Question

If \(\mathrm{f}(\mathrm{x})\) is an odd periodic function with period \(\mathrm{P}\) then \(2 p+a \int_{2 p-a} f(x) d x\) is equal to \(\ldots \ldots\) (a) \(P\) (B) \(2 \mathrm{P}\) (C) \(4 \mathrm{P}\) (d) 0

Short answer

The value of the expression \(2P + a\int_{2P-a} f(x) dx\) for an odd periodic function \(f(x)\) with period \(P\) is \(2P\).

Step-by-step solution

Recall properties of odd functions and periodic functions

For an odd function \(f(x)\), we have \(f(-x)=-f(x)\). One important property of odd functions is that their integral over a symmetric interval is always zero: \[\int_{-a}^{a} f(x) dx = 0\] For a periodic function \(f(x)\) with period \(P\), we have \(f(x+P) = f(x)\). As a result, the integral of a periodic function over an interval of length equal to its period is the same regardless of the starting point: \[\int_{x_0}^{x_0+P} f(x) dx = \int_{x_1}^{x_1+P} f(x) dx\]

Analyze the given integral

We want to find the value of \(2P + a\int_{2P-a} f(x) dx\). Let's analyze the integral part by making a substitution: Let \(y = x - (2P - a)\). Then \(x = y + (2P - a)\) and \(dx = dy\). The new integral would be: \[\int_{0}^{a} f(y + 2P - a) dy\]

Apply the periodicity of the function

Since \(f(x)\) is periodic with period \(P\), we have \(f(y+2P-a) = f(y+a-P)\). Therefore, the integral will be: \[\int_{0}^{a} f(y + a - P) dy\]

Make another substitution

Let's make another substitution: \(z = y + a - P\). Then \(y = z - a + P\) and \(dy = dz\). The integral would now be: \[\int_{-a+P}^{P} f(z) dz\]

Apply the property of odd functions

Since \(f(x)\) is an odd function, we can use the property of odd functions to evaluate the integral: \[\int_{-a+P}^{P} f(z) dz = \frac{1}{2}\int_{-(2a-2P)}^{2a-2P} f(z) dz = 0\] This is because \(f(x)\) is odd, and the interval \([-(2a-2P), 2a-2P]\) is symmetric around the origin.

Calculate the final result

Now we know that the integral part of the expression is zero. Hence, the value of the expression \(2P + a\int_{2P-a} f(x) dx\) is: \[2P + a\int_{2P-a} f(x) dx = 2P + a(0) = 2P\] Therefore, the correct option is (B), \(2P\).

Key concepts

Properties of Odd Functions

Understanding the properties of odd functions is crucial for analyzing mathematical problems involving symmetry. An odd function is defined by the property that for every input value x, it satisfies the equation f(-x) = -f(x). Examples of odd functions include f(x) = x^3 or f(x) = sin(x).

One of the most significant properties of odd functions is how they behave when integrated over symmetric intervals around the origin. Mathematically, this property is expressed as:
\[\int_{-a}^{a} f(x) dx = 0\]
This result occurs because the regions under and above the x-axis cancel each other out. This property is extremely useful in simplifying complex integrals, especially when dealing with periodic functions, as it can help reduce the problem to an analysis of a simpler interval or even eliminate the integral component altogether if the interval is symmetric.

Periodicity of Functions

The periodicity of functions is a concept that helps us understand patterns that repeat at regular intervals. A function f(x) is said to be periodic with period P if for all inputs x, the function satisfies the equation f(x+P) = f(x). The period is the length of one complete cycle of repetition, and familiar examples of periodic functions are the sine and cosine functions with a period of 2π.

A key point about periodic functions is that their integrals over any interval of length P are equivalent, regardless of the starting point. This is expressed mathematically as:
\[\int_{x_0}^{x_0+P} f(x) dx = \int_{x_1}^{x_1+P} f(x) dx\]
For odd periodic functions, when combined with their integral property, this can often result in the integral being zero over intervals of length P, which is particularly helpful in simplifying complex calculations.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the quantities where the rate of change is known. Integrals allow us to figure out properties such as the area under a curve, accumulated quantities, and other concepts linked to accumulation. The integral of a function can be thought of as the opposite of taking a derivative and is a core tool for solving a wide range of problems in physics, engineering, and beyond.

The process of integration can sometimes be streamlined by utilizing the symmetries in functions, as we see with odd functions and their result of zero over symmetric intervals. Knowing how to manipulate and substitute within integrals while preserving the limits and function behavior is crucial. The ability to transform an integral into a more familiar or simpler form allows us to solve complex problems more efficiently, which is a key skill to develop in integral calculus.

In the scenario provided in the exercise, the odd and periodic nature of the function combined with the use of substitution techniques allowed for the integration bounds to be transformed into a symmetric interval, where the integral of an odd function vanishes to zero, thus simplifying the original problem significantly.