Problem 65 - Calculus and its applications

Question

Find the area bounded by \(y=x^{3}-3 x+2,\) the \(x\) -axis, and the first coordinates of the relative maximum and minimum values of the function.

Step-by-Step Solution

Answer

The bounded area is 1 square unit.

Step 1: Find Critical Points

To find where the function has relative maxima or minima, we first find the derivative of the function. For \(y = x^3 - 3x + 2\), the derivative is \(y' = 3x^2 - 3\). Set \(y' = 0\) to find critical points: \(3x^2 - 3 = 0\). Solving this gives \(x^2 = 1\), so \(x = \pm 1\). These are our critical points.

Step 2: Determine the Nature of Critical Points

We use the second derivative test to determine if these critical points are maxima or minima. The second derivative is \(y'' = 6x\). Calculate \(y''\) at the critical points: \(y''(-1) = 6(-1) = -6\) indicating a relative maximum at \(x = -1\), and \(y''(1) = 6(1) = 6\) indicating a relative minimum at \(x = 1\).

Step 3: Sketch the Region

Based on the function and its critical points, the relative maximum occurs at \(( -1, y(-1) )\) and the relative minimum at \(( 1, y(1) )\). To find these points, calculate \(y(-1) = (-1)^3 - 3(-1) + 2 = 4\) and \(y(1) = 1^3 - 3(1) + 2 = 0\). The region bounded by the x-axis and critical points is between \(x = -1\) and \(x = 1\).

Step 4: Calculate the Area

The area under the curve between the critical points can be found by the definite integral of the function from \(-1\) to \(1\). \[\int_{-1}^{1} (x^3 - 3x + 2) \, dx\]Find the antiderivative: \[F(x) = \frac{x^4}{4} - \frac{3x^2}{2} + 2x\]Evaluate from \(-1\) to \(1\):\[\left[ \frac{x^4}{4} - \frac{3x^2}{2} + 2x \right]_{-1}^{1} = \left( \frac{1^4}{4} - \frac{3(1)^2}{2} + 2(1) \right) - \left( \frac{(-1)^4}{4} - \frac{3(-1)^2}{2} + 2(-1) \right)\]Simplifying:\[\left( \frac{1}{4} - \frac{3}{2} + 2 \right) - \left( \frac{1}{4} - \frac{3}{2} - 2 \right) = \left( \frac{3}{4} \right) - \left( \frac{1}{4} \right) = 1\]

Step 5: Final Area Calculation

Since the integral from the minimum to maximum point gave a net area of 1, this is the area of the region bounded by the function and the x-axis. For the calculation, note that the whole bounded area is above the x-axis between these points.

Key Concepts

Critical Points

To identify critical points, we first find the derivative of the given function. For the function \(y = x^3 - 3x + 2\), the derivative is \(y' = 3x^2 - 3\). By setting the derivative \(y' = 0\), we solve for \(x\), finding the critical points. This equation simplifies to \(3x^2 - 3 = 0\), resulting in \(x^2 = 1\). Thus, we find that the values \(x = 1\) and \(x = -1\) are the critical points. Critical points are essential in determining where a function's slope changes direction, indicating possible maxima or minima in a graph.

Understanding critical points helps us analyze a function's behavior and how it may transition across different intervals. These analysis steps are foundational in finding areas bounded by curves, as they help define limits or bounds for integration.

Definite Integral

The definite integral is a powerful tool for computing areas under curves. In our context, it provides the area bounded by the curve \(y = x^3 - 3x + 2\) from \(x = -1\) to \(x = 1\). To find this area, we evaluate the integral:

• First, we determine the antiderivative of the function, \(F(x) = \frac{x^4}{4} - \frac{3x^2}{2} + 2x\).
• Next, we evaluate \(F(x)\) at the bounds \(-1\) and \(1\) using the Fundamental Theorem of Calculus.
• The area is given by the difference \(F(1) - F(-1)\).

The definite integral allows us to neatly handle both positive and negative areas, providing net area over specified intervals. This concept is invaluable in integral calculus, offering not just area but also solutions to various practical and theoretical problems.

Second Derivative Test

The second derivative test is used to determine whether critical points are maxima or minima. Starting with our function's derivative, the second derivative is calculated as \(y'' = 6x\). At each critical point, \(y''\) informs the curve's concavity:

• For \(x = -1\), \(y''(-1) = 6(-1) = -6\), indicating a negative concavity (\(\cap\)), suggesting a relative maximum.
• For \(x = 1\), \(y''(1) = 6(1) = 6\), indicating positive concavity (\(\cup\)), suggesting a relative minimum.

This test is particularly useful because it provides a swift method to assess the nature of stationary points. By understanding the second derivative test, learners can quickly classify critical points without needing to sketch graphs, thus enhancing their problem-solving efficiency in calculus.