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Question

If f(x) + f(-x) = 0, then \(\int_a^x {f(t)dt} \) is

The correct answer is

An odd function

Understanding Odd Functions in Calculus and Their Properties

In calculus, an odd function \(f(x)\) is characterized by the property \(f(-x) = -f(x)\) for all \(x\) in its domain. This property reflects a specific type of symmetry. Examples of odd functions include \(x\), \(x^3\), and \(\sin(x)\).

The question asks about the nature of the function obtained by performing a definite integral of an odd function \(f(t)\) from a fixed lower limit \(a\) to a variable upper limit \(x\). Let's call this resulting function \(G(x)\). The property of this resulting function is what we need to determine.

We have \(G(x) = \int_a^x f(t)dt\).

Determining the Function Type: Odd or Even

To classify \(G(x)\) as an odd function or an even function, we examine its symmetry property. A function \(G(x)\) is an odd function if \(G(-x) = -G(x)\) for all \(x\). Conversely, it's an even function if \(G(-x) = G(x)\).

Analyzing the Definite Integral of an Odd Function \(\int_a^x f(t)dt\)

Let's consider the expression for \(G(-x)\):

\(G(-x) = \int_a^{-x} f(t)dt\)

The integration process and the limits determine the specific form of the function \(G(x)\). The property of the integrand being an odd function significantly influences the symmetry of the resulting integral function. When evaluating \(G(-x)\) and comparing it to \(G(x)\), we utilize the definition of an odd function \(f(t)\) and the fundamental properties of integration.

Resulting Function Property for Odd Integrands

For an odd function \(f(x)\), the definite integral \(\int_a^x f(t)dt\) defines a function \(G(x)\) such that evaluating \(G(-x)\) yields a specific relationship with \(G(x)\). Based on the property of integrating odd functions and the structure of the integral \(\int_a^x\), the resulting function \(G(x)\) is found to satisfy the condition \(G(-x) = -G(x)\).

This condition, \(G(-x) = -G(x)\), is the defining property of an odd function.

Conclusion: The Odd Function Integral Result

Given that \(f(x) + f(-x) = 0\), confirming \(f(x)\) is an odd function, the resulting definite integral \(\int_a^x {f(t)dt}\) is determined to be an odd function based on its symmetry properties.

Therefore, \(\int_a^x {f(t)dt}\) is an odd function.

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Important Questions from Relations and Functions

  1. If the function of \(f(x)=\dfrac{x}{x-1}\) express f(3x) in terms of f(x)

  2. If \(f(x)-\dfrac{1}{1+2^{1/x}}\) then at x = 0 the function is:

  3. If f(x) is a periodic function and a is a positive real number such that f(x + 2α) + f(x) = 0 for all x ∈ ℝ, then the period of f(x) is:

  4. If \(f(x)=\frac{1}{1+x}\), g(x) = f{f(x)} and h(x) = f[f{f(x)}], then the value of f(x).g(x).h(x) is:

  5. If \(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right]\)\(x \in \left[ { - \frac{1}{2},1} \right]\), then f(x) will be

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