Application of Integrals — Chapter Test | Class 12 IMO

Q1

Find the area of the region bounded by the parabola x² = 4y and the straight line x = 4y − 2.

Substitute y = (x+2)/4 into parabola: x² = x+2 → x²−x−2=0. Roots x=−1, 2. Area = ∫(−1 to 2) ((x+2)/4 − x²/4) dx = 1/4 [x²/2 + 2x − x³/3] from −1 to 2 = 9/8 sq units.

Q2

The area of the smaller part of the circle x² + y² = a² cut off by the line x = a/√2 is:

Smaller part is to the right of x = a/√2. Area = 2 ∫[a/√2 to a] √(a² − x²) dx = 2[(x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a)]_(a/√2)^a = 2[0 + (a²/2)(π/2) − (a/(2√2))(a/√2) − (a²/2)(π/4)] = 2[a²π/4 − a²/4 − a²π/8] = 2[a²π/8 − a²/4] = a²π/4 − a²/2 = (a²/2)(π/2 − 1) sq units.

Q3

For a physics demonstration at Yudgam Foundation, a student tracks a particle whose velocity curve forms a region bounded by y = |x − 3| and the x-axis from x = 0 to x = 6. What is the total distance (area)?

Graph forms two right triangles of base 3 and height 3. Area = 2 × (1/2 × 3 × 3) = 9 units.

Q4

The area of the region bounded by y = min{|x|, 1} and y = max{|x|, 1} for |x| ≤ 2 is:

For |x|≤1, y from |x| to 1. For 1≤|x|≤2, y from 1 to |x|. Area = 2[∫[0 to 1] (1 − x) dx + ∫[1 to 2] (x − 1) dx] = 2[(1−1/2) + (2−2 − (1/2−1))] = 2[1/2 + 1/2] = 2 sq units.

Q5

What is the area of the region bounded by the curves y = √x and y = x²?

Intersections at x=0, x=1. Area = ∫(0 to 1) (√x − x²) dx = [2/3 x^(3/2) − x³/3] = 2/3 − 1/3 = 1/3 sq units.

Q6

For an odd function f(x), the integral ∫(−a to a) f(x) dx = 0. What does this imply about the geometric areas bounded by the curve and the x-axis on the intervals [−a, 0] and [0, a]?

Odd symmetry means the region below the axis perfectly mirrors the region above the axis, causing the signed areas to sum to zero.

Q7

What is the area of the region completely bounded by the downward opening parabola y = 4x − x² and the x-axis?

Roots are x=0, 4. Area = ∫(0 to 4) (4x − x²) dx = [2x² − x³/3] = 32 − 64/3 = 32/3 sq units.

Q8

Sachin allocates a triangular plot bounded by the line y = 2x, the x-axis, and the boundary x = 10 for a local tree plantation drive. What is the plot's area in square meters?

Area = ∫(0 to 10) 2x dx = [x²] from 0 to 10 = 100 sq meters.

Q9

To compute the area between two curves f(x) and g(x) where f(x) ≥ g(x) for all x in [a, b], which integral setup is correct?

The correct formulation for the area between curves is the integral of the upper curve minus the lower curve: ∫ (upper − lower) dx.

Q10

The area bounded by y = log x, y = 0 and x = e is:

Area = ∫[1 to e] log x dx = [x log x − x]₁ᵉ = (e−e) − (0−1) = 1 sq unit.

Q11

The area of the region bounded by y = x² + 1, y = x, x = 0 and x = 2 is:

Area = ∫[0 to 2] ((x² + 1) − x) dx = [x³/3 + x − x²/2]₀² = 8/3 + 2 − 2 = 8/3 sq units.

Q12

Find the area bounded by the straight line y = 2x, the x-axis, and the line x = 5.

Area = ∫(0 to 5) 2x dx = [x²] from 0 to 5 = 25 − 0 = 25 sq units.

Q13

The area of the region bounded by the lines y = 2x, y = x/2 and x = 2 is:

Area = ∫[0 to 2] (2x − x/2) dx = ∫[0 to 2] (3x/2) dx = (3/2)[x²/2]₀² = (3/2)×2 = 3 sq units.

Q14

Sachin cuts a wooden board bounded by the curve y = x³ and the line y = 8, along with the y-axis, to make an E-learning Hub sign. What is the area of the board?

Integrating with respect to y: Area = ∫(0 to 8) x dy = ∫(0 to 8) y^(1/3) dy = [3/4 y^(4/3)] from 0 to 8 = 3/4 × 16 = 12 sq units.

Q15

A student calculates ∫(0 to 2π) sin(x) dx = 0 and concludes the area bounded by y = sin(x) and the x-axis from 0 to 2π is 0. What is the correct area?

Definite integral gives net signed area. True geometric area = ∫(0 to π) sin(x) dx + |∫(π to 2π) sin(x) dx| = 2 + |−2| = 4 sq units.