Problem 17 - Thomas Calculus in SI Units

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. $$\int(x+1) d x$$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Expression

The expression given is \( \int (x+1) \, dx \). This is an integral of a sum, which means we can integrate each term separately.

Step 2: Apply the Power Rule for Integration

The integral of \( x \) can be found using the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Since \( x \) is equivalent to \( x^1 \), integrating \( x \), we get \( \frac{x^{1+1}}{1+1} = \frac{x^2}{2} \).

Step 3: Integrate the Constant

The integral of \( 1 \) is simply \( x \), since \( \int 1 \, dx = x + C \). Thus, the overall integral becomes \( \int (x+1) \, dx = \frac{x^2}{2} + x + C \).

Step 4: Combine the Results and Include Constant of Integration

Combine the results from integrating both terms. The complete integral is \( \frac{x^2}{2} + x + C \), where \( C \) is the constant of integration.

Step 5: Validate by Differentiating the Result

Differentiate the result: \( \frac{x^2}{2} + x + C \). The derivative of \( \frac{x^2}{2} \) is \( x \) and the derivative of \( x \) is \( 1 \). The derivative of a constant is 0. So, the derivative is \( x + 1 \), which matches the original integrand, confirming our solution.

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus, particularly useful when solving integrals. It is a handy rule that allows us to integrate expressions of the form \( x^n \). The power rule states:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \( n eq -1 \) and \( C \) is the constant of integration. This rule simplifies the process of finding antiderivatives, as it directly relates the powers of \( x \) in differentiation and integration.

• For \( x^2 \), the integral becomes \( \frac{x^{2+1}}{2+1} = \frac{x^3}{3} + C \)
• For our example \( x \), seen as \( x^1 \), the integration simplifies to \( \frac{x^2}{2} \)

Using this rule, integration problems become much easier to handle, saving both time and effort.
Understanding how the power rule applies to each component of an expression is key to integrating more complex functions effectively. So, with practice, identifying when to use the power rule in your calculations will become almost second nature.

Antiderivative

An antiderivative, also known as an indefinite integral, is a reverse operation of differentiation. It seeks to find a function whose derivative is the given function. Essentially, if you have a function \( f(x) \), finding the antiderivative involves determining the function \( F(x) \) such that \( F'(x) = f(x) \).For instance, the antiderivative of a simple function like \( f(x) = x + 1 \) translates to finding the most general function that yields \( f(x) \, \) when differentiated. In this case, our antiderivative becomes \( \frac{x^2}{2} + x + C \).

• \( \int x \, dx = \frac{x^2}{2} + C \)
• \( \int 1 \, dx = x + C \)

A critical aspect of determining antiderivatives is including the constant of integration, \( C \). It accounts for any constant value that could have been part of the original function but vanished during differentiation. This constant means there are infinite functions that can be considered valid antiderivatives for a given derivative. Consequently, validating through differentiation is paramount to confirm your antiderivative is correct.

Integration by Parts

Integration by parts is another method of integration used when a simple application of the rules isn't enough. This technique is grounded in the product rule for differentiation and is used to integrate products of functions. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]Here,

• \( u \) is a function which simplifies when differentiated
• \( dv \) is a function whose integral is straightforward to find

Even though this method is not necessary for the specific problem of integrating \( \int(x+1) \, dx \), it is crucial for more complex situations involving products or compositions of functions. Understanding how to choose \( u \) and \( dv \) correctly in practice often requires a little trial and error, but with experience, your intuition improves.Lastly, although it can seem complicated at first, the integration by parts method simplifies over time, especially for functions that are products of polynomials, exponentials, or trigonometric functions.