Measuring the World: Applications of Integrals
Integrals go far beyond finding areas under curves — they calculate volumes of solids, total accumulated quantities, average values, and work done by forces. These applications make integrals indispensable tools in physics, engineering, economics, and biology for measuring things that change continuously.
Components of Applications of Integrals
This section covers the major application areas:
- Area Between Curves: The area between two functions f(x) and g(x) from a to b is ∫ from a to b of |f(x) - g(x)| dx.
- Volumes of Revolution: Rotating a region around an axis creates a solid whose volume can be found using the disk method: V = π ∫ [f(x)]² dx.
- Average Value of a Function: The average value of f(x) on [a, b] is (1/(b-a)) × ∫ from a to b of f(x) dx.
- Accumulation & Total Change: Integrating a rate function over an interval gives the total quantity accumulated during that time.
Examples of Applications of Integrals
Area Between Curves Examples
- Find the area between y = x² and y = x from 0 to 1: ∫ from 0 to 1 of (x - x²) dx = [x²/2 - x³/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
- Find the area between y = 4 and y = x² from -2 to 2: ∫ from -2 to 2 of (4 - x²) dx = [4x - x³/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 32/3 ≈ 10.67.
- The area between two revenue curves from month 1 to month 6 gives the total revenue difference between two products over that period.
Volume of Revolution Examples
- Rotate y = x from 0 to 3 around the x-axis: V = π ∫ from 0 to 3 of x² dx = π[x³/3] from 0 to 3 = π(9) = 9π ≈ 28.27 cubic units.
- Rotate y = √x from 0 to 4 around the x-axis: V = π ∫ from 0 to 4 of x dx = π[x²/2] from 0 to 4 = π(8) = 8π ≈ 25.13 cubic units.
- A bowl shape formed by rotating y = x² from 0 to 2 has volume V = π ∫ from 0 to 2 of x⁴ dx = π[x⁵/5] from 0 to 2 = 32π/5 ≈ 20.11 cubic units.
Average Value Examples
- Find the average of f(x) = x² on [0, 3]: Average = (1/3) × ∫ from 0 to 3 of x² dx = (1/3)(9) = 3.
- Find the average temperature if T(t) = 70 + 10sin(t) over [0, π]: Average = (1/π) × ∫ from 0 to π of (70 + 10sin(t)) dt = (1/π)(70π + 20) = 70 + 20/π ≈ 76.37°.
- A factory's production rate is f(t) = 50 + 4t units/hour. Average rate from t = 0 to t = 8 is (1/8) × ∫(50 + 4t) dt = (1/8)(400 + 128) = 66 units/hour.
Accumulation Examples
- Oil leaks at r(t) = 100 - 5t gallons/hour. Total leaked from t = 0 to t = 10 is ∫(100 - 5t) dt = [100t - 5t²/2] from 0 to 10 = 1,000 - 250 = 750 gallons.
- Revenue flows in at R(t) = 200e^(0.05t) dollars/day. Total revenue from day 0 to day 30 is ∫ 200e^(0.05t) dt = 4,000[e^(1.5) - 1] ≈ $12,936.
- A sprinkler delivers water at w(t) = 3 + 0.5t gallons/minute. Total water in 10 minutes is ∫(3 + 0.5t) dt = [3t + 0.25t²] from 0 to 10 = 30 + 25 = 55 gallons.