ 6𝑥 , 𝑥2 + 𝑦2 ≤ 16}

4. Sketch the region common to the circle x2 + y2 = 4 and the parabola y2 = 4 x. Also find the

area of the region by integration.

5. Find the area of the ellipse 𝑥2

𝑎2 + 𝑦2

𝑏2 = 1

6. Find the area of the smaller region bounded by the ellipse 𝑥2

9 +

𝑦2

4 = 1 and the straight line

𝑥

3+

𝑦

2= 1

7. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.

8. Find the area bounded by the lines x +2y = 2, y – x = 1 and 2x + y = 7

9. Find the area of the region bounded by the parabola 𝑦 =3

4𝑥2 and the line 3x – 2y +12 =0

10. Using integration, find the area of the triangle ABC with vertices A (-1, 0 ), B (1 ,3 ) and C

(3,2)

LEVEL III

1. Using integration find the area of the following region {(𝑥, 𝑦 ): |𝑥 + 2 | ≤ 𝑦 ≤

√20 − 𝑥2}

2. Sketch the region enclosed between the circles x2 + y2 = 1 and x2 + (y-1)2 = 1. Also find

the area of the region using integration

3. Find the area of the region lying above the x axis and included between the circle

x2 + y2 = 8x and the parabola y2 = 4x

4. Using the method of integration , find the area bounded by the curve |x| + |y| = 1

Class XII Chapter 8 – Application of Integrals Maths...

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