CBSE NCERT Solutions for Class 12 Maths Chapter 08 · 2019. 12. 6. · Class-XII-Maths Application...
Transcript of CBSE NCERT Solutions for Class 12 Maths Chapter 08 · 2019. 12. 6. · Class-XII-Maths Application...
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CBSE NCERT Solutions for Class 12 Maths Chapter 08
Back of Chapter Questions
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟖. 𝟏
1. Find the area of the region bounded by the curve 𝑦2 = 𝑥 and the lines 𝑥 = 1, 𝑥 = 4 and the 𝑥-
axis in the first quadrant.
[2 Marks]
Solution:
Step 1:
Given: Equation of the curve 𝑦2 = 𝑥 and the lines 𝑥 = 1, 𝑥 = 4 and the 𝑥-axis.
The given equations can be represented as follows:
[ 𝟏
𝟐Mark]
Step 2:
The area of the region bounded by the curve, 𝑦2 = 𝑥, the lines, 𝑥 = 1 and 𝑥 = 4, and the 𝑥-axis
is the area 𝐴𝐵𝐶𝐷.
Area of 𝐴𝐵𝐶𝐷 = ∫𝑦𝑑𝑥
4
1
= ∫ √𝑥𝑑𝑥4
1 [1 Mark]
Step 3:
= [𝑥
32
32
]
1
4
=2
3[(4)
32 − (1)
32 ]
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=2
3[8 − 1]
=14
3 square units [
𝟏
𝟐 Mark]
Hence, the required area is 14
3 square units.
2. Find the area of the region bounded by 𝑦2 = 9𝑥, 𝑥 = 2, 𝑥 = 4 and the 𝑥-axis in the first
quadrant.
[2 Marks]
Solution:
Step 1:
Given: The equations of the curve 𝑦2 = 9𝑥, 𝑥 = 2, 𝑥 = 4 and the 𝑥-axis in the first quadrant.
The given equations can be represented as follows:
[𝟏
𝟐 Mark]
The area of the region bounded by the curve, 𝑦2 = 9𝑥, 𝑥 = 2, and 𝑥 = 4, and the 𝑥-axis is the
area 𝐴𝐵𝐶𝐷.
Step 2:
Area of 𝐴𝐵𝐶𝐷 = ∫𝑦𝑑𝑥
4
2
= ∫ 3√𝑥𝑑𝑥4
2 [1 Mark]
= 3 [𝑥
32
32
]
2
4
= 2 [𝑥32]
2
4
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= 2 [(4)32 − (2)
32]
= 2[8 − 2√2]
= (16 − 4√2) square units [𝟏
𝟐 Mark]
Hence, the required area is (16 − 4√2) square units.
3. Find the area of the region bounded by 𝑥2 = 4𝑦, 𝑦 = 2, 𝑦 = 4 and the 𝑦-axis in the first
quadrant.
[2 Marks]
Solution:
Step 1:
Given: The equations of the curve 𝑥2 = 4𝑦, 𝑦 = 2, 𝑦 = 4 and the 𝑦-axis in the first quadrant.
The given equations can be represented as follows:
[𝟏
𝟐 Mark]
Step 2:
The area of the region bounded by the curve, 𝑥2 = 4𝑦, 𝑦 = 2, and 𝑦 = 4, and the 𝑦-axis is the
area 𝐴𝐵𝐶𝐷.
Area of 𝐴𝐵𝐶𝐷 = ∫𝑥𝑑𝑦
4
2
= ∫ 2√𝑦𝑑𝑦4
2 [1 Mark]
Step 3:
= 2∫√𝑦𝑑𝑦
4
2
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= 2 [𝑦
32
32
]
2
4
=4
3[(4)
32 − (2)
32]
=4
3[8 − 2√2]
= (32−8√2
3) square units [
𝟏
𝟐 Mark]
Hence, the required area is (32−8√2
3) square units.
4. Find the area of the region bounded by the ellipse 𝑥2
16+
𝑦2
9= 1. [2 Marks]
Solution:
Step 1:
Given: The equation of the ellipse: 𝑥2
16+
𝑦2
9= 1
The given equation of the ellipse can be represented as follows:
[𝟏
𝟐 Mark]
Step 2:
It is observed that the ellipse is symmetrical about 𝑥-axis and 𝑦-axis. ∴ Area bounded by ellipse
= 4 × Area of 𝑂𝐴𝐵
Area of 𝑂𝐴𝐵 = ∫ 𝑦𝑑𝑥4
0
= ∫ 3√1 −𝑥3
16𝑑𝑥
4
0 [1 Mark]
Step 3:
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=3
4∫√16 − 𝑥2
4
0
𝑑𝑥
=3
4[𝑥
2√16 − 𝑥2 +
16
2sin−1
𝑥
4]0
4
=3
4[2√16 − 16 + 18 sin−1(1) − 0 − 8 sin−1(0)]
=3
4[8𝜋
2]
=3
4[4𝜋]
= 3𝜋
Hence, area bounded by the ellipse = 4 × 3𝜋 = 12𝜋 square units. [𝟏
𝟐 Mark]
5. Find the area of the region bounded by the ellipse 𝑥2
4+
𝑦2
9= 1. [2 Marks]
Solution:
Step 1:
Given: The equation of the ellipse 𝑥2
4+
𝑦2
9= 1
The given equation of the ellipse can be represented as
𝑥2
4+
𝑦2
9= 1
⇒ 𝑦 = 3√1 −𝑥2
4 …(i) [
𝟏
𝟐 Mark]
Step 2:
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It is observed that the ellipse is symmetrical about 𝑥-axis and 𝑦-axis.
∴ Area bounded by ellipse = 4 × Area 𝑂𝐴𝐵
∴ Area of 𝑂𝐴𝐵 = ∫𝑦𝑑𝑥
2
0
= ∫ 3√1 −𝑥2
4𝑑𝑥
2
0 [Using (i)] [1 Mark]
Step 3:
=3
2∫√4 − 𝑥2𝑑𝑥
2
0
=3
2[𝑥
2√4 − 𝑥2 +
4
2sin
−𝑥
2]0
2
=3
2[2𝜋
2]
=3𝜋
2
Hence, area bounded by the ellipse = 4 ×3𝜋
2= 6𝜋 square units. [
𝟏
𝟐 Mark]
6. Find the area of the region in the first quadrant enclosed by 𝑥-axis, line 𝑥 = √3𝑦 and the circle
𝑥2 + 𝑦2 = 4. [4 marks]
Solution:
Step 1:
Given: The equations of the circle: 𝑥2 + 𝑦2 = 4 and 𝑥-axis, line 𝑥 = √3𝑦
[𝟏
𝟐 Mark]
Step 2:
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The area of the region bounded by the circle, 𝑥2 + 𝑦2 = 4, 𝑥 = √3𝑦 and the 𝑥-axis is the area
𝑂𝐴𝐵.
The point of intersection of the line and the circle in the first quadrant is (√3, 1)
Area 𝑂𝐴𝐵 = Area ∆𝑂𝐶𝐴 + Area 𝐴𝐶𝐵 [𝟏
𝟐 Mark]
Step 3:
Area of 𝑂𝐴𝐶 =1
2× 𝑂𝐶 × 𝐴𝐶 =
1
2× √3 × 1 =
√3
2 …(i) [
𝟏
𝟐 Mark]
Step 4:
Area of 𝐴𝐵𝐶 = ∫ 𝑦𝑑𝑥2
√3 [
𝟏
𝟐 Mark]
Step 5:
= ∫√4 − 𝑥2 𝑑𝑥
2
√3
= [𝑥
2√4 − 𝑥2 +
4
2sin−1 𝑥
2]√3
2 [
𝟏
𝟐 Mark]
Step 5:
= [2 ×𝜋
2−
√3
2√4 − 3 − 2 sin−1 (
√3
2)]
= [𝜋 −√3
2− 2(
𝜋
3)]
= [𝜋 −√3
2−
2𝜋
3]
= [𝜋
3−
√3
2] …(ii) [
𝟏
𝟐 Mark]
Step 6:
Hence, area enclosed by 𝑥-axis, the line 𝑥 = √3𝑦 and the circle 𝑥2 + 𝑦2 = 4 in the first
quadrant =√3𝜋
2+
𝜋
3−
3√𝜋
2=
𝜋
3 sq. units. [1 Mark]
7. Find the area of the smaller part of the circle 𝑥2 + 𝑦2 = 𝑎2 cut off by the line 𝑥 =𝑎
√2. [4
Marks]
Solution:
Step 1:
Given: The equation of the circle: 𝑥2 + 𝑦2 = 𝑎2 which is cut off by a line 𝑥 =𝑎
√2
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[𝟏
𝟐 Mark]
Step 2:
The area of the smaller part of the circle, 𝑥2 + 𝑦2 = 𝑎2, cut off by the line, 𝑥 =𝑎
√2, is the area
𝐴𝐵𝐶𝐷𝐴.
It is observed that the area 𝐴𝐵𝐶𝐷 is symmetrical about 𝑥-axis.
∴ Area 𝐴𝐵𝐶𝐷 = 2 × Area 𝐴𝐵𝐶 [𝟏
𝟐 Mark]
Step 3:
Area of 𝐴𝐵𝐶 = ∫ 𝑦𝑑𝑥
𝑎
𝑎
√2
= ∫ √𝑎2 − 𝑥2𝑎𝑎
√2
𝑑𝑥 [𝟏
𝟐 Mark]
Step 3:
= [𝑥
2√𝑎2 − 𝑥2 +
𝑎2
2sin−1 𝑥
𝑎] 𝑎
√2
𝑎
[𝟏
𝟐 Mark]
Step 4:
= [𝑎2
2(𝜋
2) −
𝑎
2√2√𝑎2 −
𝑎2
2−
𝑎2
2sin−1 (
1
√2)]
=𝑎2𝜋
4−
𝑎
2√2.
𝑎
√2−
𝑎2
2(𝜋
4)
=𝑎2𝜋
4−
𝑎2
4−
𝑎2𝜋
8
=𝑎2
4[𝜋 − 1 −
𝜋
2]
=𝑎2
4[𝜋
2− 1] [1 Mark]
Step 5:
⇒ Area 𝐴𝐵𝐶𝐷 = 2 [𝑎2
4(𝜋
2− 1)] =
𝑎2
2(𝜋
2− 1)
Hence, the area of smaller part of the circle, 𝑥2 + 𝑦2 = 𝑎2, cut off by the line, 𝑥 =𝑎
√2 is
𝑎2
2(𝜋
2− 1) sq. units. [1 Mark]
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8. The area between 𝑥 = 𝑦2 and 𝑥 = 4 is divided into two equal parts by the line 𝑥 = 𝑎, find the
value of 𝑎. [4 marks]
Solution:
Step 1:
Given: The area between 𝑥 = 𝑦2 and 𝑥 = 4 is divided into two equal parts by the line 𝑥 = 𝑎.
The line, 𝑥 = 𝑎, divides the area bounded by the parabola and 𝑥 = 4 into two equal parts.
∴ Area 𝑂𝐴𝐷 = Area 𝐴𝐵𝐶𝐷
[𝟏
𝟐 Mark]
Step 2:
It can be observed that the given area is symmetrical about 𝑥-axis.
⇒ Area 𝑂𝐸𝐷 = Area 𝐸𝐹𝐶𝐷
Area 𝑂𝐸𝐷 = ∫ 𝑦𝑑𝑥𝑎
0 [
𝟏
𝟐 Mark]
Step 3:
= ∫ √𝑥𝑎
0𝑑𝑥
= [𝑥
32
32
]
0
𝑎
=2
3(𝑎)
3
2 …(i) [𝟏
𝟐 Mark]
Step 4:
Area of 𝐸𝐹𝐶𝐷 = ∫ √𝑥 𝑑𝑥4
𝑎 [
𝟏
𝟐 Mark]
Step 5:
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= [𝑥
32
32
]
𝑎
4
=2
3[8 − 𝑎
3
2] …(ii) [𝟏
𝟐 Mark]
Step 6:
From (i) and (ii), we get
2
3(𝑎)
32 =
2
3[8 − (𝑎)
32]
⇒ 2. (𝑎)32 = 8
⇒ (𝑎)32 = 4
⇒ 𝑎 = (4)2
3 [𝟏𝟏
𝟐 Mark]
Hence, the value of 𝑎 is (4)2
3.
9. Find the area of the region bounded by the parabola 𝑦 = 𝑥2 and 𝑦 = |𝑥|. [2 Marks]
Solution:
Step 1:
Given: Equation of the parabola 𝑦 = 𝑥2 and 𝑦 = |𝑥|
The area bounded by the parabola, 𝑥2 = 𝑦 and the line, 𝑦 = |𝑥|, can be represented as
[𝟏
𝟐 Mark]
Step 2:
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The given area is symmetrical about 𝑦-axis.
∴ Area 𝑂𝐴𝐶𝑂 = Area 𝑂𝐷𝐵𝑂
The point of intersection of parabola, 𝑥2 = 𝑦 and line, 𝑦 = 𝑥 is 𝐴(1, 1).
Area of 𝑂𝐴𝐶𝑂 = Area ∆𝑂𝐴𝐵 − Area 𝑂𝐵𝐴𝐶𝑂 [𝟏
𝟐 Mark]
Step 3:
∴ Area of ∆𝑂𝐴𝐵 =1
2× 𝑂𝐵 × 𝐴𝐵 =
1
2× 1 × 1 =
1
2
Area of 𝑂𝐵𝐴𝐶𝑂 = ∫ 𝑦 𝑑𝑥 = ∫ 𝑥2 𝑑𝑥4
0= [
𝑥3
3]0
1
=1
3
1
0 [
𝟏
𝟐 Mark]
Step 4:
⇒ Area of 𝑂𝐴𝐶𝑂 = Area of ∆𝑂𝐴𝐵 − Area of 𝑂𝐵𝐴𝐶𝑂
=1
2−
1
3
=1
6
Hence, required area = 2 [1
6] =
1
3 sq. units [
𝟏
𝟐 Mark]
10. Find the area bounded by the curve 𝑥2 = 4𝑦 and the line 𝑥 = 4𝑦 − 2. [4 Marks]
Solution:
Step 1:
Given: Equation of the curve 𝑥2 = 4𝑦 and the line 𝑥 = 4𝑦 − 2
The area bounded by the curve, 𝑥2 = 4𝑦 and line, 𝑥 = 4𝑦 − 2, is represented by the shaded
area 𝑂𝐵𝐴𝑂.
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[𝟏
𝟐 Mark]
Step 2:
Let 𝐴 and 𝐵 be the points of intersection of the line and parabola.
Coordinates of point 𝐴 are (−1,1
4).
Coordinates of point 𝐵 are (2, 1).
We draw 𝐴𝐿 and 𝐵𝑀 perpendicular to 𝑥-axis. [𝟏
𝟐 Mark]
Step 3:
It is observed that,
Area 𝑂𝐵𝐴𝑂 = Area 𝑂𝐵𝐶𝑂 + Area 𝑂𝐴𝐶𝑂 …(i)
Then, Area 𝑂𝐵𝐶𝑂 = Area 𝑂𝑀𝐵𝐶 − Area 𝑂𝑀𝐵𝑂
= ∫𝑥+2
4 𝑑𝑥
2
0− ∫
𝑥2
4𝑑𝑥
2
0 [
𝟏
𝟐 Mark]
Step 4:
=1
4[𝑥2
2+ 2𝑥]
0
2
−1
4[𝑥3
3]0
2
=1
4[2 + 4] −
1
4[8
3]
=3
2−
2
3
=5
6 [
𝟏
𝟐 Mark]
Step 5:
Similarly, Area 𝑂𝐴𝐶𝑂 = Area 𝑂𝐿𝐴𝐶 − Area 𝑂𝐿𝐴𝑂
= ∫𝑥+2
4𝑑𝑥 − ∫
𝑥2
4𝑑𝑥
0
−1
0
−1 [
𝟏
𝟐 Mark]
Step 6:
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=1
4[𝑥2
2+ 2𝑥]
−1
0
−1
4[𝑥3
3]−1
0
= −1
4[(−1)2
2+ 2(−1)] − [−
1
4((−1)3
3)]
= −1
4[1
2− 2] −
1
12
=1
2−
1
8−
1
12
=7
24 [
𝟏
𝟐 Mark]
Step 7:
Hence, required area = (5
6+
7
24) =
9
8 sq. units [1 Mark]
11. Find the area of the region bounded by the curve 𝑦2 = 4𝑥 and the line 𝑥 = 3. [2 marks]
Solution:
Step 1:
Given: Equation of the curve 𝑦2 = 4𝑥 and the line 𝑥 = 3
The region bounded by the parabola, 𝑦2 = 4𝑥 and the line, 𝑥 = 3 is the area 𝑂𝐴𝐶𝑂. [𝟏
𝟐 Mark]
Step 2:
The area 𝑂𝐴𝐶𝑂 is symmetrical about 𝑥-axis
∴ Area of 𝑂𝐴𝐶𝑂 = 2 (Area of 𝑂𝐴𝐵)
Area 𝑂𝐴𝐶𝑂 = 2 [∫ 𝑦𝑑𝑥3
0] [
𝟏
𝟐 Mark]
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Step 3:
= 2∫2√𝑥𝑑𝑥
3
0
= 4 [𝑥
32
3
2
]0
3
[𝟏
𝟐 Mark]
Step 4:
=8
3[(3)
32]
= 8√3
Hence, the required area is 8√3 Square units. [𝟏
𝟐 Mark]
12. Area lying in the first quadrant and bounded by the circle 𝑥2 + 𝑦2 = 4 and the lines 𝑥 = 0 and
𝑥 = 2 is [2 Marks]
(A) 𝜋
(B) 𝜋
2
(C) 𝜋
3
(D) 𝜋
4
Solution:
(A)
Step 1:
Given: The equation of the circle 𝑥2 + 𝑦2 = 4 and the lines 𝑥 = 0 and 𝑥 = 2 is
The area bounded by the given circle and the lines, 𝑥 = 0 and 𝑥 = 2 in the first quadrant is
represented as
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[𝟏
𝟐 Mark]
Step 2:
∴ Area 𝑂𝐴𝐵 = ∫ 𝑦𝑑𝑥2
0 [
𝟏
𝟐 Mark]
Step 3:
= ∫ √4 − 𝑥2𝑑𝑥2
0 [
𝟏
𝟐 Mark]
Step 4:
= [𝑥
2√4 − 𝑥2 +
4
2sin−1
𝑥
2]0
2
= 2(𝜋
2)
= 𝜋 units
Hence, (𝐴) is the correct answer. [𝟏
𝟐 Mark]
13. Area of the region bounded by the curve 𝑦2 = 4𝑥, 𝑦-axis and the line 𝑦 = 3 is [2 Marks]
(A) 2
(B) 9
4
(C) 9
3
(D) 9
2
Solution:
(B)
Step 1:
Given: The equation of the curve 𝑦2 = 4𝑥, 𝑦-axis and the line 𝑦 = 3
The area bounded by the curve, 𝑦2 = 4𝑥, 𝑦-axis and 𝑦 = 3 is represented as
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[𝟏
𝟐 Mark]
Step 2:
∴ Area 𝑂𝐴𝐵 = ∫ 𝑥𝑑𝑦3
0 [
𝟏
𝟐 Mark]
Step 3:
= ∫𝑦3
4𝑑𝑦
3
0 [
𝟏
𝟐 Mark]
Step 4:
=1
4[𝑦3
3]0
3
=1
12(27)
=9
4 units
Hence, (𝐵) is the correct answer. [𝟏
𝟐 Mark]
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟖. 𝟐
1. Find the area of the circle 4𝑥2 + 4𝑦2 = 9 which is interior to the parabola 𝑥2 = 4𝑦. [4 Marks]
Solution:
Step 1: Given: Equation of the circle 4𝑥2 + 4𝑦2 = 9 and the equation of the parabola 𝑥2 = 4𝑦.
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The required area is represented by the shaded area 𝑂𝐵𝐶𝐷𝑂. [
𝟏
𝟐 Mark]
Step 2: Solving the given equation of circle, 4𝑥2 + 4𝑦2 = 9, and parabola, 𝑥2 = 4𝑦, we obtain the point
of intersection as 𝐵 (√2,1
2) and 𝐷 (−√2,
1
2)
It is observed that the required area is symmetrical about 𝑦-axis.
∴ Area 𝑂𝐵𝐶𝐷𝑂 = 2 × Area 𝑂𝐵𝐶𝑂 [𝟏
𝟐 Mark]
Step 3: We draw 𝐵𝑀 perpendicular to 𝑂𝐴.
Hence, the coordinates of 𝑀 are (√2, 0).
Hence, Area 𝑂𝐵𝐶𝑂 = Area 𝑂𝑀𝐵𝐶𝑂 − Area 𝑂𝑀𝐵𝑂
= ∫ √(9 − 4𝑥2)
4
√2
0
𝑑𝑥 − ∫ √𝑥2
4𝑑𝑥
√2
0
=1
2∫ √9 − 4𝑥2
√2
0
𝑑𝑥 −1
4∫ 𝑥2𝑑𝑥
√2
0
=1
4[𝑥√9 − 4𝑥2 +
9
2sin−1
2𝑥
3]0
√2
−1
4[𝑥3
3]0
√2
=1
4[√2 √9 − 8 +
9
2sin−1
2√2
3] −
1
12(√2)
3
=√2
4+
9
8sin−1
2√2
3−
√2
6
=√2
12+
9
8sin−1
2√2
3
=1
2(√2
6+
9
4sin−1 2√2
3) [2 Marks]
Step 4: Hence, the required area 𝑂𝐵𝐶𝐷𝑂 is
(2 ×1
2[√2
6+
9
4sin−1 2√2
3]) = [
√2
6+
9
4sin−1 2√2
3] square units [1 Mark]
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2. Find the area bounded by curves (𝑥 − 1)2 + 𝑦2 = 1 and 𝑥2 + 𝑦2 = 1. [4 Marks]
Solution:
Step 1: Given: The equations of the curves (𝑥 − 1)2 + 𝑦2 = 1 and 𝑥2 + 𝑦2 = 1. The area bounded by the curves, (𝑥 − 1)2 + 𝑦2 = 1 and 𝑥2 + 𝑦2 = 1 is represented by the shaded area as
[𝟏
𝟐 Mark]
Step 2: After solving the equations (𝑥 − 1)2 + 𝑦2 = 1 and 𝑥2 + 𝑦2 = 1, we get the point of
intersection as 𝐴 (1
2,√3
2) and 𝐵 (
1
2, −
√3
2).
It is observed that the required area is symmetrical about 𝑥-axis.
∴ Area 𝑂𝐵𝐶𝐴𝑂 = 2 × Area 𝑂𝐶𝐴𝑂 [𝟏
𝟐 Mark]
Step 3:
We join 𝐴𝐵, which intersects 𝑂𝐶 at 𝑀, such that 𝐴𝑀 is perpendicular to 𝑂𝐶.
The coordinates of 𝑀 are (1
2, 0).
⇒ Area 𝑂𝐶𝐴𝑂 = Area 𝑂𝑀𝐴𝑂 + Area 𝑀𝐶𝐴𝑀 [𝟏
𝟐 Mark]
Step 4:
=
[
∫√1 − (𝑥 − 1)2 𝑑𝑥 + ∫√1 − 𝑥2 𝑑𝑥
1
12
12
0]
= [𝑥 − 1
2√1 − (𝑥 − 1)2 +
1
2sin−1(𝑥 − 1)]
0
12+ [
𝑥
2√1 − 𝑥2 +
1
2sin−1 𝑥]
12
1
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= [−1
4√1 − (−
1
2)2
+1
2sin−1 (
1
2− 1) −
1
2sin−1(−1)]
+ [1
2sin−1(1) −
1
4√1 − (
1
2)2
−1
2sin−1 (
1
2)]
= [−√3
8+
1
2(−
𝜋
6) −
1
2(−
𝜋
2)] + [
1
2(𝜋
2) −
√3
8−
1
2(𝜋
6)]
= [−√3
4−
𝜋
12+
𝜋
4+
𝜋
4−
𝜋
12]
= [−√3
4−
𝜋
6+
𝜋
2]
= [2𝜋
6−
√3
4] [𝟏
𝟏
𝟐 Mark]
Step 5:
Hence, required area 𝑂𝐵𝐶𝐴𝑂 = 2 × (2𝜋
6−
√3
4) = (
2𝜋
3−
√3
2) sq. units [1 Mark]
3. Find the area of the region bounded by the curves 𝑦 = 𝑥2 + 2, 𝑦 = 𝑥, 𝑥 = 0 and 𝑥 = 3 [2
Marks]
Solution:
Step 1:
Given: The equations 𝑦 = 𝑥2 + 2, 𝑦 = 𝑥, 𝑥 = 0 and 𝑥 = 3
The area bounded by the curves 𝑦 = 𝑥2 + 2, 𝑦 = 𝑥, 𝑥 = 0 and 𝑥 = 3 is represented by the
shaded area 𝑂𝐶𝐵𝐴𝑂 as
[𝟏
𝟐 Mark]
Step 2:
Then, Area 𝑂𝐶𝐵𝐴𝑂 = Area 𝑂𝐷𝐵𝐴𝑂 − Area 𝑂𝐷𝐶𝑂
= ∫ (𝑥2 + 2)𝑑𝑥 − ∫ 𝑥𝑑𝑥3
0
3
0 [1 Mark]
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Step 4:
= [𝑥3
3+ 2𝑥]
0
3
− [𝑥2
2]0
3
= [9 + 6] − [9
2]
= 15 −9
2
=21
2 square units [
𝟏
𝟐 Mark]
Hence, the required area is 21
2 square units.
4. Using integration find the area of the region bounded by the triangle whose vertices are
(−1, 0), (1, 3) and (3, 2). [6 Marks]
Solution:
Step 1:
Given: The vertices of a triangle (−1, 0), (1, 3) and (3, 2).
Step 2:
𝐵𝐿 and 𝐶𝑀 are drawn perpendicular to 𝑥-axis.
It is observed in the figure that
Area(∆𝐴𝐶𝐵) = Area(𝐴𝐿𝐵𝐴) + Area(𝐵𝐿𝑀𝐶𝐵) − Area(𝐴𝑀𝐶𝐴) …(i) [1 Mark]
Step 3:
Equation of line segment 𝐴𝐵 is
𝑦 − 0 =3 − 0
1 + 1(𝑥 + 1)
𝑦 =3
2(𝑥 + 1) [
𝟏
𝟐 Mark]
Step 4:
∴ Area(ALBA) = ∫3
2(𝑥 + 1)𝑑𝑥 =
3
2[𝑥2
2+ 𝑥]
−1
1
=3
2[1
2+ 1 −
1
2+ 1] = 3
1
−1 sq. units. [1 Mark]
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Step 5:
Equation of line segment 𝐵𝐶 is
𝑦 − 3 =2 − 3
3 − 1(𝑥 − 1)
𝑦 =1
2(−𝑥 + 7) [
𝟏
𝟐 Mark]
Step 6:
∴ Area(𝐵𝐿𝑀𝐶𝐵) = ∫1
2(−𝑥 + 7)𝑑𝑥 =
1
2[−
𝑥2
2+ 7𝑥]
1
3
=1
2[−
9
2+ 21 +
1
2− 7] = 5
3
1 square units.
[1 Mark]
Step 7:
Equation of line segment 𝐴𝐶 is
𝑦 − 0 =2 − 0
3 + 1(𝑥 + 1)
𝑦 =1
2(𝑥 + 1) [
𝟏
𝟐 Mark]
∴ Area(𝐴𝑀𝐶𝐴) =1
2 ∫ (𝑥 + 1)𝑑𝑥 =
1
2[𝑥2
2+ 𝑥]
−1
3
=1
2[9
2+ 3 −
1
2+ 1] = 4
3
−1 square units. [1
Mark]
Step 8:
Hence, from equation (i), we obtain Area(∆𝐴𝐵𝐶) = (3 + 5 − 4) = 4 square units. [𝟏
𝟐 Mark]
5. Using integration find the area of the triangular region whose sides have the equations 𝑦 =
2𝑥 + 1, 𝑦 = 3𝑥 + 1 and 𝑥 = 4. [4 Marks]
Solution:
Step 1:
Given: The equations of the sides of a triangle: 𝑦 = 2𝑥 + 1, 𝑦 = 3𝑥 + 1 and 𝑥 = 4.
Step 2:
On solving these equations, we obtain the vertices of triangle as 𝐴(0, 1), 𝐵(4, 13) and 𝐶(4, 9).
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[1𝟏
𝟐 mark]
Step 3:
It is observed that,
Area(∆𝐴𝐶𝐵) = Area(𝑂𝐿𝐵𝐴𝑂) − Area(𝑂𝐿𝐶𝐴𝑂) [𝟏
𝟐 Mark]
Step 4:
= ∫ (3𝑥 + 1)𝑑𝑥 − ∫ (2𝑥 + 1)𝑑𝑥4
0
4
0 [1 Mark]
Step 5:
= [3𝑥2
2+ 𝑥]
0
4
− [2𝑥2
2+ 𝑥]
0
4
= (24 + 4) − (16 + 4)
= 28 − 20
= 8 square units [1 Mark]
Hence, the area of the triangular region is 8 square units.
6. Smaller area enclosed by the circle 𝑥2 + 𝑦2 = 4 and the line 𝑥 + 𝑦 = 2 is [2 Marks]
(A) 2(𝜋 − 2)
(B) 𝜋 − 2
(C) 2𝜋 − 1
(D) 2(𝜋 + 2)
Solution:
(B)
Step 1:
Given: The equation of the circle 𝑥2 + 𝑦2 = 4 and the line 𝑥 + 𝑦 = 2.
Step 2:
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The smaller area enclosed by the circle, 𝑥2 + 𝑦2 = 4 and the line 𝑥 + 𝑦 = 2 is represented by
the shaded area 𝐴𝐶𝐵𝐴. [𝟏
𝟐 Mark]
Step 3:
It is observed that,
Area 𝐴𝐶𝐵𝐴 = Area 𝑂𝐴𝐶𝐵𝑂 − Area (∆𝑂𝐴𝐵) [𝟏
𝟐 Mark]
Step 4:
= ∫√4 − 𝑥2𝑑𝑥 − ∫(2 − 𝑥)𝑑𝑥
2
0
2
0
= [𝑥
2√4 − 𝑥2 +
4
2sin−1 𝑥
2]0
2− [2𝑥 −
𝑥2
2]0
2
[𝟏
𝟐 Mark]
Step 5:
= [2.𝜋
2] − [4 − 2]
= (𝜋 − 2) square units
Hence, (𝐵) is the correct answer. [𝟏
𝟐 Mark]
7. Area lying between the curve 𝑦2 = 4𝑥 and 𝑦 = 2𝑥 is [2 Marks]
(A) 2
3
(B) 1
3
(C) 1
4
(D) 3
4
Solution:
(B)
Step 1:
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Given: The equations of the curve 𝑦2 = 4𝑥 and 𝑦 = 2𝑥
Step 2:
The area lying between the curve 𝑦2 = 4𝑥 and 𝑦 = 2𝑥 is represented by the shaded area 𝑂𝐵𝐴𝑂
as
The points of intersection of these curves are 𝑂(0, 0) and 𝐴(1, 2). [𝟏
𝟐 Mark]
Step 3:
We draw 𝐴𝐶 perpendicular to 𝑥-axis such that the coordinates of 𝐶 are (1, 0).
∴ Area 𝑂𝐵𝐴𝑂 = Area (∆𝑂𝐶𝐴) − Area (𝑂𝐶𝐴𝐵𝑂) [𝟏
𝟐 Mark]
Step 4:
= ∫ 2𝑥𝑑𝑥 − ∫ 2√𝑥1
0 𝑑𝑥
1
0 [
𝟏
𝟐 Mark]
Step 5:
= 2 [𝑥2
2]0
1
− 2 [𝑥
32
32
]
0
1
= |1 −4
3|
= |−1
3|
=1
3 square units
hence, (𝐵) is the correct answer. [𝟏
𝟐 Mark]
Miscellaneous
1. Find the area under the given curves and given lines:
(i) 𝑦 = 𝑥2, 𝑥 = 1, 𝑥 = 2 and 𝑥-axis [2 Marks]
(ii) 𝑦 = 𝑥4, 𝑥 = 1, 𝑥 = 5 and 𝑥-axis [2 Marks]
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Solution:
(i) Step 1:
Given: The equations of the curves: 𝑦 = 𝑥2, 𝑥 = 1, 𝑥 = 2 and 𝑥-axis
Step 2:
The required area is represented by the shaded area 𝐴𝐷𝐶𝐵𝐴 as
[𝟏
𝟐 Mark]
Step 3:
Area 𝐴𝐷𝐶𝐵𝐴 = ∫ 𝑦𝑑𝑥2
1 [
𝟏
𝟐 Mark]
Step 4:
= ∫𝑥2𝑑𝑥
2
1
= [𝑥3
3]1
2
=8
3−
1
3
=7
3 square units [1 Mark]
Hence, the required area is 7
3 square units.
(ii)Step 1:
Given: The equations of the curves are: 𝑦 = 𝑥4, 𝑥 = 1, 𝑥 = 5 and 𝑥-axis.
Step 2:
The required area is represented by the shaded area 𝐴𝐷𝐶𝐵𝐴
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[𝟏
𝟐 Mark]
Step 3:
Area 𝐴𝐷𝐶𝐵𝐴 = ∫ 𝑥4𝑑𝑥5
1 [
𝟏
𝟐 Mark]
Step 4:
= [𝑥5
5]1
5
=(5)5
5−
1
5
= (5)4 −1
5
= 625 −1
5
= 624.8 square units [1 Mark]
Hence, the required area is 624.8 square units
2. Find the area between the curves 𝑦 = 𝑥 and 𝑦 = 𝑥2. [4 Marks]
Solution:
Step 1:
Given: The equation of the curve are 𝑦 = 𝑥 and 𝑦 = 𝑥2
Step 2:
The required area is represented by the shaded area 𝑂𝐵𝐴𝑂 as
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[𝟏
𝟐 Mark]
Step 3:
The points of intersection of the curves, 𝑦 = 𝑥 and 𝑦 = 𝑥2 is 𝐴(1, 1) and 𝑂(0, 0).
We draw 𝐴𝐶 perpendicular to 𝑥-axis. [𝟏
𝟐 Mark]
Step 4:
∴ Area (𝑂𝐵𝐴𝑂) = Area (∆𝑂𝐶𝐴) − Area (𝑂𝐶𝐴𝐵𝑂) …(i) [1 Mark]
Step 5:
= ∫ 𝑥𝑑𝑥 −1
0 ∫ 𝑥2𝑑𝑥1
0 [1 Mark]
Step 6:
= [𝑥2
2]0
1
− [𝑥3
3]0
1
=1
2−
1
3
=1
6 square units. [1 Mark]
3. Find the area of the region lying in the first quadrant and bounded by 𝑦 = 4𝑥2, 𝑥 = 0, 𝑦 = 1 and
𝑦 = 4. [2 Marks]
Solution:
Step 1:
Given: The equations of the curves are: 𝑦 = 4𝑥2, 𝑥 = 0, 𝑦 = 1 and 𝑦 = 4.
Step 2:
The area in the first quadrant bounded by 𝑦 = 4𝑥2, 𝑥 = 0, 𝑦 = 1 and 𝑦 = 4 is represented by
the shaded area 𝐴𝐵𝐶𝐷𝐴 as
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[𝟏
𝟐 Mark]
Step 3:
∴ Area 𝐴𝐵𝐶𝐷 = ∫ 𝑥𝑑𝑦4
1 [
𝟏
𝟐 Mark]
Step 4:
= ∫√𝑦
2
4
1𝑑𝑦 [
𝟏
𝟐 Mark]
Step 5:
=1
2[𝑦
32
32
]
1
4
=1
3[(4)
32 − 1]
=1
3[8 − 1]
=7
3 square units [
𝟏
𝟐 Mark]
Hence, the required area is 7
3 square units.
4. Sketch the graph of 𝑦 = |𝑥 + 3| and evaluate ∫ |𝑥 + 3|𝑑𝑥0
−6. [4 Marks]
Solution:
Step 1:
Given: The equation is 𝑦 = |𝑥 + 3|
Step 2:
Some corresponding values of 𝑥 and 𝑦 are given in the table.
𝑥 −6 −5 −4 −3 −2 −1 0
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𝑦 3 2 1 0 1 2 3
[𝟏
𝟐 Mark]
Step 3:
After plotting these points, we obtain the graph of 𝑦 = |𝑥 + 3| as follows.
[1𝟏
𝟐 Marks]
Step 4:
As we know that (𝑥 + 3) ≤ 0 for −6 ≤ 𝑥 ≤ −3 and (𝑥 + 3) ≥ 0 for −3 ≤ 𝑥 ≤ 0
∴ ∫ |(𝑥 + 3)|𝑑𝑥 = −∫ (𝑥 + 3)𝑑𝑥 + ∫ (𝑥 + 3)𝑑𝑥 0
−3
−3
−6
0
−6 [1 mark]
Step 5:
= −[𝑥2
2+ 3𝑥]
−6
−3
+ [𝑥2
2+ 3𝑥]
−3
0
= −[((−3)2
2+ 3(−3)) − (
(−6)2
2+ 3(−6))] + [0 − (
(−3)2
2+ 3(−3))]
= −[−9
2] − [−
9
2]
= 9 [1 Mark]
Hence, the required area is 9 square units.
5. Find the area bounded by the curve 𝑦 = sin 𝑥 between 𝑥 = 0 and 𝑥 = 2𝜋. [2 Marks]
Solution:
Step 1:
Given: The equation of the curve: 𝑦 = sin𝑥 bounded between 𝑥 = 0 and 𝑥 = 2𝜋
Step 2:
The graph of 𝑦 = sin 𝑥 is drawn as
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∴ Required area = Area 𝑂𝐴𝐵𝑂 + Area 𝐵𝐶𝐷𝐵 [𝟏
𝟐 Mark]
Step 3:
= ∫ sin 𝑥 𝑑𝑥 + |∫ sin𝑥 𝑑𝑥2𝜋
𝜋|
𝜋
0 [
𝟏
𝟐 Mark]
Step 4:
= [− cos 𝑥]0𝜋 + |[− cos 𝑥]𝑥
2𝜋|
= [− cos𝜋 + cos 0] + |− cos2𝜋 + cos𝜋| [𝟏
𝟐 Mark]
Step 5:
= 1 + 1 + |(−1 − 1)|
= 2 + |−2|
= 2 + 2 = 4 square units [𝟏
𝟐 Mark]
Hence, the required area is 4 square units.
6. Find the area enclosed between the parabola 𝑦2 = 4𝑎𝑥 and the line 𝑦 = 𝑚𝑥. [6 marks]
Solution:
Step 1:
Given: The equation of parabola: 𝑦2 = 4𝑎𝑥 and the line 𝑦 = 𝑚𝑥
Step 2:
The area enclosed between the parabola, 𝑦2 = 4𝑎𝑥 and the line 𝑦 = 𝑚𝑥 is represented by the
shaded area 𝑂𝐴𝐵𝑂 as
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[𝟏
𝟐 Mark]
Step 3:
The points of intersection of both the curves are (0, 0) and (4𝑎
𝑚2 ,4𝑎
𝑚). [
𝟏
𝟐 Mark]
Step 4:
We draw 𝐴𝐶 perpendicular to 𝑥-axis.
∴ Area 𝑂𝐴𝐵𝑂 = Area 𝑂𝐶𝐴𝐵𝑂 − Area (∆𝑂𝐶𝐴) [1 Mark]
Step 5:
= ∫ 2√𝑎𝑥4𝑎
𝑚2
0 𝑑𝑥 − ∫ 𝑚𝑥 𝑑𝑥
4𝑎
𝑚2
0 [1 Mark]
Step 6:
= 2√𝑎 [𝑥
32
32
]
0
4𝑎𝑚2
− 𝑚 [𝑥2
2]0
4𝑎𝑚2
=4
3√𝑎 (
4𝑎
𝑚2)
32−
𝑚
2[(
4𝑎
𝑚2)2
]
=32𝑎2
3𝑚3−
𝑚
2(16𝑎2
𝑚4 )
=32𝑎2
3𝑚3 −8𝑎2
𝑚3 [2 Marks]
Step 7:
=8𝑎2
3𝑚3 square units [1 Mark]
Hence, the required area is 8𝑎2
3𝑚3 square units
7. Find the area enclosed by the parabola 4𝑦 = 3𝑥2 and the line 2𝑦 = 3𝑥 + 12. [4 Marks]
Solution:
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Step 1:
Given: Equation of the parabola: 4𝑦 = 3𝑥2. Equation of the line: 2𝑦 = 3𝑥 + 12
Step 2:
The area enclosed between the parabola, 4𝑦 = 3𝑥2 and the line 2𝑦 = 3𝑥 + 12 is represented
by the shaded area 𝑂𝐵𝐴𝑂 as
[1 Mark]
Step 3:
The points of intersection of the given curves are 𝐴(−2, 3) and 𝐵(4, 12). [𝟏
𝟐 Mark]
Step 4:
We draw 𝐴𝐶 and 𝐵𝐷 perpendicular to 𝑥-axis.
∴ Area 𝑂𝐵𝐴𝑂 = Area 𝐶𝐷𝐵𝐴 − (Area 𝑂𝐷𝐵𝑂 + Area 𝑂𝐴𝐶𝑂) [𝟏
𝟐 Mark]
Step 5:
= ∫1
2(3𝑥 + 12)𝑑𝑥 −
1
−2
∫3𝑥2
4𝑑𝑥
1
−2
=1
2[3𝑥2
2+ 12𝑥]
−2
4
−3
4[𝑥3
3]−2
4
=1
2[24 + 48 − 6 + 24] −
1
4[64 + 8]
=1
2[90] −
1
4[72]
= 45 − 18
= 27 square units [2 Marks]
Hence, the required area is 27 square units.
8. Find the area of the smaller region bounded by the ellipse 𝑥2
9+
𝑦2
4= 1 and the line
𝑥
3+
𝑦
2= 1.
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[6 Marks]
Solution:
Step 1:
Given: Equation of the ellipse: 𝑥2
9+
𝑦2
4= 1.
Equation of the line: 𝑥
3+
𝑦
2= 1
Step 2:
The area of the smaller region bounded by the ellipse, 𝑥2
9+
𝑦2
4= 1 and the line,
𝑥
3+
𝑦
2= 1 is
represented by the shaded region 𝐵𝐶𝐴𝐵
[1 Mark]
Step 3:
The points of intersection obtained after solving the two equations are 𝐴(0,2) and 𝐵(3,0). [𝟏
𝟐
Mark]
Step 4:
∴ Area 𝐵𝐶𝐴𝐵 = Area (𝑂𝐵𝐶𝐴𝑂) − Area (𝑂𝐵𝐴𝑂) [𝟏
𝟐 Mark]
Step 5: 𝑥2
9+
𝑦2
4= 1
Hence, 𝑦 = 2√1 −𝑥2
9= 𝑦1
𝑥
3+
𝑦
2= 1
Hence, 𝑦 = 2(1 −𝑥
3) = 𝑦2 [1 Mark]
Step 6:
= ∫ 𝑦1𝑑𝑥 − ∫ 𝑦2𝑑𝑥3
0
3
0 [1 Mark]
Step 7:
= ∫ 2√1 −𝑥2
9𝑑𝑥 − ∫ 2 (1 −
𝑥
3) 𝑑𝑥
3
0
3
0
=2
3[∫√9 − 𝑥2 𝑑𝑥
3
0
] −2
3 ∫(3 − 𝑥)𝑑𝑥
3
0
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=2
3[𝑥
2√9 − 𝑥2 +
9
2sin−1
𝑥
3]0
3
−2
3[3𝑥 −
𝑥2
2]0
3
=2
3[9
2(𝜋
2)] −
2
3[9 −
9
2]
=2
3[9𝜋
4−
9
2]
=2
3×
9
4(𝜋 − 2) [𝟏
𝟏
𝟐 Marks]
Step 8:
=3
2(𝜋 − 2) square units. [
𝟏
𝟐 Mark]
Hence, the required area is 3
2(𝜋 − 2) square units.
9. Find the area of the smaller region bounded by the ellipse 𝑥2
𝑎2 +𝑦2
𝑏2 = 1 and the line 𝑥
𝑎+
𝑦
𝑏= 1.
[6 Marks]
Solution:
Step 1:
Given: The equation of the ellipse: 𝑥2
𝑎2 +𝑦2
𝑏2 = 1 and the line 𝑥
𝑎+
𝑦
𝑏= 1
Step 2:
The area of the smaller region bounded by the ellipse, 𝑥2
𝑎2 +𝑦2
𝑏2 = 1 and the line, 𝑥
𝑎+
𝑦
𝑏= 1 is
represented by the shaded region 𝐵𝐶𝐴𝐵 as
[1 Mark]
Step 3:
The points of intersection after solving the two given equations: 𝐴(0, 𝑏) and 𝐵(𝑎, 0) [𝟏
𝟐 Mark]
Step 4:
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∴ Area 𝐵𝐶𝐴𝐵 = Area (𝑂𝐵𝐶𝐴𝑂) − Area (𝑂𝐵𝐴𝑂) [𝟏
𝟐 Mark]
Step 5:
𝑥2
𝑎2+
𝑦2
𝑏2= 1
Hence, 𝑦 = 𝑏√1 −𝑥2
𝑎2 = 𝑦1
𝑥
𝑎+
𝑦
𝑏= 1
Hence, 𝑦 = 𝑏 (1 −𝑥
𝑎) = 𝑦2 [1 Mark]
Step 6:
= ∫ 𝑦1𝑎
0 𝑑𝑥 − ∫ 𝑦2𝑑𝑥
𝑎
0 [1 Mark]
Step 7:
= ∫ 𝑏√1 −𝑥2
𝑎2
𝑎
0
𝑑𝑥 − ∫ 𝑏 (1 −𝑥
𝑎)𝑑𝑥
𝑎
0
=𝑏
𝑎∫√𝑎2 − 𝑥2𝑑𝑥 −
𝑏
𝑎∫(𝑎 − 𝑥)𝑑𝑥
𝑎
0
𝑎
0
=𝑏
𝑎[{
𝑥
2√𝑎2 − 𝑥2 +
𝑎2
2sin−1
𝑥
𝑎}0
𝑎
− {𝑎𝑥 −𝑥2
2}0
𝑎
]
=𝑏
𝑎[{
𝑎2
2(𝜋
2)} − {𝑎2 −
𝑎2
2}]
=𝑏
𝑎[𝑎2𝜋
4−
𝑎2
2]
=𝑏𝑎2
2𝑎[𝜋
2− 1]
=𝑎𝑏
2[𝜋
2− 1] [𝟏
𝟏
𝟐 Mark]
Step 8:
=𝑎𝑏
4(𝜋 − 2) square units [
𝟏
𝟐 Mark]
Hence, the required area is 𝑎𝑏
4(𝜋 − 2) square units.
10. Find the area of the region enclosed by the parabola 𝑥2 = 𝑦 the line 𝑦 = 𝑥 + 2 and the 𝑥-axis.
[6 Marks]
Solution:
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Step 1:
Given: Equation of the parabola 𝑥2 = 𝑦
Equation of the line: 𝑦 = 𝑥 + 2 and 𝑥-axis.
Step 2:
The area of the region enclosed by the parabola 𝑥2 = 𝑦 the line, 𝑦 = 𝑥 + 2 and 𝑥-axis is
represented by the shaded region 𝑂𝐴𝐵𝐶𝑂 as
[1 Mark]
Step 3:
The point of intersection of the parabola 𝑥2 = 𝑦 and the line 𝑦 = 𝑥 + 2 is 𝐴(−1, 1). [𝟏
𝟐 Mark]
Step 4:
∴ Area 𝑂𝐴𝐵𝐶𝑂 = Area (𝐵𝐶𝐴) + Area 𝐶𝑂𝐴𝐶 [𝟏
𝟐 Mark]
Step 5:
Let 𝑦1 = 𝑥2 and 𝑦2 = 𝑥 + 2
Required area = = ∫ 𝑦2𝑑𝑥 +−1
−2 ∫ 𝑦1𝑑𝑥0
−1 [1 Mark]
Step 6:
= ∫ (𝑥 + 2)𝑑𝑥 +
−1
−2
∫𝑥2𝑑𝑥
0
−1
= [𝑥2
2+ 2𝑥]
−2
−1
+ [𝑥3
3]−1
0
[1 Mark]
Step 7:
= [(−1)2
2+ 2(−1) −
(−2)2
2− 2(−2)] + [−
(−1)3
3]
= [1
2− 2 − 2 + 4 +
1
3] [1 Mark]
Step 8:
=5
6 square units [1 Mark]
Hence, the required area is 5
6 square units.
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11. Using the method of integration find the area bounded by the curve |𝑥| + |𝑦| = 1. [Hint: The
required region is bounded by lines 𝑥 + 𝑦 = 1, 𝑥 − 𝑦 = 1,−𝑥 + 𝑦 = 1 and −𝑥 − 𝑦 = 1]. [4
Marks]
Solution:
Step 1:
Given: Equation of the curve: |𝑥| + |𝑦| = 1
Step 2:
The given curve can be visualized as four different lines.
𝑥 + 𝑦 = 1 for 𝑥 > 0, 𝑦 > 0
−𝑥 + 𝑦 = 1 for 𝑥 < 0, 𝑦 > 0
𝑥 − 𝑦 = 1 for 𝑥 > 0, 𝑦 < 0
−𝑥 − 𝑦 = 1 for 𝑥 < 0, 𝑦 < 0 [1 Mark]
Step 3:
The area bounded by the curve, |𝑥| + |𝑦| = 1 is represented by the shaded region 𝐴𝐷𝐶𝐵 as
[1 Mark]
Step 4:
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 [1 Mark]
Step 5:
= 4∫(1 − 𝑥)𝑑𝑥
1
0
= 4(𝑥 −𝑥2
2)
0
1
= 4 [1 −1
2]
= 4(1
2)
= 2 square units [1 Mark]
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Hence, the required area is 2 square units.
12. Find the area bounded by curves {(𝑥, 𝑦): 𝑦 ≥ 𝑥2 and 𝑦 = |𝑥|}. [6 Marks]
Solution:
Step 1:
Given: The equations of the curves: {(𝑥, 𝑦): 𝑦 ≥ 𝑥2 and 𝑦 = |𝑥|}
Step 2:
𝑦 = |𝑥|
= 𝑥, 𝑥 ≥ 0
−𝑥, 𝑥 ≤ 0
Hence, the area bounded by the curves, {(𝑥, 𝑦): 𝑦 ≥ 𝑥2 and y = |𝑥|} is represented by the
shaded region:
[1𝟏
𝟐 Mark]
Step 3: The points of intersection obtained by solving the given equations are: 𝐴(1,1) and
𝐵(−1,1).
It is observed that the required area is symmetrical about 𝑦-axis.
Required area = 2[Area (𝑂𝐶𝐴𝑂) − Area(𝑂𝐶𝐴𝐷𝑂)] [1 Mark]
Step 4:
𝑦1 = 𝑦 = 𝑥
𝑦2 = 𝑦 = 𝑥2
Step 5:
= 2 [∫ 𝑦1𝑑𝑥 − ∫ 𝑦2𝑑𝑥1
0
1
0] [1 mark]
Step 6:
= 2 [∫𝑥𝑑𝑥 − ∫𝑥2𝑑𝑥
1
0
1
0
]
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= 2 [[𝑥2
2]0
1
− [𝑥3
3]0
1
]
= 2 [1
2−
1
3]
= 2 [1
6] =
1
3 square units [2
𝟏
𝟐 Marks]
Hence, the required area is 1
3 square units
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). [6 Marks]
Solution:
Step 1:
Given: The vertices of ∆𝐴𝐵𝐶 are 𝐴(2, 0), 𝐵(4, 5) and 𝐶(6, 3).
Step 2:
[1 Mark]
Step 2:
Equation of line segment 𝐴𝐵 is
𝑦 − 0 =5 − 0
4 − 2(𝑥 − 2)
2𝑦 = 5𝑥 − 10
𝑦 =5
2(𝑥 − 2) …(i) [
𝟏
𝟐 Mark]
Step 3:
Equation of line segment 𝐵𝐶 is
𝑦 − 5 =3 − 5
6 − 4(𝑥 − 4)
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2𝑦 − 10 = −2𝑥 + 8
2𝑦 = −2𝑥 + 18
𝑦 = −𝑥 + 9 …(ii) [𝟏
𝟐 Mark]
Step 4:
Equation of line segment 𝐶𝐴 is
𝑦 − 3 =0 − 3
2 − 6(𝑥 − 6)
−4𝑦 + 12 = −3𝑥 + 18
4𝑦 = 3𝑥 − 6
𝑦 =3
4(𝑥 − 2) …(iii) [
𝟏
𝟐 Mark]
Step 5:
Area (∆𝐴𝐵𝐶) = Area (𝐴𝐵𝐿𝐴) + Area (𝐵𝐿𝑀𝐶𝐵) − Area (𝐴𝐶𝑀𝐴) [1 Mark]
Step 6:
= ∫5
2
4
2
(𝑥 − 2)𝑑𝑥 + ∫ (−𝑥 + 9)6
4𝑑𝑥 − ∫
3
4
6
2
(𝑥 − 2)𝑑𝑥 [1 Mark]
Step 7:
=5
2[𝑥2
2− 2𝑥]
2
4
+ [−𝑥2
2+ 9𝑥]
4
6
−3
4[𝑥2
2− 2𝑥]
2
6
[1 Mark]
=5
2[8 − 8 − 2 + 4] + [−18 + 54 + 8 − 36] −
3
4[18 − 12 − 2 + 4]
= 5 + 8 −3
4(8)
= 13 − 6
Step 8:
= 7 square units [𝟏
𝟐 Mark]
Hence, the required area is 7 square units.
14. Using the method of integration find the area of the region bounded by lines: 2𝑥 + 𝑦 = 4, 3𝑥 −
2𝑦 = 6 and 𝑥 − 3𝑦 + 5 = 0 [6 Marks]
Solution:
Step 1:
Given:
The given equations of lines are
2𝑥 + 𝑦 = 4 …(i)
3𝑥 − 2𝑦 = 6 …(ii)
And 𝑥 − 3𝑦 + 5 = 0 …(iii)
Step 2:
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[1 Mark]
Step 3: The points of intersection of the three lines obtained by solving them simultaneously
are: 𝐴(1,2), 𝐵(2,0) and 𝐶(4,3)
The area of the region bounded by the lines is the area of ∆𝐴𝐵𝐶.
𝐴𝐿 and 𝐶𝑀 are the perpendiculars on 𝑥-axis. [1 Mark]
Step 4:
Area(∆𝐴𝐵𝐶) = Area(𝐴𝐿𝑀𝐶𝐴) − Area(𝐴𝐿𝐵) − Area(𝐶𝑀𝐵) [1 Mark]
Step 5:
= ∫ (𝑥+5
3)
4
2
𝑑𝑥 − ∫ (4 − 2𝑥)2
1𝑑𝑥 − ∫ (
3𝑥−6
2)
4
2
𝑑𝑥 [1 Mark]
Step 6:
=1
3[𝑥2
2+ 5𝑥]
1
4
− [4𝑥 − 𝑥2]12 −
1
2[3𝑥2
2− 6𝑥]
2
4
[1 Mark]
=1
3[8 + 20 −
1
2− 5] − [8 − 4 − 4 + 1] −
1
2[24 − 24 − 6 + 12]
Step 7:
= (1
3×
45
2) − (1) −
1
2(6)
=15
2− 1 − 3
=15
2− 4 =
15−8
2=
7
2 square units [1 Mark]
hence, the required area is 7
2 square units
15. Find the area of the region {(𝑥, 𝑦): 𝑦2 ≤ 4𝑥, 4𝑥2 + 4𝑦2 ≤ 9} [6 Marks]
Solution:
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Step 1:
Given: Equation of the curve: {(𝑥, 𝑦): 𝑦2 ≤ 4𝑥, 4𝑥2 + 4𝑦2 ≤ 9}
Step 2:
The equation 𝑦2 ≤ 4𝑥 represents the region interior to a parabola, symmetric about 𝑥-axis.
The equation 4𝑥2 + 4𝑦2 ≤ 9 represents the region interior to a circle with center at the origin
and radius 3
2.
The area bounded by the curves {(𝑥, 𝑦): 𝑦2 ≤ 4𝑥, 4𝑥2 + 4𝑦2 ≤ 9} is represented as
The points of intersection of both the curves obtained by solving the given equations
simultaneously are (1
2, √2) and (
1
2, −√2). [1 Mark]
Step 3:
The required area is given by 𝑂𝐴𝐵𝐶𝑂.
It is observed that area 𝑂𝐴𝐵𝐶𝑂 is symmetrical about 𝑥-axis.
∴ Area 𝑂𝐴𝐵𝐶𝑂 = 2 × Area 𝑂𝐵𝐶 [1 Mark]
Step 4:
Area 𝑂𝐵𝐶𝑂 = Area 𝑂𝑀𝐶 + Area 𝑀𝐵𝐶 [1 mark]
Step 5:
= ∫ 2√𝑥
12
0
𝑑𝑥 + ∫ √9
4− 𝑥2
32
12
𝑑𝑥
= 2∫ √𝑥
12
0
𝑑𝑥 + ∫ √(3
2)2 − (𝑥)2
32
12
𝑑𝑥
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= 2 [𝑥
32
3
2
]0
1
2
+ [ 𝑥
2√(
3
2)2− 𝑥2 +
(3
2)2
2sin−1 𝑥
3
2
]1
2
3
2
=4
3[𝑥
3
2]0
1
2+ [
3
2
2√(
3
2)2−
3
2
2+
(3
2)2
2sin−1
3
23
2
] − [ 1
2
2√(
3
2)2−
1
2
2+
(3
2)2
2sin−1
1
23
2
]
=4
3[(
1
2)
3
2− 0] + [
3
4√0 +
9
8sin−1 1] − [
1
4√2 +
9
8sin−1 1
3]
=√2
3+
9𝜋
16−
√2
4−
9
8sin−1 1
3 [2 Marks]
Step 6:
Hence, the required area = 2 ×[√2
3+
9𝜋
16−
√2
4−
9
8sin−1 1
3]
=1
3√2+
9𝜋
8−
9
4sin−1 1
3 [1 Mark]
16. Area bounded by the curve 𝑦 = 𝑥3, the 𝑥-axis and the ordinates 𝑥 = −2 and 𝑥 = 1 is [4
Marks]
(A) −9
(B) −15
4
(C) 15
4
(D) 17
4
Solution:
(D)
Step 1:
Given: The equations of the curve is 𝑦 = 𝑥3 and the ordinates are 𝑥 = −2 and 𝑥 = 1.
Step 2: The given equations can be represented as follows:
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[1 Mark]
Step 3:
Required area = ∫ |𝑦|𝑑𝑥1
−2
= ∫ |𝑥3|𝑑𝑥1
−2 [1 Mark]
Step 4:
= ∫ |𝑥3|𝑑𝑥0
−2+ ∫ |𝑥3|𝑑𝑥
1
0
= −∫ 𝑥3𝑑𝑥0
−2+ ∫ 𝑥3𝑑𝑥
1
0
= −[𝑥4
4]−2
0
+ [𝑥4
4]0
1
= −[0 −(−2)4
4] +
1
4− 0
= 4 +1
4=
17
4 square units
Hence, (𝐷) is the correct answer. [2 Marks]
17. The area bounded by the curve 𝑦 = 𝑥|𝑥|, 𝑥-axis and the ordinates 𝑥 = −1 and 𝑥 = 1 is given by
[Hint: 𝑦 = 𝑥2 if 𝑥 > 0 and 𝑦 = −𝑥2 if 𝑥 < 0]. [4 Marks]
(A) 0
(B) 1
3
(C) 2
3
(D) 4
3
Solution:
(C)
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Step 1:
Given: The equation of the curve: 𝑦 = 𝑥|𝑥|, 𝑥-axis and the ordinates 𝑥 = −1 and 𝑥 = 1
Step 2: The given equation can be represented as follows:
[1 Mark]
Step 3:
Required area = ∫ |𝑦|𝑑𝑥1
−1 [1 Mark]
Step 4:
= ∫|𝑥|𝑥||𝑑𝑥
1
−1
= ∫ 𝑥2𝑑𝑥 + ∫ 𝑥2𝑑𝑥1
0
0
−1 [1 Mark]
Step 5:
= [𝑥3
3]−1
0
+ [𝑥3
3]0
1
= −(−1
3) +
1
3
=2
3 square units
Hence, (𝐶) is the correct answer. [1 Mark]
18. The area of the circle 𝑥2 + 𝑦2 = 16 exterior to the parabola 𝑦2 = 6𝑥 is [6 Marks]
(A) 4
3(4𝜋 − √3)
(B) 4
3(4𝜋 + √3)
(C) 4
3(8𝜋 − √3)
Class-XII-Maths Application of Integrals
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(D) 4
3(8𝜋 + √3)
Solution:
(C)
Step 1:
Given: Equation of the circle: 𝑥2 + 𝑦2 = 16
Equation of the parabola: 𝑦2 = 6𝑥
Step 2:
The given equations is represented as follows:
The points of intersection are marked as shown.
Area bounded exterior to the parabola and the circle is as shaded above. [1 Mark]
Step 3: The area
= 2[Area(𝑂𝑀𝐶𝑂) + Area(𝐶𝑀𝐵𝐶)]
= 2 [∫ √6𝑥2
0𝑑𝑥 + ∫ √16 − 𝑥2𝑑𝑥
4
2] [1 Mark]
Step 4:
= 2 [√6{𝑥
32
32
}
0
2
] + 2 [𝑥
2√16 − 𝑥2 +
16
2sin−1
𝑥
4]2
4
= 2√6 ×2
3[𝑥
32]
0
2
+ 2 [8.𝜋
2− √16 − 4 − 8 sin−1 (
1
2)]
=4√6
3(2√2) + 2 [4𝜋 − √12 − 8
𝜋
6]
=16√3
3+ 8𝜋 − 4√3 −
8
3𝜋
=4
3[4√3 + 6𝜋 − 3√3 − 2𝜋]
=4
3[√3 + 4𝜋]
=4
3[4𝜋 + √3] square units [2 marks]
Step 5:
Class-XII-Maths Application of Integrals
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Area of circle = π(𝑟)2
= 𝜋(4)2
= 16𝜋 square units [𝟏
𝟐 Mark]
∴ Required area = 16𝜋 −4
3[4𝜋 + √3]
=4
3[4 × 3𝜋 − 4𝜋 − √3]
=4
3(8𝜋 − √3) square units
Hence, (𝐶) is the correct answer. [𝟏𝟏
𝟐 Mark]
19. The area bounded by the 𝑦-axis, 𝑦 = cos𝑥 and 𝑦 = sin𝑥 when 0 ≤ 𝑥 ≤𝜋
2 [4 Marks]
(A) 2(√2 − 1)
(B) √2 − 1
(C) √2 + 1
(D) √2
Solution:
(B)
Step 1:
The given equations are 𝑦 = cos 𝑥 …(i)
And 𝑦 = sin 𝑥 ….(ii)
Step 2: The given equations is represented as follows. The point of intersection for given curve is
as:
sin 𝑥 = cos 𝑥
⇒ tan𝑥 = 1
⇒ 𝑥 =𝜋
4
⇒ 𝑦 = sin𝜋
4=
1
√2
Class-XII-Maths Application of Integrals
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[1 Mark]
Step 3:
Required area = Area (𝐴𝐵𝐿𝐴) + area (𝑂𝐵𝐿𝑂)
= ∫ 𝑥𝑑𝑦 + ∫ 𝑥𝑑𝑦
1
√2
0
11
√2
[1 mark]
Step 4:
= ∫cos−1 𝑦𝑑𝑦 +∫ sin−1 𝑦𝑑𝑦
1
√2
0
1
1
√2
Integrating by parts, we obtain
= [𝑦 cos−1𝑦 − √1 − 𝑦2] 1
√2
1+ [𝑦 sin−1𝑦 + √1 − 𝑦2]
0
1
√2
= [cos−1(1) −1
√2cos−1 (
1
√2) + √1 −
1
2] + [
1
√2sin−1 (
1
√2) + √1 −
1
2− 1]
=−𝜋
4√2+
1
√2+
𝜋
4√2+
1
√2− 1
=2
√2− 1
= √2 − 1 square units
Hence, (𝐵) is the correct answer. [2 marks]