Grade 11

Grade 11Calculus


Applications of Integration


Integration is a fundamental concept in calculus, dealing with the accumulation of quantities, such as the area under a curve, volume, and more. Let's look at some of the major applications of integration in simple terms and with visual examples to build a strong foundation.

1. Area under the curve

The area under a curve is a common application of integration. If you have a function y = f(x) and you want to find the area between the curve and the x-axis from x = a to x = b, you use definite integration:

a = ∫[a, b] f(x) dx

You can view it as follows:

y = f(x) A B

Imagine plotting points along a curve from point A to point B and then summing the areas of the thin rectangles drawn under the curve. This idea of "summing thin rectangles" turns into a process of integration.

Example problem: Calculate the area

Consider the function f(x) = x^2. Find the area between f(x) and the x-axis from x = 0 to x = 2.

f(x) = x^2

∫[0, 2] x^2 dx = [x^3/3][0, 2] = (8/3) - (0/3) = 8/3

Thus, the area is 8/3 square units.

2. Volume of solids of revolution

Another application of integration is to find the volume of a solid of revolution. When you rotate a curve around an axis, it creates a three-dimensional shape (such as a vase or donut).

The volume can be calculated using the disk method for rotation around the x-axis:

v = π ∫[a, b] (f(x))^2 dx

Visualizer for understanding:

Revolving Area

The curve is rotated around the x-axis, producing a disc-shaped solid. Integration calculates the total volume by collecting these infinitesimal discs.

Example problem: Find the volume

Consider rotating the curve f(x) = x^2 around the x-axis from x = 0 to x = 1.

V = π ∫[0, 1] (x^2)^2 dx
  = π ∫[0, 1] x^4 dx
  = π[x^5/5][0, 1]
  = π (1/5 - 0/5)
  = π/5

Hence the volume of the solid is π/5 cubic units.

3. Average value of a function

Integration helps to determine the average value of a function over a certain interval. If a function is continuous on [a, b], then the average value f_avg is given by:

f_avg = (1/(ba)) ∫[a, b] f(x) dx

This provides the average height of the function over the interval, which is useful in a variety of practical scenarios.

Example problem: Find the average value

Find the average value of f(x) = 3x^2 on the interval [0, 2].

f_avg = (1/(2-0)) ∫[0, 2] 3x^2 dx
      = (1/2)[3x^3/3][0, 2]
      = (1/2)(24/3 - 0/3)
      = 4

The average value of the function is 4.

4. Work done by the force

In physics, integration can be used to calculate the work done by a force accelerating an object. If a force F(x) accelerates an object along the x-axis from x = a to x = b, then the work done W is given by:

w = ∫[a, b] f(x) dx

This approach is widely applied in physics and engineering, and is useful in calculating the work done by variable forces.

Example problem: Calculate the work done

If a variable force F(x) = 2x moves a particle from x = 1 to x = 3, find the work done.

w = ∫[1, 3] 2x dx
  = [x^2][1, 3]
  = (9 - 1)
  = 8

The work done is 8 units.

5. Fluid pressure and force

Integration helps to determine the pressure force exerted by a fluid on an immersed surface. The force F is calculated using:

,

where ρ is the density of the fluid, g is gravity, h is the height of the fluid, and w(y) is the width at depth y.

Example problem: Find the fluid force

Find the force on a vertical rectangular plate immersed vertically in water (ρ = 1000 kg/m³). The height is 10 m, the width is 2 m, the depth is from 0 to 10 m.

F = ∫[0, 10] (1000 kg/m³)(9.8 m/s²)(10-y)(2) dy
  = 19600 ∫[0, 10] (10-y)dy
  = 19600 [(10y - y²/2)] [0, 10]
  = 19600 (100 - 0 - 50 + 0)
  = 1960000 N

The force applied is 1960000 N

Conclusion

Integration provides a powerful toolset for solving real-world problems. Its applications are vast and varied, from calculating areas to determining volumes, finding averages of functions, and much more. Understanding these fundamental concepts prepares you well for more advanced aspects of calculus and its practical applications.


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