91Ó°ÊÓ

A quadratic function is a type of polynomial function with the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These functions produce a variety of shapes known as parabolas. The parabola can open upwards or downwards depending on the sign of the coefficient \( a \). If \( a > 0 \), the parabola opens upwards, forming a "U" shape. Conversely, if \( a < 0 \), the parabola opens downwards, resembling an "n" shape.
In this exercise, the integrand function \( f(x) = x - x^2 \) is a quadratic function with \( a = -1 \), \( b = 1 \), and \( c = 0 \). The negative coefficient \( -1 \) before \( x^2 \) indicates that the parabola opens downward. This function crosses the x-axis (the roots) when \( f(x) = 0 \). By setting the equation \( x - x^2 = 0 \), we find the roots at \( x = 0 \) and \( x = 1 \). These points divide the parabola into sections, where the sign of the quadratic function can be analyzed.

Integration is a fundamental concept in calculus, often interpreted as the process of finding the area under a curve. It can be viewed as the reverse operation of differentiation. In this context, we integrate the quadratic function \( f(x) = x - x^2 \) over a specific interval to determine the total area between the curve and the x-axis.
Before calculating the integral, finding where the function \( x - x^2 \) is positive helps to maximize this area. By integrating only over those regions, we accumulate the maximum possible value. The antiderivative of \( x - x^2 \) is \( \int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3} + C \). Evaluating this at the boundaries of the positive interval \([0, 1]\) lets us compute the definite integral:

• At \( x = 1 \): \( \frac{1^2}{2} - \frac{1^3}{3} \)
• At \( x = 0 \): \( \frac{0^2}{2} - \frac{0^3}{3} \)

Calculating these results gives \( \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \), showing that the area and thus the value of the integral over this interval is \( \frac{1}{6} \).

Interval analysis examines which sections of the domain contribute integral values. In solving integrals, understanding the behavior of the function within different intervals is key to optimal results.
The exercise focuses on the function \( f(x) = x - x^2 \), positive in the interval \( (0, 1) \). The roots at \( x = 0 \) and \( x = 1 \) create boundaries where the function changes its sign, from below the x-axis (negative) to above it (positive).
To maximize the value of the integral, we integrate only over the interval where \( f(x) \) is positive. This method captures the region where the function contributes positively to the integral's total area. By analyzing like this, we choose \( a = 0 \) and \( b = 1 \) as the limits to maximize the integral's value. This selection ensures the capture of the entire positive area under the curve of the quadratic function within the interval.