Problem 12 - Applied Calculus

Question

Decide whether the expression is a number or a family of functions. (Assume \(f(x)\) is a function.) $$\int_{1}^{3}\left(\frac{x}{3}+\frac{3}{x}\right) d x$$

Step-by-Step Solution

Answer

The expression is a number.

Step 1: Understand the Integral

The given expression is an integral: \(\int_{1}^{3}\left(\frac{x}{3}+\frac{3}{x}\right) \ dx\). This expression represents the area under the curve of the function \(\frac{x}{3}+\frac{3}{x}\) from \(x = 1\) to \(x = 3\). Our task is to determine whether this integral yields a single number or a family of functions.

Step 2: Recall Properties of Definite Integrals

By definition, a definite integral, like \(\int_{a}^{b} f(x) \ dx\), calculates the net area between the function \(f(x)\) and the x-axis from \(x = a\) to \(x = b\). This results in a single numerical value rather than a function or a family of functions.

Step 3: Evaluate the Integral

Although we are not explicitly asked to compute the integral, evaluating it can deepen understanding. Compute as follows: \[\int_{1}^{3} \left(\frac{x}{3} + \frac{3}{x}\right) \, dx = \int_{1}^{3} \frac{x}{3} \, dx + \int_{1}^{3} \frac{3}{x} \, dx.\] Evaluate these integrals one by one: \[\int \frac{x}{3} \, dx = \frac{x^2}{6} + C, \quad \int \frac{3}{x} \, dx = 3\ln|x| + C.\] Apply limits and difference: \[\left(\frac{x^2}{6} + 3\ln|x|\right) \Bigg|_{1}^{3}.\] This results in a specific number.

Step 4: Conclusion

The result of the integral \(\int_{1}^{3}\left(\frac{x}{3}+\frac{3}{x}\right) \ dx\) is a definite numerical value as it represents the area calculation over a specific interval. Therefore, it is not a family of functions.

Key Concepts

Integral Calculus

Integral Calculus deals with the concept of integration, which is essentially the reverse process of differentiation. While differentiation finds the rate of change of a function, integration is used to determine the accumulation of quantities, such as areas under curves. In this context, an integral can be classified as either definite or indefinite. A definite integral, like the one in the exercise, encompasses an interval with specified bounds, such as from 1 to 3 in the given problem.

When we work with a definite integral, we aim to calculate the total accumulation or the net area between the curve defined by the function and the x-axis over the defined interval. This process results in a specific numerical answer, not a general formula or family of functions. On the other hand, an indefinite integral results in an antiderivative of a function plus a constant \(C\), thereby defining a family of functions. For the given exercise, the definite integral yields a specific value, confirming that it results in a number.

Area Under Curve

Calculating the area under a curve is a fundamental application of definite integrals in Integral Calculus. When you have a function, like \(\frac{x}{3} + \frac{3}{x}\), the area under its curve between two points on the x-axis can provide insights into various physical phenomena or represent summation of values over time or another variable.

The process involves determining how much space is covered by the curve between these x-axis points, like from \(x=1\) to \(x=3\) as in our example. This calculation identifies the definite integral and, importantly, results in a numerical value. This value is considered the net area, as it takes into consideration areas above and possibly below the x-axis.

• Positive areas (above the x-axis) and negative areas (below the x-axis) are taken into account in this process.
• Any segments of the curve that dip below the x-axis will be subtracted from the total area.

Understanding this concept is crucial to the mastery of definite integrals and their practical application in science and engineering.

Function Evaluation

Function evaluation in the setting of definite integrals refers to the process of substituting the limits of integration into the antiderivative found during the computation. For the function \(\frac{x}{3} + \frac{3}{x}\), we first find its integral, which leads us to antiderivatives like \(\frac{x^2}{6}\) and \(3\ln|x|\).

Once these antiderivatives are determined, we apply the process of evaluating them at the specified limits, known as evaluating the integral \[\left(\frac{x^2}{6} + 3\ln|x|\right) \Bigg|_{1}^{3}.\]
This involves:

• Calculating the value of the antiderivative at the upper limit (\(x=3\)).
• Computing the same at the lower limit (\(x=1\)).
• Subtracting the lower limit result from the upper limit result gives the precise area under the curve over the interval.

This step ensures that we get a clear, definitive number, revealing not only the method but also the importance of precision in calculations.