Problem 35 - Applied Calculus

Question

Find each indefinite integral. \(\int(r-1)(r+1) d r\)

Step-by-Step Solution

Answer

The integral is \(\frac{r^3}{3} - r + C\).

Step 1: Expand the Expression

Before integrating, expand the integrand. The expression \((r-1)(r+1)\) is a difference of squares. Expanding it gives: \((r-1)(r+1) = r^2 - 1^2 = r^2 - 1\). Thus, the integral we need to solve is \(\int (r^2 - 1) \, dr\).

Step 2: Integrate Each Term Separately

Now, let's find the antiderivative of each term separately. The integral of \(r^2\) is \(\frac{r^3}{3}\) and the integral of \(1\) is \(r\). Therefore, \(\int (r^2 - 1) \, dr = \int r^2 \, dr - \int 1 \, dr = \frac{r^3}{3} - r\).

Step 3: Include the Constant of Integration

Since this is an indefinite integral, it's important to include the constant of integration \(C\). The resulting expression is: \(\frac{r^3}{3} - r + C\).

Key Concepts

Antiderivatives

When learning about indefinite integrals, a key concept is the antiderivative. An antiderivative of a function is another function whose derivative is equal to the original function. In different terms, if you have a function \( f(x) \), an antiderivative would be a function \( F(x) \) such that \( F'(x) = f(x) \). This means when you differentiate \( F(x) \) you'll obtain the original function.

• For example, if you have \( r^2 \) and want to find its antiderivative, determine a function that, when differentiated, results in \( r^2 \). That function is \( \frac{r^3}{3} \).
• Similarly, for a constant like \(-1\), the antiderivative is \(-r\) because the derivative of \(-r\) is \(-1\).

In the case of finding indefinite integrals, you are essentially finding the antiderivative of the integrand, as you did when you integrated \( r^2 - 1 \) to get \( \frac{r^3}{3} - r \). Understand that there are infinitely many antiderivatives for any given function, which leads us to our next key concept: the constant of integration.

Constant of Integration

When calculating indefinite integrals, you always conclude with the constant of integration, denoted as \( C \). This constant is crucial as it represents the infinite number of antiderivatives possible for a function. Let's break it down:

• Imagine the antiderivative of a function is a curve on a graph. Each \( C \) shifts this curve up or down. Since every constant shift results in another valid antiderivative, \( C \) acknowledges these shifts.
• In practical terms, if \( F(r) = \frac{r^3}{3} - r \) is an antiderivative for \( r^2 - 1 \), then \( F(r) + 5 \) would also be a valid antiderivative, just shifted up by 5 units.

This addition of \( C \) ensures that any particular solution of an indefinite integral includes all possible vertical shifts of the antiderivative function, thus encapsulating every possible scenario.

Difference of Squares

The term "difference of squares" refers to a specific algebraic pattern, which is pivotal in simplifying integration. This pattern arises when you multiply two binomials that differ only in sign between the terms.

• The general form is \((a - b)(a + b) = a^2 - b^2\). This identity simplifies expressions before integrating by reducing a product to a simpler subtraction.
• In our example, \((r-1)(r+1)\), each side of the expression is a simple linear term differing by a sign, making it a perfect candidate for using the difference of squares formula.

Recognizing these patterns is vital in calculus, as it allows you to reduce complex expressions into a more straightforward form. This simplification is evident in the exercise you solved, where simplifying \((r-1)(r+1)\) to \(r^2 - 1\) allowed for straightforward term-to-term integration.