91Ó°ÊÓ

Function analysis is essential for understanding how a mathematical function behaves over a specific interval. In this scenario, we are dealing with the function \( f(x) = 8x(1-x^2)^2 \) on the interval \( I = \left[-\frac{1}{2}, 1\right] \). Analyzing the function helps us understand its shape, where it increases or decreases, and the points where it may intersect the x-axis.

For the function \( f(x) \) considered here, it is built by a polynomial expression. Polynomial functions have unique properties, such as continuity and differentiability across the real numbers, making them smooth curves with no sharp turns or breaks. The function comprises a term \( 8x \), which affects the linear rate of change, and a higher degree polynomial \((1 - x^2)^2\) that controls the curvature.

By expanding and simplifying the expression, we obtain \( 8x - 16x^3 + 8x^5 \), which represents a combination of a linear term, a cubic term, and a quintic term. These terms allow us to deduce the inflections and overall behavior as the function moves from negative to positive x-values within the interval. Understanding this structure is vital before integrating to find the area under the curve.

The concept of the area under a curve is a fundamental application of definite integrals in calculus. It quantifies the region enclosed by the curve, the x-axis, and given vertical boundaries. Here, we assessed the area under the curve of \( f(x) = 8x(1-x^2)^2 \) between \( x = -\frac{1}{2} \) and \( x = 1 \).

Setting up a definite integral, \( \int_{-\frac{1}{2}}^{1} f(x) \, dx \), allows us to compute this area as a numerical value. The definite integral serves as a summation of infinitesimally small rectangles under the curve. As we evaluate \( \int_{-\frac{1}{2}}^{1} (8x - 16x^3 + 8x^5) \, dx \), each term is integrated separately, resulting in a total area based on the antiderivative evaluated at the interval's endpoints.

The final step involves subtracting the antiderivative value at the lower boundary \( x = -\frac{1}{2} \) from that at the upper boundary \( x = 1 \). The result, \( \frac{3}{4} \), gives the net area, symbolizing the total positive area between the curve and the x-axis over the specified range.

An antiderivative, or an indefinite integral, is a reverse process of differentiation. It helps us find the original function from its derivative and plays a crucial role in determining the area under a curve. For the function \( f(x) = 8x(1-x^2)^2 \), we simplify and integrate each part to find its antiderivative.

The antiderivative of a polynomial like \( f(x) = 8x - 16x^3 + 8x^5 \) is found by integrating term by term. This means:

• The integral of \( 8x \) is \( 4x^2 \).
• The integral of \( -16x^3 \) is \( -4x^4 \).
• The integral of \( 8x^5 \) is \( \frac{4}{3}x^6 \).

These antiderivatives give us a new polynomial, \( 4x^2 - 4x^4 + \frac{4}{3}x^6 \), which we can evaluate over any interval to find solutions like area or total change.

To utilize the antiderivative in calculating definite integrals, we simply substitute the boundary values into it and compute the difference. This process efficiently converts a seemingly complex integral into manageable arithmetic to solve real-world problems, such as finding the area under the curve in question.