Problem 34 - 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: Identify the function to integrate

The given function is \( g(x) = \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{3}} \). To find the antiderivative, we need to integrate each term separately.

Step 2: Find an antiderivative of \( \frac{1}{x} \)

Recall the integral of \( \frac{1}{x} \) is the natural logarithm. Therefore, the antiderivative of \( \frac{1}{x} \) is \( \ln |x| + C_1 \), where \( C_1 \) is a constant of integration.

Step 3: Find an antiderivative of \( \frac{1}{x^{2}} \)

Recognize that \( \frac{1}{x^{2}} \) can be rewritten as \( x^{-2} \). The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) for \( n eq -1 \). Thus, the antiderivative of \( x^{-2} \) is \( -x^{-1} + C_2 \), or just \( -\frac{1}{x} \).

Step 4: Find an antiderivative of \( \frac{1}{x^{3}} \)

Similarly, \( \frac{1}{x^{3}} \) can be rewritten as \( x^{-3} \). Integrating this will give \( \frac{x^{-3+1}}{-3+1} = -\frac{1}{2}x^{-2} + C_3 \).

Step 5: Combine the antiderivatives

Combine the antiderivatives from each step: \( G(x) = \ln |x| - \frac{1}{x} - \frac{1}{2x^{2}} + C \), where \( C \) is a constant of integration that combines \( C_1 \), \( C_2 \), and \( C_3 \).

Key Concepts

Integration Techniques

When we are tasked with finding an antiderivative, it involves applying different strategies and methods to accomplish the task of integration. One of the basic integration techniques is the **power rule**, which provides a straightforward way to integrate polynomial terms. If you have a function in the form of \[ x^n \] where \( n eq -1 \), you can use the power rule:- \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]Where \( C \) is the constant of integration.
Another technique used is identifying **known integrals**, such as that of \( \frac{1}{x} \), which integrates to the natural logarithm:- \[ \int \frac{1}{x} \, dx = \ln |x| + C \]Recognizing these patterns can simplify the integration process significantly.
By breaking down complex equations into simpler parts, like rewriting \( x^{-2} \) as \( \frac{1}{x^2} \), we can apply the power rule and retrieve our antiderivative conveniently.

Integration by Parts

Integration by Parts is a powerful technique used for integrating products of functions. It is based on the product rule for differentiation. The formula is given by:\[ \int u \, dv = uv - \int v \, du\]This formula is especially helpful when directly integrating a product is difficult.
Although our original function \( g(x) \) does not necessitate integration by parts, understanding it can be crucial as integration tasks grow more complicated. - Choose \( u \) and \( dv \) wisely: The part chosen as \( u \) is typically simplified upon differentiation, and \( dv \) should be easily integrated.- Apply recursively if needed: Sometimes, integration by parts needs to be used multiple times in a calculation.
This method demonstrates the interconnectedness of differentiation and integration, showing that deeper knowledge of one can assist in the other. It is a bridge when integration techniques like substitution and standard formulas don't directly solve a problem.

Definite and Indefinite Integrals

In calculus, integrals can be classified into two types: **definite** and **indefinite integrals**. Understanding their differences is key:- An **indefinite integral** represents a family of functions and includes a constant of integration \( C \).\[ \int f(x) \, dx = F(x) + C\]In this scenario, the focus is solely on finding the antiderivative of the function.
- A **definite integral** calculates the net area under the curve between two points \( a \) and \( b \):\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a)\]The result is a number, not a function.
When we solve the problem by finding \( G(x) = \ln |x| - \frac{1}{x} - \frac{1}{2x^{2}} + C \), this is an indefinite integral, as it involves integration without bounds. Such integrals allow us to understand the general behavior of accumulated quantities without specifying from where to where the accumulation occurs.