Problem 4 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 20 , find the indicated indefinite integral. \(\int\left(2 \sqrt[3]{s}+\frac{5}{s}\right) d s\)

Step-by-Step Solution

Answer

\( \int \left(2 \sqrt[3]{s} + \frac{5}{s}\right) \, ds = \frac{3}{2} s^{4/3} + 5 \ln |s| + C \)

Step 1: Rewrite the integrand

Express each term in a more convenient form to integrate. Rewrite \(2 \sqrt[3]{s} + \frac{5}{s}\) as \(2s^{1/3} + 5s^{-1}\).

Step 2: Apply the Power Rule

Use the power rule \(\int s^n \, ds = \frac{s^{n+1}}{n+1} + C\) to integrate each term separately. For \(2s^{1/3}\): \( \int 2s^{1/3} \, ds = 2 \cdot \frac{s^{1/3+1}}{1/3+1} + C_1 = 2 \cdot \frac{s^{4/3}}{4/3} = \frac{3}{2} s^{4/3} + C_1\). For \(5s^{-1}\): \( \int 5s^{-1} \, ds = 5 \int s^{-1} \, ds = 5 \ln|s| + C_2\).

Step 3: Combine the results

Add the results of the two integrations together, including the constant of integration \(C\). \( \int \left(2s^{1/3} + \frac{5}{s}\right) \, ds = \frac{3}{2} s^{4/3} + 5 \ln |s| + C \).

Key Concepts

Power Rule

The Power Rule is a fundamental principle in calculus used to find the integral of terms in the form of a variable raised to a power. This rule states that for any real number apart from -1, the indefinite integral of raised to the power is given by: \ \ \( \int s^n \, ds = \frac{s^{n+1}}{n+1} + C \) The power n can be any real number except for -1. We use this rule to handle terms like 2s^{1/3} and rewrite like 5/s as s^{-1}. These transformations make it easier to apply the Power Rule. For instance: - If you have 2 and 2 is as using the rule, it gives: \ \( \int 2s^{1/3} \, ds = 2 \cdot \frac{s^{1/3+1}}{1/3+1} \) Evaluating the exponent, \ \ \( 1/3+1 = 4/3 \) \ \( = 2 \cdot \frac{s^{4/3}}{4/3} = \frac{3}{2} s^{4/3} + C_1 \) This approach significantly simplifies the integration process.

Integration

Integration, the reverse process of differentiation, is a key concept in calculus. We use integration to find the area under curves and to solve equations that describe accumulated quantities. An indefinite integral represents a family of functions and includes an arbitrary constant (C) because the derivative of a constant is zero. Here, in our exercise, we have two functions to integrate separately: \( 2s^{1/3} \) and \( 5/s \). By splitting the integral, we simplify the process. The steps can be summarized as follows: 1. Break down the integrand: We rewrite 2s^{1/3} and 5/s in a form that is easier to integrate: \ \(2s^{1/3} \rightarrow 2s^{1/3} \ and \rightarrow \frac{5}{s} = 5s^{-1} \). 2. Apply the Power Rule and 5/s as a logarithmic function. - For : \( \int 5s^{-1} \, ds = 5 \int s^{-1} \, ds = 5 \ln |s| + C_2 \). The integral of is a well-known result, which leads us to the next core concept.

Logarithmic function

Logarithmic functions are essential in many areas of mathematics, including integration. When you integrate a function of the form 1/s, the result is a natural logarithm. In our exercise: \ \ \( \int 5s^{-1} \, ds = 5 \int \frac{1}{s} \, ds \ = 5 \ln |s| + C \) The natural logarithm, \( \ln |s| \), measures the time needed to reach a certain level of continuous growth. It is an integral part of solving differential equations and modeling exponential growth or decay. Combining the Results: After integrating each term, don't forget to combine them to form the final solution with a single constant of integration. The complete solution to our original integral is: \ \( \int \left(2s^{1/3} + \frac{5}{s}\right) \, ds = \frac{3}{2} s^{4/3} + 5 \ln |s| + C \) Always include , as it represents the family of all possible solutions. Understanding these fundamental concepts will help you tackle more complex integration problems confidently.