Problem 10 - Calculus with Concepts in Calculus

Question

Find all antiderivatives of the given function. $$ x^{n} \text { for } n \neq-1 $$

Step-by-Step Solution

Answer

The antiderivatives of \( x^n \) are \( \frac{x^{n+1}}{n+1} + C \), \( n \neq -1 \).

Step 1: Understanding the Problem

We need to find all antiderivatives, or indefinite integrals, of the function \( x^n \) where \( n eq -1 \). This involves applying the basic rules of integration.

Step 2: Applying the Power Rule for Integration

The power rule for integration states that the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). This is derived by reversing the power rule for differentiation.

Step 3: Integrating the Function

Since \( n eq -1 \), we can use the formula: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( C \) is the constant of integration, representing all possible antiderivatives.

Step 4: Concluding the Integration

All antiderivatives of the function \( x^n \) are given by the expression \( \frac{x^{n+1}}{n+1} + C \) where \( C \) is any constant. This accounts for all possible shifts vertically of the function's curve.

Key Concepts

Indefinite Integrals

An indefinite integral is a fundamental concept in calculus, used to find antiderivatives of a given function. When we talk about indefinite integrals, we are referring to integrals that do not have specified upper and lower limits. These are represented as \( \int f(x) \, dx \). The result is a family of functions that differ by a constant. Indefinite integrals are crucial because they help to determine all possible antiderivatives for a function.
They express the idea of a 'general solution' to an antiderivative problem. Instead of a single answer, indefinite integrals give a range of functions that could be the solution, which includes all shifts up or down of the basic curve.
In our specific case with \( x^n \), where \( n eq -1 \), the goal is to find its indefinite integral. This results in an expression that contains a variable term raised to a power, plus a constant.

Power Rule for Integration

The power rule for integration is a straightforward but incredibly useful tool when working with polynomials. It allows us to find the integral of any function in the form of \( x^n \) with ease. The formula is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) provided \( n eq -1 \). This comes from reversing the process of differentiation, where for differentiation the power of \( x \) decreases by one, in integration, it increases.

• The exponent of \( x \) is incremented by 1.
• The new term is divided by this increased exponent.
• A constant \( C \) is added at the end to encompass all possible antiderivatives.

Applying the power rule correctly transforms the expression, enabling us to integrate terms one by one, if dealing with a polynomial, or directly if it's a simple power of \( x \).
This rule eases the process, making polynomial integration systematic and intuitive.

Constant of Integration

The constant of integration, usually denoted as \( C \), plays a vital role in the realm of indefinite integrals. When we integrate, we reverse differentiation, a process that loses the constant information present in the original function. Thus, by adding \( C \), we acknowledge that there are infinitely many functions differing only by a constant value.
This means that the graph of the antiderivative can shift vertically, depending on the value of \( C \). Mathematically, when you differentiate any constant, the result is zero, which is why you can't see it in the differentiated expression.

• \( C \) represents any real number.
• It accounts for the initial conditions in particular solutions.
• It ensures completeness of the family of solutions.

In practical terms, whenever you integrate and obtain an expression, the \( C \) term makes sure no possible variation is left out, covering every potential starting point of the original curve. Understanding \( C \) is essential to fully grasp how integration comprehensively relates back to differentiation.