Applications of Integrals 📐

Integration ka use karo — curves ke beech area calculate karo!

🌾 Khet ki Area

Ek farmer ka khet ek parabola y=x² aur ek straight line y=x ke beech bounded hai. Woh jaanna chahta hai — uska khet ka area kya hai? Yahan standard shapes ke formulas kaam nahi karenge. Lekin integration se hum exactly calculate kar sakte hain! Yahi hai applications of integrals — curves ke bounded regions ka area nikalna.

1. Area Under a Curve

f(x) ≥ 0 on [a,b] ke liye, x-axis ke upar curve ka area:

Area = ∫[a to b] f(x) dx

Agar f(x) ≤ 0, toh area = |∫f(x)dx| (absolute value lo)

2. Area Between Two Curves

If f(x) ≥ g(x) on [a,b]: Area = ∫[a to b] [f(x) - g(x)] dx = ∫[a to b] [upper - lower] dx Finding limits: Solve f(x) = g(x) for intersection points

✅ Area between y=x² and y=x (line)

Intersection: x²=x → x=0, x=1

On [0,1]: y=x is above y=x²

Area = ∫[0 to 1](x - x²)dx = [x²/2 - x³/3]₀¹ = 1/2 - 1/3 = 1/6 sq units

3. Area of Standard Curves

Circle: x²+y²=a²

Area = πa² (by integration or formula)

Using integration: 4∫[0 to a] √(a²-x²)dx = πa²

Parabola: y²=4ax

Area between parabola and latus rectum = (8/3)a²

Ellipse: x²/a² + y²/b² = 1

Total area = πab

✅ Area of circle x²+y²=4 (using integration)

Area = 4∫[0 to 2]√(4-x²)dx

= 4[x√(4-x²)/2 + 2·sin⁻¹(x/2)]₀²

= 4[0 + 2·sin⁻¹(1)] = 4[2·π/2] = 4π sq units ✓ (πr²=π×4=4π ✓)

4. Practice Questions

Q1. Find area enclosed between y=x² and y=√x.

Solution

Intersection: x²=√x → x⁴=x → x=0,1. On [0,1] √x ≥ x². Area=∫[0,1](√x-x²)dx=[2x^(3/2)/3 - x³/3]₀¹ = 2/3-1/3=1/3 sq units.

Q2. Find area bounded by y=sin x between x=0 and x=π.

Solution

∫[0 to π]sin x dx = [-cos x]₀^π = 1+1=2 sq units.

Q3. Area of region bounded by y=x+1, y=0, x=0, x=2.

Solution

∫[0 to 2](x+1)dx = [x²/2+x]₀² = 2+2=4 sq units.