91Ó°ÊÓ

To start understanding the concept of antiderivatives, think of it as a function that works in reverse to differentiation. When you have a function, let's say \( f(x) \), its antiderivative gives you another function \( F(x) \) such that the derivative of \( F \) brings you back to your original function \( f(x) \). In mathematical terms, if \( F'(x) = f(x) \), then \( F(x) \) is called an antiderivative of \( f(x) \).

Here’s how this ties into our exercise: Based on the Second Fundamental Theorem of Calculus, when we define \( F(x) = \int_a^x f(t) \, dt \), \( F \) acts as an antiderivative over the interval. In our example, with \( f(t) = t^2 \), the integral function \( F(x) \) is \( \frac{x^3}{3} \) and satisfying the condition \( F'(x) = x^2 = f(x) \).

By choosing \( f(t) \) to be continuous, this ensures that \( F(x) \) happens naturally as the antiderivative across the given interval.

The definite integral is a powerful mathematical tool that allows you to calculate the area under the curve of a function from one point to another. When specifying limits, these boundaries are denoted as \( a \) and \( b \) within the integral symbol. For instance, \( \int_a^b f(x) \, dx \) represents the exact area under the curve \( f(x) \) from \( x = a \) to \( x = b \).

In the exercise, \( F(x) \) is found using a definite integral. Specifically, it’s defined by \( F(x) = \int_0^x t^2 \, dt \). Here, \( a = 0 \) is our starting point, and \( b = x \) changes and represents any endpoint up to which we calculate. The process of solving this integral allowed us to find that \( F(x) = \frac{x^3}{3} \), which gives an expression for the accumulated area under \( t^2 \) starting from \( t = 0 \).

The integral provides both the exact value of this area, as a function of \( x \), and the antiderivative necessary for solving our problem.

A non-decreasing function is one where, as you go from left to right along the x-axis, the function does not decrease in value. In other words, if \( y \) values of the function either stay the same or increase. This doesn't necessarily mean the function must continuously rise; it can stay flat for intervals.

The function \( F(x) = \frac{x^3}{3} \) from our exercise is non-decreasing because its derivative \( F'(x) = x^2 \) is non-negative. Since \( x^2 \) always results in a value \( \geq 0 \) for all values of \( x \), we can be assured that \( F(x) \) doesn’t decrease over its domain.

By showing that \( F'(x) \) remains non-negative ensures that as \( x \) increases, \( F(x) \) either stays the same or it increases, hence the term "non-decreasing function."

A function is considered continuous if you can draw it without lifting your pen from paper; it has no breaks, gaps, or jumps. In more rigorous terms, a function \( f(x) \) is continuous at a point \( x = c \) if the following three conditions hold:

• \( f(c) \) is defined,
• the limit of \( f(x) \) as \( x \) approaches \( c \) exists,
• the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

For the interval \([a, b]\), \( f(x) \) must be continuous at every point to ensure the integral \( \int_a^b f(x) \, dx \) is valid.

In our example, the function \( f(t) = t^2 \) is continuous over all real numbers because it's a polynomial, which by nature is always continuous everywhere. This continuity is crucial for the application of the Second Fundamental Theorem of Calculus, allowing \( F(x) = \int_0^x t^2 \, dt \) to properly exist and be differentiable, giving us the antiderivative we need.