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## NCERT Class 12 Maths Solutions Chapter 8 – Application of Integrals

**The topics and sub-topics included in the Integrals chapter are the following:**

Section Name | Topic Name |

8 | Application of Integrals |

8.1 | Introduction |

8.2 | Area under Simple Curves |

8.3 | Area between Two Curves |

### NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

#### Page No 365:

#### 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.

#### 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.

#### Page No 366:

#### 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 × 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:

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

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.

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

Area OAB = Area ΔOCA + Area ACB

Area of OAC

Area of ABC

Therefore, required area enclosed =

32 + π3 – 32 = π3 square units

#### 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, *x*^{2} +* y*^{2} = *a*^{2}, 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

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.

âˆ´ Area OAD = Area ABCD

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

â‡’ 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.

âˆ´ 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 Î”OAM â€“ Area OMACO

Area of Î”OAM

Area of OMACO

â‡’ Area of OACO = Area of Î”OAM â€“ Area of OMACO

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, *y*^{2} = 4*x*, *y*-axis, and *y* = 3 is represented as

Thus, the correct answer is B.

#### Page No 371:

#### 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 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are.

Therefore, Area OBCO = Area OMBCO – Area OMBO

=∫02(9-4×2)4dx-∫02x24dx

=∫02322-x2dx-14∫02x2dx

=x2322-x2+98sin-12×302-14×3302

=24+98sin-1223-11223

=122+98sin-1223-132

=162+98sin-1223

=1226+94sin-1223Therefore, 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 Aand 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* = *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 (ΔACB) = 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 (ΔABC) = (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 (ΔACB) = Area (OLBAO) –Area (OLCAO)

#### Page No 372:

#### 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 (ΔOAB)

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

âˆ´ Area OBAO = Area (OCABO) â€“ Area (Î”OCA)

square units

Thus, the correct answer is B.

#### Page No 375:

#### 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:

- The required area is represented by the shaded area ADCBA as

- 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 (ΔOCA) – 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

Area of ABCDA = ∫14 x dy =∫14 y2 dy as, y = 4×2 =12∫14y dy =12×23y3/214 =1343/2 – 13/2 =138 – 1 =13×7 =73 square units

#### 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 *y* = sin *x* can be drawn as

∴ 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 (ΔOCA)

#### 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 OACO as

The point of intersection of the parabola, *x*^{2} = *y*, and the line, *y* = *x* + 2, is A (–1, 1) and C(2, 4).

Area of OACO = ∫-12x + 2 dx – ∫-12 x2 dx⇒Area of OACO = x22 + 2x-12 – 13×3-12⇒Area of OACO = 222+22 – -122+2-1 – 1323 – -13⇒Area of OACO = 2 + 4 – 12-2 – 138 + 1⇒Area of OACO = 6 + 32 – 3⇒Area of OACO = 3 + 32 = 92 square units

#### 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

#### Page No 376:

#### 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 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)

#### 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, *x* – 3*y *+ 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)

#### 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

Therefore, the required area is 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.**

**C.**

**D. **

#### Answer:

Required Area =

∫-2 0ydx+∫01ydx

=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. unitsThus, the correct answer is D.

#### 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)

*y*^{2} = 6*x*… (2)

Area bounded by the circle and parabola

=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23×3202+2×216-x2+162sin-1×424 =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=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square 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.

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