91Ó°ÊÓ

When you hear the term definite integral, think about finding the "total quantity" within a specific range. A definite integral helps calculate things like area, volume, and other important quantities within a given interval.
For a function, the definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of the interval. The interval marks the range over which you're measuring the "total quantity."
In our exercise, we integrate the function \( f(x) = x - x^2 \) from \( x = 0 \) to \( x = 1 \), capturing the area between the curve and the x-axis over this interval.
It’s important to know that the definite integral doesn't care about the constant term \( C \) that appears in indefinite integrals because we're calculating a specific net value from \( a \) to \( b \). This net area may include both above and below the x-axis, where areas below the axis reduce the total quantity calculated by the integral.

The power rule of integration is a handy tool for integrating functions that are powers of \( x \). It states that if you need to integrate \( x^n \), the result is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is an arbitrary constant.
This only applies when \( n \) isn't equal to -1, because in that special case, we'd use a different method, resulting in a logarithmic function.
For our function \( f(x) = x - x^2 \), we applied the power rule separately for \( x^1 \) and \( x^2 \).
Here's how:

• Integrate \( x \): becomes \( \frac{x^2}{2} \)
• Integrate \( -x^2 \): becomes \( -\frac{x^3}{3} \)

After integrating, we evaluate these at the bounds from \( a = 0 \) to \( b = 1 \) to find the definite integral.

Graphing a function can tell us a lot about its behavior, especially how it changes across an interval. For our function \( f(x) = x - x^2 \), which is a quadratic function, the graph forms a parabola.
Quadratic functions typically appear as a U-shaped or an inverted U-shaped curve. In this case, because the coefficient of \( x^2 \) is negative, our parabola opens downwards.
The vertex of this parabola is at the point \( (0.5, 0.25) \), which represents the highest point on the curve over the interval from \( 0 \) to \( 1 \). We can use the graph to visually confirm the average value \( y = \frac{1}{6} \) intersects the parabola at two points, showing the balance in the area under the curve.
This average value represents a constant horizontal line, providing a visual reference for how the "weighted center" of the area underneath the curve is distributed across the interval.