Problem 114 - University Calculus: Early Transcendentals

Question

a. Show that if \(f\) is odd on \([-a, a],\) then \(x\) $$\int_{-a}^{a} f(x) d x=0$$ b. Test the result in part (a) with \(f(x)=\sin x\) and \(a=\pi / 2\)

Step-by-Step Solution

Answer

The integral of an odd function over a symmetric interval is zero.

Step 1: Understanding Odd Functions

A function \(f(x)\) is called odd if it satisfies the condition \(f(-x) = -f(x)\) for all \(x\). This means that the function has point symmetry about the origin.

Step 2: Setting Up the Integral

We need to evaluate the integral \(\int_{-a}^{a} f(x) \, dx\). Since \(f(x)\) is odd, \(f(-x) = -f(x)\) holds within the interval \([-a, a]\).

Step 3: Symmetric Properties of the Integral

For an odd function \(f(x)\), the definite integral over a symmetric interval about the origin (like \([-a, a]\)) cancels out, because for every part of the area under \(f(x)\) from \(-a\) to \(0\), there is an equal and opposite area from \(0\) to \(a\). Therefore, \(\int_{-a}^{a} f(x) \, dx = 0\).

Step 4: Testing with \(f(x)=\sin x\) and \(a=\pi/2\)

The function \(f(x) = \sin x\) is odd, because \(\sin(-x) = -\sin(x)\). We need to compute \(\int_{-\pi/2}^{\pi/2} \sin x \, dx\).

Step 5: Evaluate the Integral

Calculate \(\int \sin x \, dx = -\cos x + C\). So, evaluate the integral from \(-\pi/2\) to \(\pi/2\):\[-\cos(\pi/2) + \cos(-\pi/2) = 0 - 0 = 0\].

Step 6: Conclusion

Both theoretically, from the property of odd functions, and practically, by calculating the integral of \(\sin x\) over \([-\pi/2, \pi/2]\), we find \(\int_{-\pi/2}^{\pi/2} \sin x \, dx = 0\).

Key Concepts

Odd Functions

An odd function is a mathematical concept that describes a certain type of symmetry in functions. For a function to be classified as odd, it must satisfy the equation \(f(-x) = -f(x)\) for every value of \(x\) in its domain. This means that if you plot the function on a coordinate system, it will exhibit symmetry about the origin.
This type of symmetry implies that for every point \((x, y)\) on the function, there is a corresponding point \((-x, -y)\).

• Examples include functions like \(f(x) = \sin(x)\) and \(f(x) = x^3\).
• These functions have a distinctive property when integrated over symmetric intervals, such as \([-a, a]\).

Understanding odd functions is crucial when dealing with integrals over symmetric intervals, particularly because they simplify calculations dramatically.

Symmetric Interval

A symmetric interval in the context of mathematics generally refers to an interval of the form \([-a, a]\), where \(a\) is a positive real number. Such intervals are perfectly balanced around zero, making them of particular interest when evaluating certain types of functions.
For odd functions, symmetric intervals play a key role in simplifying the evaluation of definite integrals.

• The interval \([-a, a]\) ensures that every positive portion of the interval has a corresponding negative portion.
• This balance is a foundation for understanding why the integral of an odd function over a symmetric interval results in zero.

Symmetric intervals are an important tool in calculus, especially when assessing the behavior of functions that are either odd or even.

Properties of Integrals

The properties of integrals are numerous, but when focusing on definite integrals involving odd functions over symmetric intervals, there's a specific property that stands out. It states that if you integrate an odd function over a symmetric interval, the result is always zero.

• This is because the area under the curve from \(-a\) to 0 will cancel out with the area from 0 to \(a\).
• For example, consider \(f(x) = \sin(x)\) over \([-\pi/2, \pi/2]\): the positive area under \(\sin(x)\) from 0 to \(\pi/2\) is exactly mirrored and canceled by the negative area from \(-\pi/2\) to 0.

Recognizing these properties allows for simpler computation in many integrals without having to perform complex calculations directly. This property is especially beneficial in both theoretical and practical applications of calculus.