Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by

(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
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Question 3.
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Question 5.
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Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
Answer:
(d) 4 sq units

Question 7.
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Question 9.
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Question 11.
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Question 13.
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Question 15.
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Question 16.
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Question 18.
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Question 20.
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(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = π3
Solution:

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(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.
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Two bounded regions  and  are obtained. Both the regions have equal areas.
∴ required area =
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Question 2.
Find the area of the circle  = 9, using integration.
Solution:

By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.
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∴ area of the circle = 4 (area of the region OABO)
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Question 3.

Solution:

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By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
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∴ area of the ellipse = 4(area of the region OABO)
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Question 4.
Find the area of the region lying between the parabolas:

Solution:

(i)
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For finding the points of intersection of the two parabolas, we equate the values of  from their equations.
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(ii)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii)
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Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii).1
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Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii).2

(iii)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii)
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Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii).1
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Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii).2

Question 5.
Find the area of the region in the first quadrant bounded by the circle  = 4 and the X-axis and the line x = y√3.
Solution:

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Question 6.
Find the area of the region bounded by the parabola  = x and the line y = x in the first quadrant.
Solution:

To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.
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∴  = y
∴ – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola  = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q6.1
Area of the region OBADO = area under the line y = x between x = 0 and x = 1
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Question 7.
Find the area enclosed between the circle +  = 1 and the line x + y = 1, lying in the first quadrant.
Solution:

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Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle +  =1 between x = 0 and x = 1
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Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1
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Question 8.
Find the area of the region bounded by the curve (y – 1 = 4(x + 1) and the line y = (x – 1).
Solution:

The equation of the curve is (y – 1 = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
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When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).
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To find the points where the parabola (y – 1 = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1 = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1 = 4(x + 1), Y-axis from y = -1 to y = 3
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Since, the area cannot be negative,
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Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG
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Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:

The equation of the line is
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Question 10.
Find the area of the region bounded by the curve y = 4, Y-axis and the lines y = 1, y = 4.
Solution:

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By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
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