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An antiderivative of a function is a function whose derivative is the original function. It is the reverse process of differentiation, helping us to determine the original function from its derivative. To compute the integral of a function, you often need to find its antiderivative.
In our exercise, we begin by finding the antiderivative of the given function, which is a polynomial expression. The function provided is \[ y = x^3 - 3x^2 + 2x.\]To find its antiderivative, apply the power rule of integration which involves incrementing the exponent by one and dividing by the new exponent.

• For \(x^3\), the antiderivative is \(\frac{x^4}{4}\).
• For \(-3x^2\), the antiderivative is \(-x^3\).
• For \(2x\), the antiderivative is \(x^2\).

The antiderivative for the whole function becomes:
\[ \int (x^3 - 3x^2 + 2x)\, dx = \frac{x^4}{4} - x^3 + x^2. \]This function will be used to calculate the definite integral over each interval.

The roots of a polynomial are the values of \(x\) for which the polynomial equation equals zero. Finding the roots helps us locate where the function intersects the x-axis. This is a crucial step in finding the area between the curve and the x-axis.
In our example, the polynomial given is:\[ x^3 - 3x^2 + 2x = 0. \]By factoring out an \(x\), the equation becomes:

• \(x(x^2 - 3x + 2) = 0.\)

From this, we have:

• First root is \(x = 0.\)

To factor \(x^2 - 3x + 2\), look for two numbers that multiply to 2 (the constant term) and add to -3 (the linear coefficient). These numbers are -1 and -2. Thus, the factors are \((x - 1)(x - 2)\) leading to the roots:

• \(x = 1\)
• \(x = 2\)

These roots, \(x = 0, 1, 2\), are the critical points determining the intervals for calculating the area.

Finding the area between a curve and the x-axis involves evaluating the definite integral of the function over specified intervals. This process allows us to measure the enclosed regions, whether they lie above or below the x-axis.
In our situation, we are concerned with the interval from 0 to 2, which we've broken into sub-intervals based on the roots:

• \([0, 1]\)
• \([1, 2]\)

For the interval \([0, 1]\), the function remains above the x-axis entirely, and so the area is simply the value of the integral:\[ \int_{0}^{1} (x^3 - 3x^2 + 2x)\, dx, \]which evaluates to \(\frac{1}{4}\).
In contrast, over the interval \([1, 2]\), part of the curve dips below the x-axis, impacting the integral:\[ \int_{1}^{2} (x^3 - 3x^2 + 2x)\, dx. \]This computes to \(-\frac{1}{4}\). In terms of area, however, we take the absolute value, yielding \(\frac{1}{4}\).
By summing these areas, \(\frac{1}{4} + \frac{1}{4}\), we obtain the total area between the curve and the x-axis over the interval [0, 2], which is \(\frac{1}{2}\). This approach ensures we always count positive area, regardless of the curve's position relative to the x-axis.