Question 1
Find the enclosed area of x²/16 + y²/9 = 1.
Steps only
a = 4; b = 3
A = 4(3/4)∫[0, 4] √(16 − x²) dx
= πab = 12π
Answer
12π square units.
CLASS 12 · MATHEMATICS PART II · CHAPTER 8
Application of Integrals
Use definite integrals to find geometric areas. Short calculations show the limits and signs; the explanation button helps you choose strips and split a region correctly.
10 worked answersAll 1 exercise setsAll 5 end-of-chapter questions
10 worked answers · short steps first, explanations when you need them
Exercise 8.1
Textbook pages 296
Question 2
Find the enclosed area of x²/4 + y²/9 = 1.
Steps only
a = 2; b = 3
A = 4(3/2)∫[0, 2] √(4 − x²) dx
= πab = 6π
Answer
6π square units.
Question 3
Find the first-quadrant area inside x² + y² = 4, from x = 0 to x = 2.
Steps only
r = 2
A = (1/4)π(2)² = π
Answer
π square units; option A.
Question 4
Find the area between y² = 4x, the y-axis and y = 3.
Steps only
x = y²/4
A = ∫[0, 3] (y²/4) dy
= [y³/12] from 0 to 3
= 27/12 = 9/4
Answer
9/4 square units; option B.
Miscellaneous Exercise
Textbook pages 298
Question 1(i)
Find the area below y = x² and above the x-axis from x = 1 to x = 2.
Steps only
A = ∫[1, 2] x² dx
= [x^3/3] from 1 to 2
= (8 − 1)/3
= 7/3
Answer
7/3 square units.
Question 1(ii)
Find the area below y = x⁴ and above the x-axis from x = 1 to x = 5.
Steps only
A = ∫[1, 5] x⁴ dx
= [x^5/5] from 1 to 5
= (3125 − 1)/5
= 3,124/5
Answer
3,124/5 square units.
Question 2
Sketch y = |x + 3| and evaluate its integral from −6 to 0.
Steps only
|x + 3| = −x − 3 for x ≤ −3
|x + 3| = x + 3 for x ≥ −3
A = (1/2)(3)(3) + (1/2)(3)(3)
= 9
Left(-6, 3) · Vertex(-3, 0) · Right(0, 3)
Answer
9 square units.
Question 3
Find the geometric area between y = sin x and the x-axis from 0 to 2π.
Steps only
A = ∫[0, π] sin x dx − ∫[π, 2π] sin x dx
= 2 − (−2)
= 4
Answer
4 square units.
Question 4
Find the area between y = x³ and the x-axis from x = −2 to x = 1.
Steps only
A = −∫[−2, 0] x³ dx + ∫[0, 1] x³ dx
= 4 + 1/4
= 17/4
Answer
17/4 square units; option D.
Question 5
Find the geometric area between y = x|x| and the x-axis from −1 to 1.
Steps only
y = −x² for x < 0; y = x² for x ≥ 0
A = ∫[−1, 0] x² dx + ∫[0, 1] x² dx
= 1/3 + 1/3
= 2/3
Answer
2/3 square units; option C.