www.ncrtsolutions.blogspot.com

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.

Page 1 of 53

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.

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.

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

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

Question 5:

Find the area of the region bounded by the ellipse  Answer

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

∴ Area bounded by ellipse = 4 × Area OAB

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 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

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

∴ Area ABCD = 2 × Area ABC

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.

∴ Area OAD = Area ABCD

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

⇒ Area OED = Area EFCD

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

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

Therefore, required area =                 units

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

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.

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

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.

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

Thus, the correct answer is B.

Exercise 8.2

Question 1:

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

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

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

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             .

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

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)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain

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.

It can be observed that,

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)

B.  π – 2

C.  2π – 1

D. 2 (π + 2)

Answer

It can be observed that,

Area ACBA = Area OACBO – Area (∆OAB)

Thus, the correct answer is B.

Question 7:

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

A.

B.

C.

D.

Answer

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)

Thus, the correct answer is B.

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

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 (∆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

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

The given equation is

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

It is known that,

Question 5:

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

∴ 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 (∆OCA)

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:

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:

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 OABCO as

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

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 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

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

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment CA is

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)

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

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.

C.

D.

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]

A. 0

B.

C.

D.

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)

Area bounded by the circle and parabola

Area of circle = π (r)²

= π (4)²

= 16π 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.

units