Class 12 - Maths - Application of Integrals

**Exercise 8.1**

Question 1:

Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis.

Answer:

The area of the region bounded by the curve, y^{2} = x, the lines, x = 1 and x = 4, and the x-axis is

the area ABCD.

So, area of ABCD = ʃ_{1}^{4 }y dx

= ʃ_{1}^{4 }√x dx

= [x^{3/2}/(3/2)]_{1}^{4}

= (2/3)[4^{3/2} - 1^{3/2}]

= (2/3)[8 – 1]

= (2/3) * 7

= 14/3 units

Question 2:

Find the area of the region bounded by the curve y^{2} = 9x and the lines x = 2, x = 4 and the x-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, y^{2} = 9x, the lines, x = 2 and x = 4, and the x-axis is

the area ABCD.

So, area of ABCD = ʃ_{2}^{4 }y dx

= ʃ_{2}^{4 }3√x dx

= 3[x^{3/2}/(3/2)]_{2}^{4}

= 3[4^{3/2} - 2^{3/2}]

= 2[8 – 2√2]

= (16 – 4√2) units

Question 3:

Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, x^{2} = 4y, y = 2, and y = 4, and the y-axis is the area

ABCD.

So, area of ABCD = ʃ_{2}^{4 }x dy

= ʃ_{2}^{4 }2√y dy

= 2[y^{3/2}/(3/2)]_{2}^{4}

= (4/3)[4^{3/2} - 2^{3/2}]

= (4/3)[8 – 2√2]

= (32 – 8√2)/2 units

Question 4:

Find the area of the region bounded by the ellipse x^{2}/16 + y^{2}/9 = 1

Answer:

Given, equation if ellipse is x^{2}/16 + y^{2}/9 = 1

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

Area bounded by ellipse = 4 * Area of OAB

So, area of OAB = ʃ_{0}^{4 }y dx

= ʃ_{0}^{4 }3√(1 – x^{2}/16) dx

= (3/4)ʃ_{0}^{4 }√(16 – x^{2}) dx

= (3/4)[x√(16 – x^{2})/2 + (16/2) * sin^{-1} x/4]_{0}^{4}

= (3/4) [4√(16 – 4^{2})/2 + (16/2) * sin^{-1} 4/4] – (3/4) [0 + (16/2) * sin^{-1} 0/4]

= (3/4)[8 * π/2]

= 3π units

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

Question 5:

Find the area of the region bounded by the ellipse x^{2}/4 + y^{2}/9 = 1

Answer:

Given, equation if ellipse is x^{2}/4 + y^{2}/9 = 1

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

Area bounded by ellipse = 4 * Area of OAB

So, area of OAB = ʃ_{1}^{2 }y dx

= ʃ_{1}^{2 }3√(1 – x^{2}/4) dx

= (3/2)ʃ_{1}^{2 }√(4 – x^{2}) dx

= (3/2)[x√(4 – x^{2})/2 + (4/2) * sin^{-1} x/2]_{1}^{2}

= (3/2) [2√(4 – 2^{2})/2 + (4/2) * sin^{-1} 2/2] – (3/2)[1√(4 – 1^{2})/2 + (4/2) * sin^{-1} 1/2]

= (3/2)[2π/2]

= 3π/2 units

Therefore, area bounded by the ellipse = 4 * 3π/2 = 6π units.

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x^{2} + y^{2} = 4

Answer:

The area of the region bounded by the circle x^{2} + y^{2} = 4, x = √3y, and the x-axis is the

area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1).

Area OAB = Area ∆OCA + Area ACB

Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2

Area ACB = ʃ_{√}_{3}^{2 }y dx

= ʃ_{√}_{3}^{2 }√(4 – x^{2}) dx

= [x√(4 – x^{2})/2 + (4/2) * sin^{-1} x/2]_{√3}^{2}

= [2 * π/2 - (√3/2) * √(4 - 3) - 2sin^{-1} √3/2]

= [π - √3/2 – 2(π/3)]

= [π/3 - √3/2]

Therefore, area enclosed by x-axis, x = √3y and the circle x^{2} + y^{2} = 4 in the first quadrant

= √3/2 + [π/3 - √3/2] = π/3 units

Question 7:

Find the area of the smaller part of the circle x^{2} + y^{2} = a^{2} cut off by the line x = a/√2

Answer:

The area of the smaller part of the circle, x^{2} + y^{2} = a^{2} cut off by the line x = a/√2 is the area

ABCDA.

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

So, Area ABCD = 2 * Area ABC

Now, Area of ABC = ʃ_{a/√}_{2}^{a }y dx

= ʃ_{a/√}_{2}^{a }√(a^{2} – x^{2}) dx

= [x√( a^{2} – x^{2})/2 + (a^{2}/2) * sin^{-1} x/a]_{a/√2}^{a}

= [a^{2}/2 * π/2 - (a/2√2) * √( a^{2} - a^{2}/2) - a^{2}/2 * sin^{-1} 1/√2]

= a^{2}/2 * π/2 - (a/2√2) * (a/√2) - a^{2}/2 * (π/4)

= a^{2}π/4 - a^{2}/4 - a^{2}π/8

= (a^{2}/4)(π - 1 - π/2)

= (a^{2}/4)(π/2 - 1)

So, area of ABCD = 2[(a^{2}/4)(π/2 - 1)] = (a^{2}/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x^{2} + y^{2} = a^{2} cut off by the line x = a/√2 is

(a^{2}/2)(π/2 - 1) units.

Question 8:

The area between x = y^{2} 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.

So, Area OAD = Area ABCD

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

⇒ Area OED = Area EFCD

Now, Area of OED = ʃ_{0}^{a }y dx

= ʃ_{0}^{a }√x dx

= [x^{3/2}/(3/2)]_{0}^{a}

= (2/3)[a^{3/2}] ………………1

Again Area of EFCD = ʃ_{a}^{4 }√x dx

= [x^{3/2}/(3/2)]_{a}^{4}

= (2/3)[4^{3/2 }- a^{3/2}]

= (2/3)[8 - a^{3/2}] ………………2

From equation 1 and 2, we get

(2/3)[a^{3/2}] = (2/3)[8 - a^{3/2}]

=> [a^{3/2}] = [8 - a^{3/2}]

=> 2a^{3/2} = 8

=> a^{3/2} = 4

=> a = 4^{2/3}

Therefore, the value of a is 4^{2/3}

Question 9:

Find the area of the region bounded by the parabola y = x^{2} and y = |x|.

Answer:

The area bounded by the parabola, x^{2} = y and the line y = |x| can be represented as:

The given area is symmetrical about y-axis.

So, Area OACO = Area ODBO

The point of intersection of parabola x^{2} = 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

Now, area of ∆OAB = (1/2) * OB * AB = (1/2) * 1 * 1 = 1/2

and area of OBACO = ʃ_{0}^{1 }y dx

= ʃ_{0}^{1 }x^{2} dx

= [x^{3}/3]_{0}^{1}

= 1/3

Hence, Area of OACO = Area of ∆OAB – Area of OBACO

= 1/2 - 1/3

= 1/6

Therefore, the required are = 2 * 1/6 = 1/3 units.

Question 10:

Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2.

Answer:

The area bounded by the curve, x^{2} = 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 A is (-1, 1/4).

Coordinates of point B is (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

= ʃ_{0}^{2 }(x + 2)/4 dx - ʃ_{0}^{2 }x^{2}/4 dx

= (1/4)[x^{2}/2 + 2x]_{0}^{2} – (1/4)[x^{3}/3]_{0}^{2}

= (1/4)[2 + 4] – (1/4)[8/3]

= 3/2 - 2/3

= 5/6

Similarly, Area OACO = Area OLAC – Area OLAO

= ʃ_{-1}^{0 }(x + 2)/4 dx - ʃ_{-1}^{1 }x^{2}/4 dx

= (1/4)[x^{2}/2 + 2x]_{-1}^{0} – (1/4)[x^{3}/3]_{-1}^{0}

= -(1/4)[(-1)^{2}/2 + 2(-1)] – (-1/4)[(-1)^{3}/3]

= -(1/4)[1/2 - 2] – 1/12

= 1/2 - 1/8 – 1/12

= 7/24

Therefore, required area = 5/6 + 7/24 = 9/8 units.

Question 11:

Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3.

Answer:

The region bounded by the parabola, y^{2} = 4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-axis.

So, Area of OACO = 2 (Area of OAB)

Now, Area of OACO = 2ʃ_{0}^{3 }y dx

= 2ʃ_{0}^{3 }2√x dx

= 4[x^{3/2}/(3/2)]_{0}^{3}

= (8/3)[3^{3/2}]

= 8√3

Therefore, the required area is 8√3 units.

Question 12:

Area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is

A. π B. π/2 C. π/3 D. π/4

Answer:

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

represented as:

So, area of OAB = ʃ_{0}^{2 }y dx

= ʃ_{0}^{2 }√(4 – x^{2}) dx

= [x√(4 – x^{2})/2 + (4/2) * sin^{-1} x/2]_{0}^{2}

= 2(π/2)

= π units

Hence, the correct answer is option A.

Question 13:

Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is

- 2 B. 9/4 C. 9/3 D. 9/2

Answer:

The area bounded by the curve, y^{2} = 4x, y-axis, and y = 3 is represented as:

Now, Area of OAB = ʃ_{0}^{3 }x dy

= ʃ_{0}^{3 }(y^{2}/4) dx

= (1/4)[y^{3}/3]_{0}^{3}

= (1/12)[3^{3}]

= 27/12

= 9/4 units

Thus, the correct answer is B.

**Exercise 8.2**

Question 1:

Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y.

Answer:

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x^{2} + 4y^{2} = 9, and parabola, x^{2} = 4y, we get the point of

intersection as B(√2, 1/2) and D(-√2, 1/2).

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

So, Area OBCDO = 2 * Area OBCO

Now, draw BM perpendicular to OA.

Therefore, the coordinates of M is (√2, 0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

= ʃ_{0}^{√2 }{√(9 – 4x^{2})/4} dx - ʃ_{0}^{√2 }x^{2}/4 dx

= (1/2)ʃ_{0}^{√2 }√(9 – 4x^{2}) dx – (1/4)ʃ_{0}^{√2 }x^{2} dx

= (1/4)[x√(9 – 4x^{2}) + (9/2) * sin^{-1} 2x/3]_{0}^{√2 } - (1/4)[x^{3}/3]_{0}^{√2}

= (1/4)[√2 * √(9 – 8) + (9/2) * sin^{-1} 2√2/3] - (1/12)(√2)^{3}

= (√2/4) + (9/8) * sin^{-1} 2√2/3 - √2/6

= (√2/12) + (9/8) * sin^{-1} 2√2/3

= (1/2)[√2/6 + (9/4) * sin^{-1} 2√2/3]

Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin^{-1} 2√2/3]

= [√2/6 + (9/4) * sin^{-1} 2√2/3] units

Question 2:

Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1

Answer:

The area bounded by the curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1, is represented by the shaded

area as:

On solving the equations (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1, we get the point of intersection as

A(1/2, √3/2) and B(1/2, -√3/2).

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

So, Area of OBCAO = 2 * Area of OCAO

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

The coordinates of M is (1/2, 0)

Now, area of OCAO = Area of OMAO + Area of MCAM

= ʃ_{0}^{1/2 }{√{1 - (x – 1)^{2}} dx - ʃ_{1/2}^{1 }√(1 – x^{2}) dx

= [(x - 1)/2 * √{1 - (x – 1)^{2}} + (1/2) * sin^{-1} (x - 1)]_{0}^{√2}

+ [x/2 * √(1 - x^{2}) + (1/2) * sin^{-1} x]_{1/2}^{1}

= [(-1)/4 * √{1 - (-1/2)^{2}} + (1/2) * sin^{-1} (1/2 - 1) - (1/2) * sin^{-1} (-1)]

+ [(1/2) * sin^{-1} 1 - 1/4 * √{1 – (1/2)^{2}} - (1/2) * sin^{-1} 1/2]

= [-√3/8 + (1/2)(-π/6) - (1/2)(-π/2)] + [(1/2)(π/2) - √3/8 -(1/2)(π/6)]

= [-√3/4 - π/6 + π/2]

= [2π/6 - √3/4]

Therefore, required area OBCAO = 2 * [2π/6 - √3/4] = [2π/3 - √3/2] units.

Question 3:

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

Answer:

The area bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3 is represented by the shaded

area OCBAO as:

Then, Area OCBAO = Area ODBAO – Area ODCO

= ʃ_{0}^{3 }(x^{2} + 2) dx + ʃ_{0}^{3 }x dx

= [x^{3}/3 + 2x]_{0}^{3} + [x^{2}/2]_{0}^{3}

= (9 + 6) – 9/2

= 15 – 9/2

= 21/2 units

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)

Equation of line segment AB is

y - 0 = {(3 - 0)/(1 + 1)}(x + 1)

=> y = (3/2)(x + 1)

So, area of ALBA = ʃ_{-1}^{1 }3(x + 1)/2 dx

= (3/2)[x^{2}/2 + x]_{-1}^{1}

= (3/2)[1/2 + 1 – 1/2 + 1]

= 3 units

Equation of line segment BC is

y - 3 = {(2 - 3)/(3 - 1)}(x - 1)

=> y = (1/2)(-x + 7)

So, area of BLMCB = ʃ_{1}^{3 }(1/2)(-x + 7) dx

= (1/2)[-x^{2}/2 +7 x]_{1}^{3}

= (1/2)[-9/2 + 21 + 1/2 - 7]

= 5 units

Equation of line segment AC is

y - 0 = {(2 - 0)/(3 + 1)}(x + 1)

=> y = (1/2)(x + 1)

So, area of ALBA = ʃ_{-1}^{3 }(1/2)(x + 1) dx

= (1/2)[x^{2}/2 + x]_{-1}^{3}

= (1/2)[9/2 + 3 – 1/2 + 1]

= 4 units

Therefore, from equation (1), we get

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).

It can be observed that,

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

= ʃ_{0}^{4 }(3x + 1) dx - ʃ_{0}^{1 }(2x + 1) dx

= [3x^{2}/2 + x]_{0}^{4} - [2x^{2}/2 + x]_{0}^{4}

= (24 + 4) – (16 + 4)

= 28 – 20

= 8 units

Question 6:

Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is

A. 2 (π – 2) B. π – 2 C. 2π – 1 D. 2 (π + 2)

Answer:

The smaller area enclosed by the circle, x^{2} + y^{2} = 4, and the line, x + y = 2, is represented by the

shaded area ACBA as

It can be observed that,

Area ACBA = Area OACBO – Area (∆OAB)

= ʃ_{0}^{2 }{√(4 – x^{2})} dx - ʃ_{0}^{2 }(2 - x) dx

= [x√(4 – x^{2})/2 + (4/2) * sin^{-1} x/2]_{0}^{2 } - [2x – x^{2}/2]_{0}^{2}

= 2 * π/2 – (4 - 2)

= π – 2 units

Thus, the correct answer is B.

Question 7:

Area lying between the curve y^{2} = 4x and y = 2x is

A. 2/3 B. 1/3 C. 1/4 D. 3/4

Answer:

The area lying between the curve, y^{2} = 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)

= ʃ_{0}^{1 }2x dx - ʃ_{0}^{2 }2√x dx

= 2[x^{2}/2]_{0}^{1 } - [x^{3/2}/(3/2)]_{0}^{1}

= |1 – 4/3|

= |-1/3|

= 1/3 units

Thus, the correct answer is B.

**Miscellaneous Exercise on Chapter 8**

Question 1:

Find the area under the given curves and given lines:

(i) y = x^{2}, x = 1, x = 2 and x-axis (ii) y = x^{4}, x = 1, x = 5 and x –axis

Answer:

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

Area ADCBA = ʃ_{1}^{2 }y dx

= ʃ_{1}^{2 }x^{2} dx

= [x^{3}/3]_{1}^{2 }

= 8/3 – 1/3

= 7/3 units

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

Area ADCBA = ʃ_{1}^{5 }x^{4} dx

= [x^{5}/5]_{1}^{5 }

= 5^{5}/5 – 1/5

= 5^{4} – 1/5

= 625 – 1/5

= 625 – 0.2

= 624.8 units

Question 2:

Find the area between the curves y = x and y = x^{2}

Answer:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x^{2} is A (1, 1).

Now, draw AC perpendicular to x-axis.

So, Area (OBAO) = Area (∆OCA) – Area (OCABO)

= ʃ_{0}^{1 }x dx - ʃ_{0}^{1 }x^{2} dx

= [x^{2}/2]_{0}^{1} – [x^{3}/3]_{0}^{1}

= 1/2 - 1/3

= 1/6 units

Question 3:

Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y= 1 and y = 4.

Answer:

The area in the first quadrant bounded by y = 4x^{2}, x = 0, y = 1, and y = 4 is represented by the

shaded area ABCDA as:

So, area of ABCD = ʃ_{1}^{4 }x dy

= (1/2)ʃ_{1}^{4 }√y dy

= (1/2)[y^{3/2}/(3/2)]_{1}^{4 }

= (1/3)[ 4^{3/2} - 1]

= (1/3)[8 - 1]

= 7/3 units

Question 4:

Sketch the graph of y = |x + 3|and evaluate ʃ_{-6}^{0 }|x + 3| dx

Answer:

The given equation is y = |x + 3|

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 y = |x + 3|as follows:

It is known that (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

Now, ʃ_{-6}^{0 }|x + 3| dx = -ʃ_{-6}^{-3 }(x + 3) dx + ʃ_{-3}^{0 }(x + 3) dx

= -[x^{2}/2 + 3x]_{-6}^{-3} - [x^{2}/2 + 3x]_{-3}^{0}

= -[{9/4 - 6} – {36/2 - 18}] – [0 – 9/2 + 9]

= -[-9/2] – [-9/2]

= 9/2 + 9/2

= 9 units

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

Now, required area = Area OABO + Area BCDB

= ʃ_{0}^{π }sin x dx + |ʃ_{π}^{2π }sin x dx|

= [-cos x]_{0}^{π} + [-cos x]_{π}^{2π}

= [-cos π + cos 0] + |-cos 2π + cos π|

= 1 + 1 + |-1 - 1|

= 2 + |-2|

= 2 + 2

= 4 units

Question 6:

Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx.

Answer:

The area enclosed between the parabola, y^{2} = 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 (4a/m^{2}, 4a/m).

Now, draw AC perpendicular to x-axis.

So, Area OABO = Area OCABO – Area (∆OCA)

= ʃ_{0}^{4a/m2 }2√(ax) dx + ʃ_{0}^{4a/m2 }mx dx

= 2√a[x^{3/2}/(3/2)]_{0}^{4a/m2} – m[x^{2}/2]_{0}^{4a/m2}

= (4√a/3) * (4a/m^{2})^{3/2} – (m/2) * (4a/m^{2})^{2}

= 32a^{2}/3m^{3} - (m/2) * (16a^{2}/m^{4})

= 32a^{2}/3m^{3} - 8a^{2}/m^{3}

= 8a^{2}/3m^{3}

Question 7:

Find the area enclosed by the parabola 4y = 3x^{2} and the line 2y = 3x + 12.

Answer:

The area enclosed between the parabola, 4y = 3x^{2} 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).

Now, draw AC and BD perpendicular to x-axis.

So, Area OBAO = Area CDBA – (Area ODBO + Area OACO)

= ʃ_{-2}^{4 }(3x + 12)/2 dx + ʃ_{-2}^{4 }(3x^{2}/4) dx

= (1/2)[3x^{2}/2 + 12x]_{-2}^{4} + (3/4)[x^{3}/3]_{-2}^{4}

= (1/2)[24 + 48 – 6 + 24] - (1/4)[64 + 8]

= 90/2 – 72/4

= 45 - 18

= 27 units

Question 8:

Find the area of the smaller region bounded by the ellipse x^{2}/9 + y^{2}/4 = 1and the line x/3 + y/2 = 1.

Answer:

The area of the smaller region bounded by the ellipse x^{2}/9 + y^{2}/4 = 1and the line x/3 + y/2 = 1

is represented by the shaded region BCAB.

Now, Area BCAB = Area (OBCAO) – Area (OBAO)

= ʃ_{0}^{3 }2 √(1 - x^{2}/9) dx - ʃ_{0}^{3 }2(1 – x/3) dx

= (2/3)[ ʃ_{0}^{3 }√(9 - x^{2}) dx] - (2/3)ʃ_{0}^{3 }(3 – x) dx

= (2/3)[(x/2)√(9 - x^{2}) + (9/2)sin^{-1} x/3]_{0}^{3} - (2/3)[(3x – x^{2}/2)]_{0}^{3}

= (2/3)[9/2 * π/2] – (2/3)[9 - 9/2]

= (2/3)[9π/4 – 9/2]

= (2/3)[(9/4)(π – 2)]

= 3(π – 2)/2 units

Question 9:

Find the area of the smaller region bounded by the ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1 and the line x/a + y/b = 1.

Answer:

The area of the smaller region bounded by the ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1 and the line

x/a + y/b = 1 is represented by the shaded region BCAB.

Now, Area BCAB = Area (OBCAO) – Area (OBAO)

= ʃ_{0}^{a }b√(1 - x^{2}/a^{2}) dx - ʃ_{0}^{a }b(1 – x/a) dx

= (b/a)[ ʃ_{0}^{a }√(a^{2} - x^{2}) dx] - (b/a)ʃ_{0}^{a }(a – x) dx

= (b/a)[(x/2)√( a^{2} - x^{2}) + (a^{2}/2) * sin^{-1} x/a]_{0}^{a} - (b/a)[(ax – x^{2}/2)]_{0}^{a}

= (b/a)[a^{2}/2 * π/2 – (a^{2} - a^{2}/2)]

= (b/a)[a^{2}π/4 – a^{2}/2]

= (ba^{2}/2a)(π/2 – 1)

= (ab/2)(π/2 – 1)

= (ab/4)(π – 2) units

Question 10:

Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and x-axis.

Answer:

The area of the region enclosed by the parabola x^{2} = y, the line, y = x + 2, and x-axis is

represented by the shaded region OABCO as:

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

Now, Area OABCO = Area (BCA) + Area COAC

= ʃ_{-2}^{-1 }(x + 2) dx + ʃ_{-1}^{0 }x^{2} dx

= [x^{2}/2 + 2x]_{-2}^{-1} + [x^{3}/3]_{-1}^{0}

= [1/2 - 2 – 4/2 + 4] + 1/3

= 5/6 units

Question 11:

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

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

Answer:

The area bounded by the curve |x| + |y| = 1 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.

Now, Area ADCB = 4 * Area OBAO

= 4ʃ_{0}^{1 }(1 - x) dx

= 4[x - x^{2}/2]_{0}^{1}

= 4[1 – 1/2]

= 4 * 1/2

= 2 units

Question 12:

Find the area bounded by curves {(x, y): y ≥ x^{2} and y = |x|}.

Answer:

The area bounded by the curves {(x, y): y ≥ x^{2} and y = |x|} is represented by the shaded region

as shown below:

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

Now, Required area = 2[Area(OCAO) - Area(OCADO)]

= 2[Area(OCAO) - Area(OCADO)]

= 2[ʃ_{0}^{1 }x dx + ʃ_{0}^{1 }x^{2} dx]

= 2[x^{2}/2 – x^{3}/3]_{0}^{1}

= 2[1/2 – 1/3]

= 2 * 1/6

= 1/3 units

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 ∆ABC are A (2, 0), B (4, 5), and C (6, 3).

Equation of line segment AB is

(y - 0) = {(5 - 0)/(4 - 2)}(x - 2)

=> y = 5(x - 2)/2 ………….1

Equation of line segment BC is

(y - 5) = {(3 - 5)/(6 - 4)}(x - 4)

=> (y - 5) = -(x - 4)

=> y – 5 = -x + 4

=> y = -x + 9 ………….2

Equation of line segment CA is

(y - 3) = {(0 - 3)/(2 - 6)}(x - 6)

=> (y - 3) = 3(x - 4)/4

=> y – 3 = 3x/4 - 3

=> y = 3x/4 - 6

=> y = 3(x - 2)/4 ………….3

So, Area (∆ABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

= ʃ_{2}^{4 }5(x - 2)/2 dx + ʃ_{4}^{6 }(-x + 9) dx - ʃ_{2}^{6 }3(x - 2)/4 dx

= (5/2)[x^{2}/2 – 2x)]_{2}^{4} + [-x^{2}/2 + 9x]_{4}^{6 }- (3/4)[x^{2}/2 – 2x)]_{2}^{6}

= (5/2)[8 – 8 – 2 + 4] + [-18 + 54 + 8 - 36]^{ }- (3/4)[18 – 12 – 2 + 4]

= 5 + 8 – 6

= 13 – 6

= 7 units

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

x – 3y + 5 = 0 …….3

The area of the region bounded by the lines is the area of ∆ABC.

AL and CM are the perpendiculars on x-axis.

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

= ʃ_{1}^{4 }(x + 5)/3 dx - ʃ_{1}^{2 }(4 - 2x) dx - ʃ_{2}^{4 }(3x - 6)/2 dx

= (1/3)[x^{2}/2 + 5x)]_{1}^{4} - [4x - x^{2}]_{1}^{2 }- (1/2)[3x^{2}/2 – 6x)]_{2}^{4}

= (1/3)[8 +20 – 1/2 - 5] - [8 - 4 - 4 + 1]^{ }- (1/2)[24 – 24 – 6 + 12]

= (1/3) * (45/2) - 1 – 6/2

= 15/2 – 3

= 7/2 units

Question 15:

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

Answer:

The area bounded by the curves {(x, y): y^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 9} is represented as:

The points of intersection of both the curves are (1/2, √2) and (1/2, -√2)

The required area is given by OABCO.

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

So, Area OABCO = 2 * Area OBC

Now, Area OBCO = Area OMC + Area MBC

= ʃ_{0}^{1/2 }2√x dx + (1/2)ʃ_{1/2}^{3/2 }√(9 - 4x^{2}) dx

= ʃ_{0}^{1/2 }2√x dx + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= 2[x^{3/2}/(3/2)]_{0}^{1/2 }+ [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= (4/3)* (1/2)^{3/2} + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= (4/3)* (1/2√2) + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= (4/3)* (1/2√2) * (√2/√2) + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= (4√2)/(4 * 3) + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= √2/3 + [9π/16 - √2/4 – (9/8) * sin^{-1} 1/3]

= 9π/16 - (9/8) * sin^{-1} 1/3 + √2/12

So, Area OABCO = 2 * Area OBC

= 2 * [9π/16 - (9/8) * sin^{-1} 1/3 + √2/12]

= 9π/8 - (9/4) * sin^{-1} 1/3 + √2/6

= 9π/8 - (9/4) * sin^{-1} 1/3 + (√2/6) * (√2/√2)

= 9π/8 - (9/4) * sin^{-1} 1/3 + 2/6√2

= 9π/8 - (9/4) * sin^{-1} 1/3 + 1/3√2 units

Question 16:

Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = –2 and x = 1 is

A. – 9 B. -15/4 C. 15/4 D. 17/4

Answer:

From the figure,

Required area = ʃ_{-2}^{1 }x^{3} dx

= [x^{4}/4]_{-2}^{1}

= 1/4 – 16/4

= 1/4 - 4

= -15/4 units

Hence, the correct answer is option B.

Question 17:

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

[Hint: y = x^{2} if x > 0 and y = –x^{2} if x < 0]

A. 0 B. 1/3 C. 2/3 D. 4/3

Answer:

From the figure,

Required area = ʃ_{-1}^{1 }x|x| dx

= ʃ_{-1}^{0 }x^{2} dx + ʃ_{0}^{1 }x^{2} dx

= [x^{3}/3]_{-1}^{0} + [x^{3}/3]_{0}^{1}

= -(-1/3) + 1/3

= 1/3 + 1/3

= 2/3 units

Hence, the correct answer is option C.

Question 18:

The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 6x is

A. 4(4π - √3)/3 B. 4(4π + √3)/3 C. 4(8π - √3)/3 D. 4(8π + √3)/3

Answer:

The given equations are

x^{2} + y^{2} = 16 …………….1

y^{2} = 6x ………………..2

Area bounded by the circle and parabola

= 2[Area OADO + Area (ADBA)]

= 2[ʃ_{0}^{2 }√(6x) dx + ʃ_{2}^{4 }{√(16 – x^{2})} dx

= 2[√6{x^{3/2}/(3/2)]_{0}^{2} + 2[x/2 * √(16 – x^{2}) + (16/2) * sin^{-1} x/4]_{2}^{4 }

= 2 * √6 * (2/3) * [x^{3/2}]_{0}^{2} + 2[8 * π/2 - √(16 – 4) - 8 * sin^{-1} 1/2]_{2}^{4}

= 4√6 * (2√2) + 2[4π - √(12) - 8π/6]

= 16√3/3 + 8π - 4√3 - 8π/3

= (4/3)[4√3 + 6π - 3√3 - 2π]

= (4/3)[√3 + 4π]

= (4/3)[4π + √3]

Area of circle = π(r)^{2}

= π(4)^{2}

= 16π units

Now, required area = 16π - (4/3)[4π + √3]

= (4/3)[4 * 3π - 4π - √3]

= (4/3)[8π - √3] units

Thus, the correct answer is C.

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2

A. 2(√2 - 1) B. √2 – 1 C. √2 + 1 D. √2

Answer:

The given equations are y = cos x ........1

and y = sin x .........2

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

= ʃ_{1/√2}^{1 }x dy + ʃ_{1}^{1/√2 }x dy

= ʃ_{1/√2}^{1 }cos^{-1} y dy + ʃ_{1}^{1/√2 }sin^{-1} y dy

Integrating by parts, we get

= [y * cos^{-1} y - √(1 – y^{2})]_{1/√2}^{1} + [y * sin^{-1} y - √(1 – y^{2})]_{1}^{1/√2}

= [cos^{-1} 1 – (1/√2) * cos^{-1} (1/√2) + √(1 – 1/2)] + [(1/√2) * sin^{-1} (1/√2) + √(1 – 1/2) - 1]

= -π/4√2 + 1/√2 + π/4√2 + 1/√2 – 1

= 2/√2 – 1

= (√2 – 1) units

Thus, the correct answer is option B.

.