Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5
I. Choose the correct option from the given alternatives:
Question 1
Maharashtra Board Solution
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
Solution & Step-by-Step Answer:
(a) 12 sq units
Question 2
Maharashtra Board Solution
The area of the region enclosed by the curve y = , and the lines x = e, x = e2 is given by (a) 1 sq unit (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(a) 1 sq unit
Question 3
Maharashtra Board Solution
The area bounded by the curve y = x3, the X-axis and the lines x = -2 and x = 1 is (a) -9 sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(c) sq units
Question 4
Maharashtra Board Solution
The area enclosed between the parabola y2 = 4x and line y = 2x is (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(b) sq units
Question 5
Maharashtra Board Solution
The area of the region bounded between the line x = 4 and the parabola y2 = 16x is (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(a) sq units
Question 6
Maharashtra Board Solution
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
Solution & Step-by-Step Answer:
(d) 4 sq units
Question 7
Maharashtra Board Solution
The area bounded by the parabola y2 = 8x, the X-axis and the latus rectum is (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(b) sq units
Question 8
Maharashtra Board Solution
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is (a) 4 sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(d) sq units
Question 9
Maharashtra Board Solution
The area of the circle x2 + y2 = 25 in first quadrant is (a) sq units (b) 5π sq units (c) 5 sq units (d) 3 sq units
Solution & Step-by-Step Answer:
(a) sq units
Question 10
Maharashtra Board Solution
The area of the region bounded by the ellipse is (a) ab sq units (b) πab sq units (c) sq units ab (d) πa2 sq units
Solution & Step-by-Step Answer:
(b) πab sq units
Question 11
Maharashtra Board Solution
The area bounded by the parabola y2 = x and the line 2y = x is (a) sq units (b) 1 sq unit (c) sq unit (d) sq unit
Solution & Step-by-Step Answer:
(a) sq units
Question 12
Maharashtra Board Solution
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ and the X-axis is (a) sq unit (b) 1 sq unit (c) sq unit (d) sq unit
Solution & Step-by-Step Answer:
(d) sq unit
Question 13
Maharashtra Board Solution
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is (a) 2√3 sq units (b) 9 sq units (c) sq units (d) 18 sq units
Solution & Step-by-Step Answer:
(b) 9 sq units
Question 14
Maharashtra Board Solution
The area bounded by the ellipse and the line is (a) (πab – 2ab) sq units (b) sq units (c) (πab – ab) sq units (d) πab sq units
Solution & Step-by-Step Answer:
(b) sq units
Question 15
Maharashtra Board Solution
The area bounded by the parabola y = x2 and the line y = x is (a) sq unit (b) sq unit (c) sq unit (d) sq unit
Solution & Step-by-Step Answer:
(c) sq unit
Question 16
Maharashtra Board Solution
The area enclosed between the two parabolas y2 = 4x and y = x is (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(c) sq units
Question 17
Maharashtra Board Solution
The area bounded by the curve y = tan x, X-axis and the line x = is (a) log 2 sq units (b) log 2 sq units (c) 2 log 2 sq units (d) 3 log 2 sq units
Solution & Step-by-Step Answer:
(a) log 2 sq units
Question 18
Maharashtra Board Solution
The area of the region bounded by x2 = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(d) sq units
Question 19
Maharashtra Board Solution
The area of the region included between the parabolas y2 = 4ax and x2 = 4ay, (a > 0) is given by (a) sq units (b) sq units (c) sq units (d) sq units
Solution & Step-by-Step Answer:
(a) sq units
Question 20
Maharashtra Board Solution
The area of the region included between the line x + y = 1 and the circle x2 + y2 = 1 is (a) sq units (b) π – 2 sq units (c) sq units (d) π – sq units
Solution & Step-by-Step Answer:
(c) sq units
(II) Solve the following:
Question 1
Maharashtra Board Solution
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 =
Solution & Step-by-Step Answer:
(i) Required area = , where y = 2 = = = 2 × 5 – 0 = 10 sq units.
(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.
Two bounded regions A1and A2are obtained. Both the regions have equal areas.
∴ required area = A1+ A2= 2A1
(iii) Required area = , where y = sin x
Question 2
Maharashtra Board Solution
Find the area of the circle x2 + y2 = 9, using integration.
Solution & Step-by-Step Answer:
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. From the equation of the circle, y2 = 9 – x2. In the first quadrant, y > 0 ∴ y = ∴ area of the circle = 4 (area of the region OABO)
Question 3
Maharashtra Board Solution
Find the area of the ellipse using integration.
Solution & Step-by-Step Answer:
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 ∴ y2 = (25 – x2) In the first quadrant y > 0 ∴ y = ∴ area of the ellipse = 4(area of the region OABO)
Question 4
Maharashtra Board Solution
Find the area of the region lying between the parabolas: (i) y2 = 4x and x2 = 4y (ii) 4y2 = 9x and 3x2 = 16y (iii) y2 = x and x2 = y.
Solution & Step-by-Step Answer:
(i) For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations. From the equation x2 = 4y, y = y = = 4x ∴ x4 – 64x = 0 ∴ x(x3 – 64) = 0 ∴ x = 0 or x3 = 64 i.e. x = 0 or x = 4 When x = 0, y = 0 When x = 4, y = = 4 ∴ the points of intersection are 0(0, 0) and A(4, 4). Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO] Now, area of the region ODACO = area under the parabola y2 = 4x, i.e. y = 2√x between x = 0 and x = 4
(ii)
For finding the points of intersection of the two parabolas, we equate the values of 4y2from their equations.
From the equation 3x2= 16y, y =
∴ y =
∴ = 9x
∴ 3x4– 2304x = 0
∴ x(x3– 2304) = 0
∴ x = 0 or x3= 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y =
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2= 4x,
i.e. y = 2√x between x = 0 and x = 4
Area of the region ODABO = area under the rabola x2= 4y,
i.e. y = between x = 0 and x = 4
(iii)
For finding the points of intersection of the two parabolas, we equate the values of y2from their equations.
From the equation x2= y, y =
∴ y =
∴ = x
∴ x2– y = 0
∴ x(x3– y) = 0
∴ x = 0 or x3= y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2= 4x,
i.e. y = 2√x between x = 0 and x = 4
Area ofthe region ODABO = area under the rabola x2= 4y,
i.e. y = between x = 0 and x = 4
Question 5
Maharashtra Board Solution
Find the area of the region in the first quadrant bounded by the circle x2 + y2 = 4 and the X-axis and the line x = y√3.
Solution & Step-by-Step Answer:
For finding the points of intersection of the circle and the line, we solve x2 + y2 = 4 ………(1) and x = y√3 ……..(2) From (2), x2 = 3y2 From (1), x2 = 4 – y2 3y2 = 4 – y2 4y2 = 4 y2 = 1 y = 1 in the first quadrant. When y = 1, r = 1 × √3 = √3 ∴ the circle and the line intersect at A(√3, 1) in the first quadrant Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED Now, area of the region OCADO = area under the line x = y√3, i.e. y = between x = 0 and x = √3
Question 6
Maharashtra Board Solution
Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.
Solution & Step-by-Step Answer:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations. ∴ y2 = y ∴ y2 – 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 y2 = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1 Area of the region OBADO = area under the line y = x between x = 0 and x = 1
Question 7
Maharashtra Board Solution
Find the area enclosed between the circle x2 + y2 = 1 and the line x + y = 1, lying in the first quadrant.
Solution & Step-by-Step Answer:
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 x2 + y2 = 1 between x = 0 and x = 1 Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1 ∴ required area = sq units.
Question 8
Maharashtra Board Solution
Find the area of the region bounded by the curve (y – 1)2 = 4(x + 1) and the line y = (x – 1).
Solution & Step-by-Step Answer:
The equation of the curve is (y – 1)2 = 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 (x – 1 – 1)2 = 4(x + 1) ∴ x2 – 4x + 4 = 4x + 4 ∴ x2 – 8x = 0 ∴ x(x – 8) = 0 ∴ x = 0, x = 8 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). To find the points where the parabola (y – 1)2 = 4(x + 1) cuts the Y-axis. Put x = 0 in the equation of the parabola, we get (y – 1)2 = 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)2 = 4(x + 1), Y-axis from y = -1 to y = 3 Since, the area cannot be negative, Area of the region BFAB = sq units. Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG Since, area cannot be negative, area of the region = sq units. ∴ required area = = = = sq units.
Question 9
Maharashtra Board Solution
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution & Step-by-Step Answer:
The equation of the line is 2y = 5x + 7, i.e., y = Required area = area of the region ABCDA = area under the line y = between x = 2 and x = 5
Question 10
Maharashtra Board Solution
Find the area of the region bounded by the curve y = 4x2, Y-axis and the lines y = 1, y = 4.
Solution & Step-by-Step Answer:
By symmetry of the parabola, the required area is 2 times the area of the region ABCD. From the equation of the parabola, x2 = In the first quadrant, x > 0 ∴ x = ∴ required area =
