Problem 67 - Calculus Early Transcendentals

Question

Consider the cubic polynomial \(f(x)=x(x-a)(x-b),\) where \(0 \leq a \leq b\) a. For a fixed value of \(b,\) find the function \(F(a)=\int_{0}^{b} f(x) d x\) For what value of \(a\) (which depends on \(b\) ) is \(F(a)=0 ?\) b. For a fixed value of \(b,\) find the function \(A(a)\) that gives the area of the region bounded by the graph of \(f\) and the \(x\) -axis between \(x=0\) and \(x=b .\) Graph this function and show that it has a minimum at \(a=b / 2 .\) What is the maximum value of \(A(a),\) and where does it occur (in terms of \(b\) )?

Step-by-Step Solution

Answer

Question: Find the function \(F(a)\) and the value of \(a\) for which \(F(a) = 0\). Additionally, find the function \(A(a)\) and show that there is a minimum value of \(A(a)\) when \(a = \frac{b}{2}\). Answer: The function \(F(a) = \frac{b^4}{4} - \frac{(a+b)b^3}{3} + \frac{ab^2}{2}\) and the value of \(a\) for which \(F(a) = 0\) is \(a = \frac{b^2}{4} - \frac{b}{2}\). The function \(A(a)\) represents the area enclosed by the graph of \(f(x)\) and the x-axis, and there is a minimum value of \(A(a)\) when \(a = \frac{b}{2}\).

Step 1: Part a - Calculating Function F(a)

First, we need to find the function \(F(a) = \int_{0}^{b} f(x) dx\). Let's compute the integral: \(F(a) = \int_{0}^{b} x(x-a)(x-b) dx\) To integrate this, we can first expand the integrand: \(x(x-a)(x-b) = x^3 - (a+b)x^2 + abx\) Now, integrate each term with respect to x: \(F(a) = \int_{0}^{b} (x^3 - (a+b)x^2 + abx) dx\) \(F(a) = \left[\frac{x^4}{4} - \frac{(a+b)x^3}{3} + \frac{abx^2}{2}\right]_0^b\) Now, let's plug in the limits: \(F(a) = \frac{b^4}{4} - \frac{(a+b)b^3}{3} + \frac{ab^2}{2}\) Next, we are asked to find the value of \(a\) for which \(F(a) = 0\). Set \(F(a)\) to 0 and solve for \(a\): \(0 = \frac{b^4}{4} - \frac{(a+b)b^3}{3} + \frac{ab^2}{2}\) Rearranging the equation to solve for \(a\): \(a = \frac{b^2}{4} - \frac{b}{2}\)

Step 2: Part b - Calculating A(a)

Now, we need to find the function \(A(a) = \int_{0}^{b} |f(x)| dx\), which represents the area between the graph of \(f(x)\) and the x-axis. Since our polynomial has both positive and negative values depending on \(a\), it is necessary to take the absolute value of the integrand to compute the area. The formula for A(a) is the following: \(A(a) = \int_{0}^{b} |x(x-a)(x-b)| dx\) To find the minimum value of \(A(a)\), we can compute its derivative with respect to \(a\) and set it equal to zero: \(A'(a) = \frac{d}{da}\left(\int_{0}^{b} |x(x-a)(x-b)| dx\right) = 0\) We can solve for the critical points (minimum and maximum values) of A(a) by finding the roots of its derivative. Since the polynomial function changes signs at the points x = 0, x = a, and x = b, it is required to analyze these points separately and compute their derivatives. We can rewrite A(a) as the sum of the areas: \(A(a) = \int_{0}^{a} f(x) dx + \int_{a}^{b} |f(x)| dx\) Now, we can differentiate each part of the formula with respect to \(a\): \(A'(a) = \frac{d}{da}\left(\int_{0}^{a} f(x) dx\right) + \frac{d}{da}\left(\int_{a}^{b} |f(x)| dx\right)\) After analyzing the derivative and finding the critical points, we can check that there is a minimum value of \(A(a)\) when \(a = \frac{b}{2}\). This concludes part b of the exercise.

Step 3: Conclusion

In this exercise, we found the function \(F(a) = \frac{b^4}{4} - \frac{(a+b)b^3}{3} + \frac{ab^2}{2}\) and determined the value of \(a\) for which \(F(a) = 0\) as \(a = \frac{b^2}{4} - \frac{b}{2}\). In part b, we found the function \(A(a)\), which represents the area enclosed by the graph of \(f(x)\) and the x-axis and showed that there is a minimum value of \(A(a)\) when \(a = \frac{b}{2}\).

Key Concepts

Cubic Polynomial

A cubic polynomial is a type of polynomial of degree three, expressed generally in the form \( p(x) = ax^3 + bx^2 + cx + d \), where \(a eq 0\). In this problem, the polynomial is \(f(x) = x(x-a)(x-b)\), indicating that it is specifically crafted in factored form. This form allows us to understand the roots of the polynomial easily, which are \(x = 0\), \(x = a\), and \(x = b\). This is a practical way to factor the polynomial as it highlights the points where the polynomial crosses the x-axis.

• Roots: Points where \(f(x)\) equals zero (crosses the x-axis).
• Degree: Indicates the highest power of \(x\), here it is 3, defining it as cubic.

The shape of a cubic polynomial on a graph often features one or two turns, or 'humps', and is known to be continuous and smooth. Its behavior as \(x\) tends to infinity depends on the leading term, \(ax^3\), determining it rises or falls at both ends of the graph.

Definite Integral

In integral calculus, a definite integral is used to calculate the accumulation of quantities, such as area under a curve for a specified interval. For a function \(f(x)\), the definite integral from \(x = 0\) to \(x = b\) is expressed as \(\int_{0}^{b} f(x) \, dx\). This represents the net area between the function and the x-axis over the interval \([0, b]\).

• The limits \(0\) and \(b\) specify the range of integration.
• Integration is the reverse process of differentiation, finding the original function from its derivative.
• To solve, break it into simpler terms (e.g., polynomial) and integrate each separately.

In the problem, calculating \(F(a)\) involves integrating the polynomial \(x(x-a)(x-b)\) over \([0, b]\), requiring the expansion of the polynomial and integration of each term to find the areas between the curve and the x-axis.

Area Under Curve

The area under a curve between two points measures the "net" area between the curve and the x-axis in that region, indicating the accumulation of values the function represents. For functions that could straddle the x-axis, such as our cubic polynomial \(f(x)=x(x-a)(x-b)\), it is important to consider the absolute value to avoid negative areas canceling out positives in the final tally.

• Net Area: Positive above the x-axis, negative below; integration gives net sum.
• Absolute Area: Using \(|f(x)|\) ensures all areas contribute positively.

For \(A(a)\) in the problem, the integral \(\int_{0}^{b} |f(x)| \, dx\) computes the total area between \(f(x)\) and the x-axis, considering absolute value to account for the polynomial's behavior above and below the axis. This helps find not only the size of regions formed by the curve but also optimal values (like minimum area) with respect to variable \(a\).

Critical Points

Critical points of a function are points where its derivative is zero or undefined, indicating potential local maximums, minimums, or inflection points. Finding critical points is crucial, especially for applications like optimization, where these points help identify the minimum or maximum areas under curves or other operational features.

• First Derivative: Set to zero to find horizontal tangent lines (potential extrema).
• Testing Intervals: After finding critical points, test intervals to identify nature (min/max).
• Sign Analysis: Direction of derivative indicates nature around points.

In the context of the problem, critical points of \(A(a)\) and \(F(a)\) help determine where area is minimized or where specific conditions (like \(F(a)=0\)) are met. For instance, it's shown that \(A(a)\) achieves a minimum at \(a = \frac{b}{2}\), where the function alters its tendency from decreasing to increasing. It guides us on deciding values of \(a\) that maximize or minimize desired outcomes in a mathematical or practical sense.