Important Questions Class 12 Maths Chapter 8-Applications of Integrals

Important For Questions Class 12 Maths Chapter 8 Application of Integrals are provided at BYJU’S along with the detailed explanations. These questions are given as per the new guidelines of CBSE for standard 12th 2019-2020 board examination. Students can also practice all the problems of NCERT book along with these questions of Application of Integrals for 12th.

Integrals application is the chapter for which students can score good marks after getting thorough with the formulas. We have provided formulas of Class 12 Maths syllabus to help the students revise during their examination. Students can also refer to the important questions for all chapters of Class 12 maths here to prepare well for the CBSE-2020 exam.

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Important Questions & Answers For Class 12 Maths Chapter 8 Application of Integrals

Q. No.1: Find the area enclosed by the ellipse x2/a2 + y2/b2 =1.

Solution:

Given,

Class 12 Chapter 8 Imp Ques 1 figure 1

Class 12 Chapter 8 Imp Ques 1 figure 2

We know that,

Ellipse is symmetrical about both x-axis and y-axis.

Class 12 Chapter 8 Imp Ques 1 figure 3

Area of ellipse = 4 × Area of AOB

Class 12 Chapter 8 Imp Ques 1 figure 4

Substituting the positive value of y in the above expression since OAB lies in the first quadrant.

Class 12 Chapter 8 Imp Ques 1 figure 5

= 2ab × sin-1(1)

= 2ab × π/2

= πab

Hence, the required area is πab sq.units.

Q. No. 2: Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Solution:

We can draw the figure od y2 = 9x; x = 2, x = 4 and the x-axis in the first curve as below.

Class 12 Chapter 8 Imp Ques 2 figure 1

y2 = 9x

y = ±√(9x)

y = ±3√x

We can consider the positive value of y since the required area is in the first quadrant.

The required area is the shaded region enclosed by ABCD.

Class 12 Chapter 8 Imp Ques 2 figure 2

= 2 [(2)3 – (√2)3]

= 2[8 – 2√2]

= 16 -2√2

Hence, the area of the required area is 16 – 2√ sq.units.

Q. No. 3: Find the area of the curve y = sin x between 0 and π.

Solution:

Given,

y = sin x

Class 12 Chapter 8 Imp Ques 3 figure 1

Area of OAB

Class 12 Chapter 8 Imp Ques 3 figure 2

= – [cos π – cos 0]

= -(-1 -1)

= 2 sq. units

Q. No. 4: Find the area of the region bounded by the two parabolas y = x2 and y2 = x.

Solution:

Given two parabolas are y = x2 and y2 = x.

The point of intersection of these two parabolas is O (0, 0) and A (1, 1) as shown in the below figure.

Class 12 Chapter 8 Imp Ques 4 figure 1

Now,

y2 = x

y = √x = f(x)

y = x2 = g(x), where, f (x) ≥ g (x) in [0, 1].

Area of the shaded region

Class 12 Chapter 8 Imp Ques 4 figure 2

= (⅔) – (⅓)

= ⅓

Hence, the required area is ⅓ sq.units.

Q. No. 5: 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)

Solution:

Option (B) is the correct answer.

Explantion:

Given,

Equation of circle is x2+ y2 = 4……….(i)

x2+ y2 = 22

y = √(22 – x2) …………(ii)

Equation of a lines is x + y = 2 ………(iii)

y = 2 – x

x 0 2
y 2 0

Class 12 Chapter 8 Imp Ques 5 figure 1

Therefore, the graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).

From the graph of a circle (i) and straight-line (iii), it is clear that points of intersections of circle

(i) and the straight line (iii) is A (2, 0) and B (0.2).

Area of OACB, bounded by the circle and the coordinate axes is

Class 12 Chapter 8 Imp Ques 5 figure 2

= [ 1 × √0 + 2 sin-1(1) – 0√4 – 2 × 0]

= 2 sin-1(1)

= 2 × π/2

= π sq. units

Area of triangle OAB, bounded by the straight line and the coordinate axes is

Class 12 Chapter 8 Imp Ques 5 figure 3

= 4 – 2 – 0 + 0

= 2 sq.units

Hence, the required area = Area of OACB – Area of triangle OAB

= (π – 2) sq.units

Practice Questions For Class 12 Maths Chapter 8 Application of Integrals

  1. Find the area of the region bounded by y =√a and y = a.
  2. Find the area bounded by the curve y = 4 – x2 and the lines y = 0 and y = 3.
  3. Find the area including between the parabolas y2 = 4ax and x2 = 4by.
  4. What is the area bounded by the parabola y2 = 8x and x2 = 8y?
  5. Find the area bounded by the parabola y2 = 4ax, latus rectum and the x-axis.
  6. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x- axis and the lines x = 2 and x = 8.
  7. Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2.

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