NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Application of Integrals |

Exercise |
Ex 8.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

**Ex 8.2 Class 12 Maths Question 1.**

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

**Solution:**

Area is bounded by the circle 4x² + 4y² = 9 and interior of the parabola x² = 4y.

Putting x² = 4y in x² + y² = \(\frac { 9 }{ 4 }\)

We get 4y + y² = \(\frac { 9 }{ 4 }\)

**Ex 8.2 Class 12 Maths Question 2.**

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

**Solution:**

Given circles are x² + y² = 1 …(i)

and (x – 1)² + y² = 1 …(ii)

Centre of (i) is O (0,0) and radius = 1

**Ex 8.2 Class 12 Maths Question 3.**

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

**Solution:**

Equation of the parabola is y = x² + 2 or x² = (y – 2)

Its vertex is (0,2) axis is y-axis.

Boundary lines are y = x, x = 0, x = 3.

Graphs of the curve and lines have been shown in the figure.

Area of the region PQRO = Area of the region OAQR-Area of region OAP

**Ex 8.2 Class 12 Maths Question 4.**

Using integration find the area of region bounded by the triangle whose vertices are (-1,0), (1,3) and (3,2).

**Solution:**

The points A (-1,0), B( 1,3) and C (3,2) are plotted and joined.

Area of ∆ABC = Area of ∆ ABL + Area of trap. BLMC – Area of ∆ACM …(i)

The equation of the line joining the points

**Ex 8.2 Class 12 Maths 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.

**Solution:**

The given lines are y = 2x + 1 …(i)

y = 3x + 1 …(ii)

x = 4 …(iii)

Subtract (i) from eq (ii) we get x = 0, Putting x = 0 in eq(i) y = 1

∴ Lines (ii) and (i) intersect at A (0,1) putting x = 4 in eq. (2) =>y = 12 + 1 = 13

The lines (ii) and (iii) intersect at B (4,13) putting x=4ineq. (i):y = 8 + 1 = 9

∴ Lines (i) and (ii); Intersect at C (4,9),

**Ex 8.2 Class 12 Maths Question 6.**

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

(a) 2 (π – 2)

(b) π – 2

(c) 2π – 1

(d) 2(π + 2)

**Solution:**

(b) A circle of radius 2 and centre at O is drawn.The line AB: x + y = 2 is passed through (2,0) and (0,2). Area of the region ACB

= Area of quadrant OAB – Area of ∆OAB …(i)

**Ex 8.2 Class 12 Maths Question 7.**

Area lying between the curves y² = 4x and y = 2x.

(a) \(\frac { 2 }{ 3 }\)

(b) \(\frac { 1 }{ 3 }\)

(c) \(\frac { 1 }{ 4 }\)

(d) \(\frac { 3 }{ 4 }\)

**Solution:**

(b) The curve is y² = 4x …(1)

and the line is y = 2x …(2)

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