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Chapter 6: Problem 27

Find the general antiderivative. $$g(x)=\frac{5}{x^{3}}$$

Short Answer

Expert verified
The general antiderivative is \(-\frac{5}{2x^2} + C\).

Step by step solution

01

Identify the Function Form

The function given is \( g(x) = \frac{5}{x^3} \). We can rewrite this function in a form that's easier to integrate: \( g(x) = 5x^{-3} \).
02

Apply the Power Rule for Integration

The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) is the constant of integration. For \( g(x) = 5x^{-3} \), apply this rule: Each power of \( x \) will increase by 1 and divide the coefficient by this new power.
03

Calculate the Antiderivative

Integrate \( 5x^{-3} \) to get \( \int 5x^{-3} \, dx = 5 \cdot \frac{x^{-3+1}}{-3+1} + C = 5 \cdot \frac{x^{-2}}{-2} + C = -\frac{5}{2}x^{-2} + C \).
04

Rewrite the Expression

The general antiderivative can be expressed in a simpler form: \(-\frac{5}{2}x^{-2} + C = -\frac{5}{2x^2} + C \). Thus, the general antiderivative is expressed in terms of a more conventional fraction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve. There are various techniques employed to find integrals, depending on the form of the function. Some common techniques include:
  • **Substitution method**: Useful when the integrand contains a function and its derivative.
  • **Integration by parts**: Employed when the integrand is a product of two functions.
  • **Partial fraction decomposition**: Helps in integrating rational functions by breaking them into simpler parts.
In this context, we focus on a specific technique suitable for functions of the form \( x^n \) which is the Power Rule for Integration. This technique allows us to handle polynomials and functions that can be manipulated into polynomial-like forms. By knowing these methods, students can choose the most efficient approach based on the problem type.
Power Rule for Integration
The Power Rule for Integration is a straightforward method to find the antiderivative of functions in the form \( x^n \). It provides a simple formula:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]where \( n eq -1 \) and \( C \) is the constant of integration.
To utilize this rule effectively:
  • Ensure the function is expressed as \( x^n \) by manipulating algebraically if necessary. For example, rewrite \( \frac{5}{x^3} \) as \( 5x^{-3} \).
  • Apply the rule by increasing the exponent by one and dividing the term by the new exponent.
  • Don't forget to add the constant of integration at the end.
In the given exercise, this rule transforms the expression allowing an easy computation: \[5 \cdot \int x^{-3} \, dx = 5 \cdot \frac{x^{-2}}{-2} = -\frac{5}{2}x^{-2}\]Ultimately, understanding how and when to apply the power rule is crucial for solving basic integration problems efficiently.
Constant of Integration
In integration, particularly when finding antiderivatives, we often introduce a term called the "constant of integration," denoted by \( C \). This constant is essential because:
  • When you differentiate a function, any constant added to the function vanishes. Thus, when integrating to find the original function, adding \( C \) accounts for any possible constants that might have been differentiated away.
  • It represents the family of all possible antiderivatives. Since integration is the reverse process of taking a derivative, multiple functions can share the same derivative but differ by a constant.
In the given solution, \( C \) signifies that for each possible original function, the general form \(-\frac{5}{2x^2} + C\) captures all of them. Therefore, including the constant of integration in the solution is indispensable for completeness and mathematical accuracy.

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Most popular questions from this chapter

In drilling an oil well, the total cost, \(C,\) consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are $$ C^{\prime}(x)=4000+10 x \text { riyals/meter } $$ where \(x\) is the depth in meters. Find the total cost of drilling a well \(x\) meters deep.

Let \(F(x)=\int_{0}^{x} \sin (2 t) d t\) (a) Evaluate \(F(\pi)\) (b) Draw a sketch to explain geometrically why the answer to part (a) is correct. (c) For what values of \(x\) is \(F(x)\) positive? negative?

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Explain what is wrong with the statement. $$\frac{d}{d x} \int_{0}^{5} t^{2} d t=x^{2}$$
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