Problem 19 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int_{1}^{-1}(r+1)^{2} d r$$

Step-by-Step Solution

Answer

The integral evaluates to \( -\frac{8}{3} \).

Step 1: Identify Integral Bounds and Limits

First, notice that the integral is evaluated from \( r = 1 \) to \( r = -1 \). This means the limits of the integral are reversed. When the upper limit is smaller than the lower limit, you can reverse the limits by changing the sign of the integral. Hence, \( \int_{1}^{-1}(r+1)^{2} \, dr = -\int_{-1}^{1}(r+1)^{2} \, dr \).

Step 2: Expand the Integrand

Expand \( (r+1)^{2} \) using the binomial theorem: \( (r+1)^{2} = r^{2} + 2r + 1 \). Rewrite the integral as: \[ - \int_{-1}^{1} (r^{2} + 2r + 1) \, dr \]

Step 3: Integrate Each Term Separately

Integrate each term in the expanded form:1. \( \int r^{2} \, dr = \frac{r^{3}}{3} \)2. \( \int 2r \, dr = r^{2}\)3. \( \int 1 \, dr = r \)Thus: \[ - \left[ \frac{r^{3}}{3} + r^{2} + r \right]_{-1}^{1} \]

Step 4: Evaluate the Integrals at the Bounds

Calculate the integrated expression at the upper and lower bounds:1. **At \( r = 1 \):** \( \frac{1^{3}}{3} + 1^{2} + 1 = \frac{1}{3} + 1 + 1 = \frac{7}{3} \)2. **At \( r = -1 \):** \( \frac{(-1)^{3}}{3} + (-1)^{2} + (-1) = -\frac{1}{3} + 1 - 1 = -\frac{1}{3} \)

Step 5: Subtract and Find the Result

Subtract the result of the lower bound evaluation from the upper bound evaluation:\[- \left( \frac{7}{3} - (-\frac{1}{3}) \right) = - \left( \frac{7}{3} + \frac{1}{3} \right) = - \left( \frac{8}{3} \right) = -\frac{8}{3} \]

Key Concepts

Integral Bounds

When dealing with definite integrals, one of the first things to consider is the integral bounds. These are the numbers at the bottom and the top of the integral sign, which tell you the interval over which you need to evaluate the function. In our example, the integral bounds are from 1 to -1. However, an important property of definite integrals is that if the upper bound is smaller than the lower bound, you simply reverse the bounds and change the sign of the integral. This simplifies the process and ensures consistent results.

• Original bounds: \(1 \, \text{to} \, -1\)
• Reversed bounds: \(-1 \, \text{to} \, 1\) and negate the integral.

Understanding this not only helps in tackling integrals involving reverse limits but also expands your comprehension of how integrals function fundamentally.

Binomial Theorem

The binomial theorem is a powerful tool used to expand expressions that are raised to a power, like \((r+1)^2\). It provides a straightforward way to expand such expressions into a sum of terms. This is especially useful when integrating polynomials.
In our solution, expanding \((r+1)^2\) was essential, turning it to \(r^2 + 2r + 1\). Once expanded, each term can be individually integrated, simplifying the process remarkably.

• The original expression: \( (r+1)^2 \).
• Expanded form: \( r^2 + 2r + 1 \).

By expanding, we ensure that the integration process becomes simple and straightforward, avoiding any unnecessary complications.

Integration Techniques

Integrating involves finding the antiderivative of a function. In this exercise, after expanding the function \((r+1)^2\) into \(r^2 + 2r + 1\), each term was integrated separately. These basic integration techniques are foundational to solving integrals efficiently.
Let’s break down the integration steps:

• Integrate \(r^2 \: \int r^2 \, dr = \frac{r^3}{3}\).
• Integrate \(2r \: \int 2r \, dr = r^2\).
• Integrate \(1 \: \int 1 \, dr = r\).

These integrations were then evaluated at the bounds calculated earlier to find the cumulative value over the interval. This methodical process ensures each part of the polynomial is correctly evaluated, leading to the precise solution of the integral.