Question

A periodic function with period 1 is integrable over any finite interval. Also for two real numbers \(\mathrm{a}, \mathrm{b}\) and for two unequal non-zero postive integers \(\mathrm{m}\) and \(\mathrm{n}, \int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{b}}^{b+\mathrm{m}} \mathrm{f}(\mathrm{x}) \mathrm{dx} .\) Calculate the value of \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)

Short answer

Answer: \((n - m) \int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\).

Step-by-step solution

Using the given property of the periodic function

We are given the condition that for two real numbers \(\mathrm{a}, \mathrm{b}\) and for two unequal non-zero positive integers \(\mathrm{m}\) and \(\mathrm{n}\): \(\int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} = \int_{b}^{b+\mathrm{m}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\).

Expressing the integral \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\) using the property

Since the function has a period of 1, we can express the desired integral \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\) as a sum of integrals of the form \(\int_{i}^{i+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\), where \(i\) varies from \(m\) to \(n - 1\). Therefore, \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx} = \sum\limits_{i=m}^{n-1} \int_{i}^{i+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\)

Using the given property to simplify our expression

Now, we can use the property given in the problem to replace each integral in the sum with an equivalent integral. In particular, \(\int_{i}^{i+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) will be equal to \(\int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\), so our expression becomes: \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx} = \sum\limits_{i=m}^{n-1} \int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\)

Simplifying the sum

The sum in our expression has the same integral (\(\int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\)) repeated \((n - m)\) times. We can simplify this by taking the integral out of the sum and multiplying it by the number of repetitions, which is \(n - m\): \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx} = (n - m) \int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) Our final answer for the value of the integral \(\int_{\mathrm{m}}^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\) is: \((n - m) \int_{a}^{a+1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\).

Key concepts

Periodic Function Integration

Periodic functions have a characteristic that allows us to simplify integrals across intervals of their period. A periodic function repeats its values at regular intervals, and this period is denoted as the length of the smallest interval that accomplishes this repetition.

When integrating a periodic function over a full period, it doesn't matter what starting point we choose. This is because the area under one period is the same, regardless of where we start the integration. For example, a periodic function with period 1 means that \( \int_{a}^{a+1} f(x) dx \) is the same for any value of 'a'. This concept is particularly useful for simplifying integrals over larger intervals, as we can break them down into smaller intervals of one period each.

Definite Integrals

Definite integrals calculate the net area between the function and the x-axis over a specific interval. The integral bounds give us a clear limit, the lower bound 'a' and the upper bound 'b', within which we evaluate the function's behavior.

Properties of definite integrals are handy in solving problems more easily and quickly. One such property is additivity, meaning we can break down a larger region into smaller subregions and sum their areas. For instance, if a function is known to have a regular pattern every interval, then integrating over any such interval yields the same result. This makes calculating the area under a curve over large intervals significantly less daunting if we understand these properties adequately.

Properties of Periodic Functions

Periodic functions exhibit several useful properties that are advantageous in integral calculus. One of the key properties is that the integral of one period of the function is constant; it does not depend on the starting point of the interval as long as the length of the interval equals one period.

Another property is that if the period is 'P', then \( \int_{a}^{a+P} f(x) dx \) remains constant for any 'a', which means we can manipulate the integral's bounds to simplify the problem, as long as we're integrating over a whole number of periods. This understanding is crucial, especially for functions that are complex and difficult to integrate directly.

Integral Calculus Problem Solving

Solving integral calculus problems, particularly involving periodic functions, involves recognizing patterns and applying the appropriate properties. Breaking down complex integrals into smaller and easier to manage pieces relies on a clear grasp of these periodic properties.

In problems where we have an integral over a range that includes multiple periods, we can use the periodicity to express the integral as a sum of integrals over single periods. This approach can transform a seemingly difficult problem into a much simpler one, particularly when combined with the periodic function's property that its integral over one period is constant. Moreover, an understanding of these concepts not only simplifies calculations but also provides profound insights into the function's behavior across its domain.