Question

Extending the ldeas Writing to Learn If \(f\) is an odd continuous function, give a graphical argument to explain why \(\int_{0}^{x} f(t) d t\) is even.

Short answer

The integral of an odd function from 0 to x is an even function because for any real number x, the value of the integral from 0 to -x will be the same as the integral from 0 to x, which is the property of even functions.

Step-by-step solution

Understanding Odd functions and Even functions

Firstly, the definition of an odd function, \(f(x)\), is that for all \(x\) in its domain, \(f(-x) = -f(x)\). This property means this function is symmetrical with respect to the origin in a graphical sense. An even function, \(g(x)\), is defined as a function that for all \(x\) in its domain, \(g(-x) = g(x)\). This means this function is symmetrical with respect to the y-axis graphically.

Understanding properties of definate integrals

Now let's integrate the odd function \(f(t)\) from 0 to x, meaning, \(\int_{0}^{x} f(t) d t\). If the integral turns out to be an even function, for any real number x, the value of the integral from 0 to -x should be the same as the integral from 0 to x.

Applying the properties on the function

Consider the integral from 0 to -x: \(\int_{0}^{-x} f(t) dt = \) - \(\int_{0}^{x} f(-t) dt\). Because \(f\) is an odd function, \(f(-t)\) is \(-f(t)\), therefore, \(\int_{0}^{x} f(-t) dt =\) -\(\int_{0}^{x} f(t) dt\). This demonstrates that the integral of an odd function from 0 to x is an even function.