Problem 101 - College Algebra and Calculus: An Applied Approach

Question

Use a symbolic integration utility to evaluate the definite integral. \(r^{6}\). $$ \int_{2}^{5}\left(\frac{1}{x^{2}}-\frac{1}{x^{3}}\right) d x $$

Step-by-Step Solution

Answer

The value of the definite integral is -0.03.

Step 1: Identify the form

The integrand is \(1/x^2 - 1/x^3\). These terms are of the form \(1/x^n\), and easily integrable.

Step 2: Integrate Term by Term

We integrate each term separately. So, the integral \(\int x^{-n} dx\) will be \(-x^{-(n-1)}/(n-1)\). Applying this rule, we get:\(\int_{2}^{5} x^{-2} dx - \int_{2}^{5} x^{-3} dx\) will become \(-x^{-1}|_{2}^{5} + (x^{-2}/2) |_{2}^{5}\)

Step 3: Compute the Definite Integrals

Next step is to find the definite integral by substituting the range values. So, we subtract the value at x=2 from the value at x=5 for both integrals. Evaluation of these yields -\((1/5 - 1/2) + (1/2*1/5^2 - 1/2*1/2^2)\)

Step 4: Simplify the Result

Simplifying this, we get -0.3 + 0.02 + 0.25 = -0.03.

Key Concepts

Symbolic Integration

Symbolic integration is a process used to find the exact antiderivative of a function, in a way that we can express the answer as a mathematical expression involving numbers and symbols. This contrasts with numerical integration, where we seek a numerical approximation to the value of the integral.

Symbolic integration often involves recognizing patterns or standard forms in the integrand that correspond with known integrals. For example, in the exercise given, the integrand is a difference of two terms: \(\frac{1}{x^2} - \frac{1}{x^3}\). Each of these terms is a power function of x, which can be integrated using known formulas and rules of integration. The ability to transform and simplify the given integral into a combination of simpler, standard forms, is central to symbolic integration and is frequently used in problems involving integral calculus.

Integral Calculus

Integral calculus is one of the two principal branches of calculus, the other being differential calculus. Integral calculus is focused on the concept of integration, which essentially involves finding the area under a curve or, more generally, finding the total accumulation of a quantity.

In the context of the problem at hand, integrating the function \(\frac{1}{x^2} - \frac{1}{x^3}\) over the interval from x=2 to x=5 can be thought of as measuring the net area between the function and the x-axis over that interval. The 'definite integral' evaluates this expression to give a real number, which represents this total accumulated value. Integral calculus includes techniques and theorems such as the Fundamental Theorem of Calculus, which connects integration with differentiation and provides a method for evaluating definite integrals.

Power Rule for Integration

The power rule for integration is a fundamental rule used to integrate powers of x. In its general form, it states that the integral of \(x^n\) with respect to x is \(\frac{x^{n+1}}{n+1}\), provided that \(n \eq -1\).

However, if \(n = -1\), we have a special case and the integral of \(x^{-1}\), which is \(\ln|x|\). The power rule simplifies the process of integration by providing a straightforward formula to apply when the integrand is a power function, as in the step-by-step solution provided for the exercise, where the integrand is in the form of \(x^{-2}\) and \(x^{-3}\). By utilizing the power rule for each term separately, the problem becomes more manageable, allowing for efficient symbolic integration.

Indefinite Integrals

Indefinite integrals, sometimes called antiderivatives, are integrals that do not have specified limits of integration. In other words, when we find an indefinite integral, we are finding a function (or family of functions) that would differentiate to give the original integrand.

In practice, an indefinite integral is represented by the integral symbol without bounds and a constant of integration, usually denoted by C, is added at the end. The indefinite integral of a function f(x) with respect to x is written as \(\int f(x) \, dx\).

The indefinite integral is fundamental to the process of evaluating definite integrals, as seen in the exercise. First, we found the indefinite integrals of \(x^{-2}\) and \(x^{-3}\), included constants, and then applied these results within the limits of integration to find the definite integral value, which in this case was -0.03.