Problem 142 - Calculus Volume 1

Question

If \(f\) is 1 -periodic \((f(t+1)=f(t)), \quad\) odd, and integrable over \([0,1], \quad\) is it always true that \(\int_{0}^{1} f(t) d t=0 ?\)

Step-by-Step Solution

Answer

Yes, the integral of a periodic and odd function over one period is always zero due to symmetry.

Step 1: Understanding Periodicity

First, note that the function \( f \) is 1-periodic, which means \( f(t+1) = f(t) \) for all \( t \). This implies that the function repeats its values every unit interval \([n, n+1)\) for any integer \( n \).

Step 2: Understanding Odd Functions

The function \( f \) is odd, which means \( f(-t) = -f(t) \) for all \( t \). This property tells us that the function is symmetric about the origin in a way that makes the graph of the function reflect through the origin.

Step 3: Symmetry and Integration Bounds

Given the periodic nature and odd symmetry of \( f \), consider its behavior over a symmetric interval about 0. However, since we are given the integral over \([0,1]\), it is necessary to relate this to the properties of the function despite the integration limits not centering around zero.

Step 4: Graphical Insight

If you graph one period of an odd and periodic function, recall that the area under the positive part and the area over the negative part will often cancel each other out due to symmetry, especially when the function's average value over one period tends to balance positive and negative values.

Step 5: Evaluating the Integral by Symmetry

Although the function is periodic at \( t+1 \), the fact that it is odd suggests symmetry about the midpoint of the interval when the function is appropriately centered (i.e., by taking translations like \([-0.5, 0.5]\) centered around zero). Over one full period \([0,1]\), for an odd function which is not centered for evaluation, explicitly calculating might be required to note any presence of symmetry-driven zero result.

Step 6: Conclusion by Property Validation

Despite \( [0,1] \) not being symmetric like other intervals typically used for odd functions (e.g., \([-a, a]\)), the properties of periodic and odd functions inherently balance out over each period to result in integrating to zero. This is because one integral over a perfectly symmetric subset will produce zero.

Key Concepts

Odd Functions

Odd functions have a unique symmetry that makes them special. For any function \( f(t) \) to be classified as an odd function, it must satisfy the property \( f(-t) = -f(t) \) for all values of \( t \). This means that the function reflects across the origin of a coordinate plane. Imagine flipping and rotating a plot of the function around the origin: an odd function will look identical before and after the transformation.

This symmetry has key implications for integration, especially across intervals

• Reflexive Symmetry: The areas of the graph on opposite sides of the y-axis are mirror images but have opposite signs.
• Zero Net Area: When an odd function is integrated over a symmetric interval about the origin (like \([-a, a]\)), the result is zero. This is because the positive and negative areas cancel out each other.

Understanding these properties allows us to predict the integral of odd functions quickly over symmetric limits, even without evaluation. In our exercise case, maintaining the symmetry around the origin is key even when dealing with the different limits like \([0,1]\).

Integration Properties

Integration is a crucial concept in calculus, allowing the summation of infinitely small areas under a curve. When considering integration properties of periodic and odd functions, specific rules and observations tend to simplify what initially seems complex.

Two notable properties distinguish the integration of such functions:

• Linearity: The integration process is linear, i.e., \( \int (af(t) + bg(t)) dt = a \int f(t) dt + b \int g(t) dt \) for constants \( a \) and \( b \).
• Symmetry: If a periodic function is also odd, integrating it over a symmetric interval about the origin yields zero, a property heavily beneficial in calculations.

These properties can also apply graphed aspects of odd periodic functions. By being aware of the function’s even or odd nature, and its symmetry, we can often predict the output of integration over specific intervals without detailed computations.

Symmetry in Integrals

Symmetry is a powerful tool in mathematics, especially in the context of integrals. For periodic functions, symmetry allows us to intelligently simplify otherwise complicated integrals. This holds especially true when integrating odd functions.

When you integrate an odd function over an interval that's symmetric about the origin, such as \([-0.5, 0.5]\) or \([-1, 1]\), the result is always zero. This is because:

• The left side of the y-axis cancels out the right side, as they are mirror images with equal but opposite signed areas.
• The integral over a symmetric interval accounts for this cancellation perfectly, leading directly to zero.

Even when not centered exactly around zero, understanding and locating this symmetry can provide surprising insights, predicting the integral of odd and periodic functions over their period. This critical property is not immediately obvious when integrating over non-centered boundaries like \([0, 1]\), but it holds due to the periodic nature allowing these features to "reset" and repeat. Hence, regardless of the asymmetry in the interval from 0 to 1, the periodic reset alone contributes to a balanced integral.