Question

Let \(F(x)=\int_{0}^{x} f(t) d t .\) Show that if \(f\) is even, then \(F\) is odd, and that if \(f\) is odd, then \(F\) is even.

Short answer

Question: Prove that if a function \(f\) is even, then its antiderivative \(F(x) = \int_0^x f(t) dt\) is odd, and if a function \(f\) is odd, then its antiderivative \(F(x) = \int_0^x f(t) dt\) is even.

Step-by-step solution

Part 1: If \(f\) is even, then \(F\) is odd

Given that \(f\) is an even function, we have the property \(f(-x) = f(x)\). To show that \(F\) is odd, we need to show that \(F(-x) = -F(x)\). We will do this by evaluating \(F(-x)\) using the given definition of \(F(x)\): $$F(-x)=\int_{0}^{-x} f(t) d t$$ Now, we'll use the property of even functions and a substitution trick to evaluate this integral. Let \(u = -t\), so \(du = -dt\). Then, substituting \(u\) into the expression: $$\int_{0}^{-x} f(t) dt = -\int_{0}^{x} f(-u) du$$ Since \(f\) is even, we can replace \(f(-u)\) with \(f(u)\): $$-\int_{0}^{x} f(-u) du = -\int_{0}^{x} f(u) du$$ This integral is the same as \(F(x)\), but with the negation sign: $$F(-x)=-F(x)$$ So, we can conclude that if \(f\) is even, then \(F\) is odd.

Part 2: If \(f\) is odd, then \(F\) is even

Given that \(f\) is an odd function, we have the property \(f(-x) = -f(x)\). To show that \(F\) is even, we need to show that \(F(-x) = F(x)\). We will do this by evaluating \(F(-x)\) using the given definition of \(F(x)\): $$F(-x)=\int_{0}^{-x} f(t) d t$$ Now, we'll use the property of odd functions and a substitution trick to evaluate this integral. Let \(u = -t\), so \(du = -dt\). Then, substituting \(u\) into the expression: $$\int_{0}^{-x} f(t) dt = -\int_{0}^{x} f(-u) du$$ Since \(f\) is odd, we can replace \(f(-u)\) with \(-f(u)\): $$-\int_{0}^{x} f(-u) du = \int_{0}^{x} f(u) du$$ This integral is the same as \(F(x)\): $$F(-x)=F(x)$$ So, we can conclude that if \(f\) is odd, then \(F\) is even.

Key concepts

Even and Odd Functions

In calculus, we come across the concepts of even and odd functions, which are essential in understanding symmetry in graphs and their impact on integrals. An **even function** is one where the function satisfies the relation \(f(-x) = f(x)\). What this means is that the graph of an even function is symmetric with respect to the \(y\)-axis. Examples of even functions include \(f(x) = x^2\) and \(f(x) = \cos(x)\).

On the other hand, an **odd function** satisfies the relation \(f(-x) = -f(x)\). This means that the graph of an odd function is symmetric with respect to the origin, meaning a 180-degree rotation about the origin produces the same graph. Common examples are \(f(x) = x^3\) and \(f(x) = \sin(x)\).

When dealing with integrals, these properties become very useful. For example, when we compute definite integrals over symmetric intervals \([-a, a]\) of even functions, the integration can sometimes be simplified since the contributions from the negative \(x\) values mirror those from the positive \(x\) values. In contrast, for odd functions, the contributions from the interval \([-a, a]\) cancel each other out, often simplifying to zero.

Integration by Substitution

Integration by substitution is a powerful technique, akin to differentiation's chain rule, helping simplify complex integrals by changing variables. This is particularly useful when the integrand (the function being integrated) can be simplified by substituting part of it with a new variable.

The general strategy involves:

• Choosing a substitution, often \(u = g(x)\), where \(g(x)\) is part of the integral that makes the integration simpler when replaced.
• Determining the differential \(du\), which requires differentiating \(u\) with respect to \(x\) and then solving for \(dx\) in terms of \(du\).
• Replacing \(dx\), \(x\), and possibly \(f(x)\) in the integral with \(du\), \(u\), and other relevant expressions.
• Solving the integral in terms of \(u\) and then reverting back to \(x\) using the initial substitution relation.

Substitution helps further, especially with even and odd functions, as shown in the exercise where negations and symmetries exploit the property alterations of these functions. Using substitution, a complex evaluation becomes straightforward, transforming it by leveraging function symmetry.

Properties of Definite Integrals

Definite integrals, distinguished by their upper and lower limits, offer insight into the accumulation of quantities and the area under curves. Their properties can significantly simplify computations and evaluations.

• **Additivity**: The integral from \(a\) to \(b\) of a function \(f\) can be split into the sum of integrals from \(a\) to an intermediate point \(c\), and from \(c\) to \(b\). Formulaically: \(\int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx\).
• **Reversing Limits**: Reversing the limits of integration changes the sign of the integral: \(\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx\).
• **Zero from Symmetry**: For an odd function integrated over a symmetric interval \([-a, a]\), the result is zero: \(\int_{-a}^{a} f(x)\,dx = 0\).
• **Mean Value**: The definite integral of a continuous function over \([a, b]\) is closely related to finding the average value of the function: \(\frac{1}{b-a} \int_{a}^{b} f(x)\,dx\).

These properties allow us to understand how the nature of the function and the limits of integration interact, particularly in the context of symmetry provided by even and odd functions. Definite integrals help reinforce understanding of how such symmetries alter outcomes across varied intervals.