Class XII Chapter 8 Application of Integrals Maths

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.

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Question 2:

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

Answer

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

Question 3:

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

Answer

The area of the region bounded
by the curve, *x*^{2} = 4*y*, *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 Χ Area of OAB

Therefore, area bounded by the ellipse = 4 Χ 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 Χ 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 area of the region bounded by the circle, , and the

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
*x*^{2} + *y*^{2} = *a*^{2} cut off by the line Answer

The area of the smaller part of the circle,

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

∴ Area ABCD = 2 Χ Area ABC

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

∴ 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 *= *x*^{2} and Answer

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

The given area is symmetrical about
*y*-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola,

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

Therefore, required area = units

Question 10:

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

The area bounded by the curve,
*x*^{2} = 4*y*, and line, *x *= 4*y * 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 *y*^{2} = 4*x *and
the line *x *= 3 Answer

The region bounded by the parabola,
*y*^{2} = 4*x*, 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 *x*^{2} + *y*^{2} = 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 *y*^{2} = 4*x*, *y*-axis and the line *y *= 3 is

A. 2

B.

C.

D.

Answer

The area bounded by the curve,

Thus, the correct answer is B.

Exercise 8.2

Question 1:

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

The required area is represented by the shaded area OBCDO.

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

point of intersection as

.

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

∴ Area OBCDO = 2 Χ 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} + *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 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 Χ 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 *= *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

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*

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

The equations of sides of the triangle
are *y *= 2*x *+1,
*y *= 3*x
*+ 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)

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,

It can be observed that,

Area ACBA = Area OACBO Area (∆OAB)

Thus, the correct answer is B.

Question 7:

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

A.

B.

C.

D.

Answer

The area lying between the curve,

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 *= *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

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

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). 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 *= 4*x*^{2}, *x *= 0, *y*

= 1 and *y *= 4

Answer

The area in the first quadrant
bounded by *y *= 4*x*^{2}, *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.

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 *x *between *x *= 0 and *x *= 2π
Answer

The graph of

∴ Required area = Area OABO + Area BCDB

Question 6:

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

Answer

The area enclosed between
the parabola, *y*^{2} = 4*ax*, 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 4*y *=
3*x*^{2} and the line 2*y *= 3*x *+ 12 Answer

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

∴ 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 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 Χ 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

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

Area (∆ABC) = Area (ABLA) + Area (BLMCB) Area (ACMA)

Question 14:

Using the method of integration find the area of the region bounded
by lines: 2*x
*+ *y *= 4, 3*x
* 2*y *= 6
and *x * 3*y
*+ 5 = 0

Answer

The given equations of lines are 2*x *+ *y *= 4
(1)

3*x * 2*y
*= 6
(2)

And,

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 Χ Area OBC

Area OBCO = Area OMC + Area MBC

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.

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 *= *x*^{2} if *x *> 0 and *y *= *x*^{2} if *x
*< 0]

A. 0

B.

C.

D.

Answer

Thus, the correct answer is C.

Question 18:

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

A.

B.

C.

D.

Answer

The given equations are

*x*^{2} + *y*^{2} = 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