91Ó°ÊÓ

A definite integral is a powerful tool in calculus used to compute the accumulation of quantities, typically representing the area under a curve or between curves over a specific interval.

• Boundaries: In the case of the exercise, the definite integral is taken from 1 to some value of \(x\) within \([1, 3]\). These are your boundaries or limits of integration.
• Expression Understanding: The given function \(f(x)\) in the exercise is defined as the definite integral of the polynomial \(x(x^2 - 3x + 2)\). This means we accumulate all the tiny pieces of the area under the curve from 1 up to \(x\).

To solve the integral:- First, express the integral in terms of a simplified polynomial so it can be easily integrated.- Perform the integration process step-by-step to get a general antiderivative.- Finally, apply the Fundamental Theorem of Calculus by substituting the upper and lower bounds into the antiderivative to get a particular solution, \(f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4}\). This function describes the accumulated area as \(x\) varies over its interval, leading to insights about the function's range.

The process of polynomial integration involves integrating terms like \(x^n\) individually and combining them to find the integral of the entire polynomial.

• Power Rule: The power rule states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). This rule is fundamental when transforming individual terms of a polynomial integrand.
• Breaking Down the Polynomial: In our case, the integrand \(x^3 - 3x^2 + 2x\) is simplified by applying the power rule to each term separately:
  • \(\int x^3 \, dx = \frac{x^4}{4}\)
  • \(\int -3x^2 \, dx = -\frac{3x^3}{3} = -x^3\)
  • \(\int 2x \, dx = x^2\)
  Combine these to form the antiderivative \(\frac{x^4}{4} - x^3 + x^2 + C\).

Pay close attention to each step in polynomial integration, as correctly handling the coefficients and signs will ensure the resulting function is accurate.

Critical points of a function are where its derivative is zero or undefined, indicating potential maximums, minimums, or inflection points. Analyzing these points helps determine the range of a function, especially when dealing with integrals over an interval.

• Derivative Calculation: For the given function \(f(x)\), the derivative is the original polynomial \(x^3 - 3x^2 + 2x\). Setting \(f'(x)\) equal to zero yields the equation \(x(x-1)(x-2) = 0\).
• Solving for Critical Points: Solve \(x(x-1)(x-2) = 0\) to find the critical values of \(x\), which are \(x = 0, 1, 2\). But relevant within the given interval are \(x = 1\) and \(x = 2\).
• Function Evaluation at Critical Points: Evaluate \(f(x)\) at \(x = 1, 2, \) and other boundaries to find the range:
  • \(f(1) = -\frac{1}{4}\)
  • \(f(2) = 2\)
  • Looking at these values, the range of \(f(x)\) over the interval is \([ -\frac{1}{4}, 2 ]\).

Boundaries, along with values at critical points, provide important insights into how the function behaves across the given domain. This method ensures comprehensive understanding and validation of the solution.