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Definite Integrals


The definite integral is a fundamental concept in calculus, used to calculate accumulations of quantities such as area, volume, work, and other physical properties. Unlike the indefinite integral, which represents a family of functions, a definite integral returns a single number. It essentially calculates the net area under a curve over an interval, providing a very powerful tool for analysis.

a b f(x) dx

Basic concepts of definite integrals

When we talk about definite integrals, we are interested in finding the area under a curve described by a function between two points on the x-axis, say a and b.

The notation for the definite integral of a function f(x) from a to b is:

a b f(x) dx

You can visualize that gap like this: A B

Understanding integrals

To calculate a definite integral, we essentially divide the area under the curve into thin vertical strips and sum their areas. Each strip is approximately a rectangle and the more strips we use, the better our estimate will be. The thickness of each strip approaches zero over the range of our calculations.

Consider the following function, for simplicity let's take f(x) = x^2. We are interested in the interval from a = 0 to b = 2.

    f(x) = x^2
The integral setup is:
0 2 x^2 dx

Example calculation

Let's calculate the definite integral of f(x) = x^2 from 0 to 2 step by step:

Step 1: Find the antiderivative of f(x)

The antiderivative of f(x) = x^2 is found by raising the power by one and dividing by the new power.

        f(x) = (1/3)x^3 + c

Step 2: Evaluate the antiderivative from a to b

Using the fundamental theorem of calculus, we evaluate the antiderivative at the boundaries of the interval and reduce:

        f(b) - f(a) = [(1/3)(2)^3 + c] - [(1/3)(0)^3 + c]
                    = [(1/3)(8)] - [0]
                    = 8/3

Fundamental theorem of calculus

The fundamental theorem of calculus connects the concept of differentiation to integration. It states that if F is the antiderivative of f on the interval [a, b], then

a b f(x) dx = f(b) - f(a)

This theorem not only provides a way to compute a definite integral by evaluating the antiderivative, but also establishes a deep connection between the geometry of a function (represented by the integral) and its calculus description.

Geometrical interpretation

Suppose the function f(x) is a simple curve on the xy-plane, and we are finding the area under this curve from point a to point b.

In this illustration, the blue shaded area represents the area under the curve: A B

Working with practical examples

Example 1: Finding the area under a curve

Suppose you have a function f(x) = 3x^2 and you want to find the area under the curve from x = 1 to x = 4.

Step 1: Find the antiderivative of f(x)

         f(x) = ∫ 3x^2 dx = x^3 + c

Step 2: Evaluate the antiderivative from 1 to 4

        f(4) - f(1) = [(4)^3] - [(1)^3]
                    = [64] - [1]
                    = 63

Example 2: Total distance travelled

If a chemical solution is being pumped into a tank at the rate of R(t) = 5t - 2 liters per hour, how much solution is pumped into the tank from t = 1 to t = 5 hours?

Step 1: Find the antiderivative of R(t)

        ∫ (5t - 2)dt = (5/2)t^2 - 2t + c

Step 2: Rate from 1 to 5

        f(5) - f(1) = [(5/2)(5)^2 - 2(5)] - [(5/2)(1)^2 - 2(1)]
                    = [(5/2)(25) – 10] – [(5/2) – 2]
                    = [62.5 – 10] – [2.5 – 2]
                    = 52.5 - 0.5
                    = 52

Pythagorean tuning: Frequency ratios

Even though this may extend beyond traditional calculus, consider how frequency intervals are related through ratios. If you work with a frequency ratio function with intervals:

    r(f) = 440 * 2^(x/12)
where x is the number of semitones from 440 Hz. To find the accumulated "increase" in frequency from x = 0 to x = 12 (one octave), calculate:

Step 1: Integral setup

0 12 440 * 2^(x/12) dx

Step 2: Solve the integral

Applying the Fundamental Theorem and Rules of Calculus:

Let y = 2^(x/12) \
Then, we differentiate d/dx[y] = ln(2)/12 * 2^(x/12) \
∫ 440 * y dy = 440 * 12/ln(2) * [y^2/2 - y/2] = 5280/ln(2) * [2 - 1]

The solution describes the evolution under a musical interval, providing fascinating insight!

Conclusion

Definite integrals are an essential component of calculus that provide us with tools to calculate areas under curves, total changes in quantities, and a vast number of other applications that have profound implications in both theoretical mathematics and practical scenarios in a variety of fields including physics, engineering, economics, and beyond. They connect us deeply to geometric intuition and algebraic computation.

As you further explore this rich field, you will dive deeper into various techniques such as numerical integration, improper integration, and applications in differential equations, as well as more advanced integration techniques and real-world applications, enriching your understanding and mathematical toolkit for solving complex problems.


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