Problem 32 - University Calculus: Early Transcendentals

Question

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(\frac{1}{5}-\frac{2}{x^{3}}+2 x\right) d x$$

Step-by-Step Solution

Answer

The most general antiderivative is \( \frac{x}{5} + \frac{1}{x^2} + x^2 + C \).

Step 1: Integrate Each Term Separately

The given integral is \( \int \left( \frac{1}{5} - \frac{2}{x^3} + 2x \right) \, dx \). We integrate each term separately: 1. \( \int \frac{1}{5} \, dx = \frac{x}{5} \), because the antiderivative of a constant \( c \) is \( cx \).2. \( \int -\frac{2}{x^3} \, dx = \int -2x^{-3} \, dx = -2 \left( \frac{x^{-2}}{-2} \right) = \frac{1}{x^2} \), because the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), given \( n eq -1 \).3. \( \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 \), using the power rule for antiderivatives.

Step 2: Combine Results and Add Constant of Integration

Combine the results from Step 1 to find the most general antiderivative: \[ \frac{x}{5} + \frac{1}{x^2} + x^2 + C \]where \( C \) is the constant of integration.

Step 3: Check the Solution by Differentiation

Differentiate the obtained antiderivative \( \frac{x}{5} + \frac{1}{x^2} + x^2 + C \) to verify it:1. The derivative of \( \frac{x}{5} \) is \( \frac{1}{5} \).2. The derivative of \( \frac{1}{x^2} \) is \( -2x^{-3} \).3. The derivative of \( x^2 \) is \( 2x \).4. The derivative of a constant \( C \) is zero.Combine these: \( \frac{1}{5} - \frac{2}{x^3} + 2x \), which matches the original integrand.

Key Concepts

Antiderivative

In calculus, finding an antiderivative is like solving a puzzle where you search for a function whose derivative will yield the original function given in the problem. The antiderivative, also known as an "indefinite integral," for a function is not unique because of the constant of integration. An antiderivative reverses the process of differentiation.

• For example, if you know the derivative of a function is a constant, say 5, then its antiderivative could be something like 5x, where x is a variable because when you differentiate 5x, you get back to 5.
• The term "indefinite" comes into play because when you compute the indefinite integral, you're not bound to an endpoint or limit like in definite integrals.
• It's vital to correctly identify the basic rules of antiderivatives to solve the problems. Functions could be simple like constants or more complex like polynomials.

The integral of a constant, polynomial, or even some rational functions requires applying different rules and creativity to deconstruct them into manageable parts.

Constant of Integration

When you find an antiderivative, the constant of integration, denoted by \( C \), plays a crucial role. Every indefinite integral includes this constant because when differentiating any constant, it results in zero; thereby staying "hidden" if not included in the solution.

• The constant of integration ensures that all possible antiderivatives of a function are captured.
• No matter how complicated the function is, when you integrate without limits, you add this constant to reflect that there are infinitely many antiderivatives.
• For example, if \( \int 3x^2 \,dx = x^3 + C \), the \( C \) stands for any constant value. Differentiating back will always lead to the original function \( 3x^2 \) irrespective of \( C \).

Always remember to add \( C \) to your answer in indefinite integrals, as it is a symbol of completeness in your solution.

Power Rule

The power rule is a handy tool in calculus, allowing you to find antiderivatives or derivatives of functions in a straightforward way. When you apply the power rule for integration, any function of the form \( x^n \) where \( n eq -1 \) is transformed by increasing the exponent by 1 and then dividing by the new exponent.Here's the formula:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

• This rule simplifies the integration of polynomials directly.
• It's applicable across various integration problems and helps break them down into smaller, solvable bits.
• For instance, the integral \( \int 2x \, dx \) becomes \( \frac{2x^2}{2} = x^2 + C \).

However, note that you cannot apply it when \( n = -1 \). In such cases, the integral turns into \( \ln|x| + C \). The power rule aids in quick solutions and is foundational for tackling more complex calculus problems.