Problem 22 - Applied Calculus

Question

Find an antiderivative. $$ g(x)=\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}} $$

Step-by-Step Solution

Answer

The antiderivative is \( G(x) = \ln|x| - \frac{1}{x} - \frac{1}{2x^2} + C \).

Step 1: Understand the Exercise

We need to find an antiderivative of the function \( g(x) = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \). This means we need to find a function \( G(x) \) such that \( G'(x) = g(x) \).

Step 2: Apply the Power Rule for Antiderivatives

Recall that the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) for any real number \( n eq -1 \). If we have \( \frac{1}{x^n} \), we rewrite it as \( x^{-n} \).

Step 3: Find the Antiderivative of \( \frac{1}{x} \)

For \( \frac{1}{x} = x^{-1} \), the antiderivative is \( \ln|x| \), due to the nature of the specific power rule for \( n = -1 \).

Step 4: Find the Antiderivative of \( \frac{1}{x^2} \)

Rewrite \( \frac{1}{x^2} \) as \( x^{-2} \). The antiderivative is \( \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} \).

Step 5: Find the Antiderivative of \( \frac{1}{x^3} \)

Rewrite \( \frac{1}{x^3} \) as \( x^{-3} \). The antiderivative is \( \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} \).

Step 6: Combine the Results with a Constant of Integration

Combine the antiderivatives: \( G(x) = \ln|x| - \frac{1}{x} - \frac{1}{2x^2} + C \), where \( C \) is the constant of integration.

Key Concepts

Understanding Calculus and Antiderivatives

Calculus is a branch of mathematics that deals with studying rates of change and accumulation. It consists of two main parts: differentiation and integration. In this context, an **antiderivative** is a function whose derivative gives the original function. For example, if the derivative of a function is \( g(x) \), then an antiderivative of \( g(x) \) is a function \( G(x) \) so that \( G'(x) = g(x) \). Finding an antiderivative is essentially a reverse process of differentiation.b. It results in a family of functions with a constant of integration \( C \), which acknowledges any constant could have been added to the original function prior to differentiation.

Utilizing the Power Rule for Antiderivatives

The power rule is a fundamental tool in calculus for finding derivatives and antiderivatives. For antiderivatives, the power rule states that the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), unless \( n = -1 \). In such a case, it doesn't apply directly, as the antiderivative of \( x^{-1} \) is \( \ln|x| \). This rule is particularly useful when dealing with functions in the form of powers of \( x \), and converting expressions like \( \frac{1}{x^n} \) into \( x^{-n} \) simplify the process.

• Rewrite expressions to fit the power rule format.
• Care for exceptions like \( n = -1 \) where the logarithmic function comes into play.
• Add the constant \( C \) to account for the family of antiderivatives.

Integration by Parts: A Different Approach

While not used in this exercise, **integration by parts** is a useful technique for finding antiderivatives of products of functions. It stems from the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]This technique typically comes in handy when dealing with integrals that involve a product of a polynomial and an exponential or trigonometric function. It transforms the original integral into a more straightforward one.
However, integration by parts requires the proper identification of functions \( u \) and \( dv \) to facilitate simpler integration. This exercise didn't require integration by parts due to its straightforward application of the power rule and properties of logarithmic functions.

Logarithmic Functions in Integration

Logarithmic functions often appear in integration, especially when dealing with expressions like \( \frac{1}{x} \). The antiderivative of \( \frac{1}{x} \) is particularly significant because it results in the natural logarithm: \( \ln|x| \). This occurs because the substitution \( x^{-1} \) doesn't fit the regular power rule which leads to \( \ln|x| \) due to the undefined nature of power rule at \( n = -1 \).
Besides,

• Logarithmic integration is crucial for transforming division into a simpler form, particularly when simplifying power expressions.
• The logarithmic function represents the concept of area under the curve for functions like \( \frac{1}{x} \).

Understanding how logarithmic functions work alongside other basic rules is essential for mastering integral calculus.