Problem 143 - Calculus Volume 1

Question

If \(f\) is 1 -periodic and \(\int_{0}^{1} f(t) d t=A, \quad\) is it necessarily true that \(\int_{a}^{1+a} f(t) d t=A\) for all \(A\) ?

Step-by-Step Solution

Answer

Yes, \( \int_{a}^{1+a} f(t) dt = A \) for all \( a \), due to the periodicity of \( f(t) \).

Step 1: Understand the Concept of Periodicity

A function is said to be periodic with period 1 if it satisfies the condition that for all values of the variable, the function repeats itself every 1 unit. Thus, for a 1-periodic function \( f(t) \), the relationship \( f(t) = f(t+1) \) holds true across its entire domain.

Step 2: Apply Periodicity to Integrals

Given that \( f \) is 1-periodic, the integral from any point \( a \) to \( a+1 \) should be the same as the integral from 0 to 1. This is because the function repeats every 1 unit, so the total area under the curve over this interval should also repeat. Mathematically, this implies \( \int_{a}^{1+a} f(t) dt = \int_{0}^{1} f(t) dt = A \).

Step 3: Verify with the Definition of the Definite Integral

The function \( f \) is 1-periodic, thus \( \int_{a}^{a+1} f(t) dt = \int_{0}^{1} f(t) dt \) holds because shifting the integration interval by a complete period shouldn’t change the value of the integral since the function values repeat identically across periods. Hence, no matter where the interval \( [a, 1+a] \) starts, the integral will always sum up to \( A \).

Key Concepts

Definite Integrals

Definite integrals are vital in understanding how areas under the curve of a function can be calculated within a given interval. When you take the definite integral of a function, you're essentially summing up all the infinitesimally small areas under that function's curve between two points.
This intrinsic property makes definite integrals valuable in various applications, like physics and engineering. One of the distinctive characteristics of a definite integral is that it doesn't depend on the antiderivative, or primitive function, but only on the values of the function at the boundaries of integration. This means when you compute \( \int_{a}^{b} f(t) \, dt \), you're looking at the difference \( F(b) - F(a) \), where \( F(t) \) is an antiderivative of \( f(t) \).
In the context of periodic functions, definite integrals become even more interesting because the periodicity can simplify calculations by exploiting repetitive patterns. If a function has a period that matches the interval of integration, the integral becomes predictable and often much easier to compute.

Function Periodicity

Function periodicity is a fascinating concept that tells us how a function behaves over its domain. A periodic function repeats values at regular intervals, known as the period.
If you know a function's period, you have a powerful tool to predict and simplify various calculations, especially integrals. For instance, in the case of a 1-periodic function such as our original exercise, the function satisfies \( f(t) = f(t + 1) \) for all \( t \).
Thus, every time you shift \( t \) by one unit, the function's values repeat. This quality is leveraged in computing integrals because it allows us to predict the integral over any interval that matches the period. By understanding a function's periodic nature, we can extend the properties over different domains and ensure reproducibility of results.

Integral Properties

Integral properties are crucial in calculus, providing essential rules that make integration more manageable. These properties include linearity, additivity, and how integrals behave under certain transformations and conditions.
The linearity property states that the integral of a sum of functions is the sum of the integrals, and the integral of a constant multiplied by a function can be moved outside the integral: \[ \int (af(t) + bg(t)) \, dt = a \int f(t) \, dt + b \int g(t) \, dt \]. Additivity allows an integral over a contiguous interval to be split into parts: \( \int_{a}^{b} f(t) \, dt = \int_{a}^{c} f(t) \, dt + \int_{c}^{b} f(t) \, dt \).
When dealing with periodic functions, an essential property is that the integral over one period can be extended over any interval composed of whole periods without changing the integral's value over that specific period. Thus, for a 1-periodic function, integrating over any interval exactly one period long will yield the same result as integrating over the base period [0,1].
These properties greatly simplify integration problems and are especially useful when applied to periodic functions.