Problem 20 - Thomas Calculus

Question

In Exercises \(15-22,\) graph the integrands and use areas to evaluate the integrals. $$ \int_{-1}^{1}(1-|x|) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: Understand the Integrand

The integrand is given as \(1 - |x|\). This expression represents a piecewise linear function. Here, \(|x|\) indicates the absolute value of \(x\). This function will change its form at \(x=0\).

Step 2: Break the Integral into Piecewise Components

Since \(|x|\) changes at \(x=0\), divide the integral into two intervals: from \(-1\) to \(0\) and from \(0\) to \(1\). Thus, the integral becomes: \[ \int_{-1}^{1}(1-|x|) \, dx = \int_{-1}^{0}(1+x) \, dx + \int_{0}^{1}(1-x) \, dx.\]

Step 3: Evaluate the First Integral

For the interval \([-1, 0]\), the function is \(1 + x\). The integral is: \[ \int_{-1}^{0}(1+x) \, dx = \left[ x + \frac{x^2}{2} \right]_{-1}^{0}. \] Evaluating this gives:\[ \left(0 + \frac{0^2}{2}\right) - \left((-1) + \frac{(-1)^2}{2}\right) = 0 - (-1 + 0.5) = 0.5.\]

Step 4: Evaluate the Second Integral

For the interval \([0, 1]\), the function is \(1 - x\). The integral is: \[ \int_{0}^{1}(1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1}. \] Evaluating this gives:\[ \left(1 - \frac{1^2}{2}\right) - \left(0 - \frac{0^2}{2}\right) = 0.5. \]

Step 5: Sum the Areas

The total area is the sum of the areas found in the previous steps: \[ 0.5 + 0.5 = 1. \]

Step 6: Conclusion

The integral \(\int_{-1}^{1}(1-|x|) \, dx\) evaluates to 1, which is the area under the curve from \(x = -1\) to \(x = 1\).

Key Concepts

Absolute Value

The absolute value of a number, denoted by \(|x|\), is the distance of that number from zero on the number line. **Absolute value** is always non-negative, so for any real number \(x\):
\[%|x| = \begin{cases} x, & \text{if} \ x \geq 0 \-x, & \text{if} \ x < 0 \end{cases}\]
This piecewise definition allows us to break functions involving absolute value into simpler linear functions in specific intervals of \(x\). For example, for the function \(f(x) = 1 - |x|\), we have different expressions depending on whether \(x\) is negative or non-negative. Understanding absolute value is crucial for dividing broader functions into manageable parts.

Piecewise Function

A **piecewise function** is a function that has different expressions for different intervals of its independent variable. When working with integrals that involve absolute values, recognizing the piecewise nature of the function is essential to correctly evaluate them.
For \(f(x) = 1 - |x|\), the function behaves differently on either side of \(x=0\):

• When \(x < 0\), \(f(x) = 1 + x\)
• When \(x \geq 0\), \(f(x) = 1 - x\)

Being able to split such functions into simpler components allows us to handle each part easily when integrating, which simplifies the process for solving the integral over a specific range.

Area Under Curve

The **area under a curve** is a concept in calculus used to find the total area between a given curve and the x-axis. This helps us understand the "total value" that a function represents over an interval. To find the area under the curve for a function that is composed of different segments, like a piecewise function, we calculate the area separately for each segment and then sum up these areas.
In our particular exercise, we evaluated the area under the curve \( f(x) = 1 - |x| \) from \( x = -1 \) to \( x = 1 \). We found that splitting the integral into two segments made the calculation straightforward, yielding an area of 1 square unit in total. Recognizing how to break down functions helps in accurately determining the total area.

Definite Integral

A **definite integral** represents the net area under a curve between two specific points, providing the total accumulation of the function between these limits.
The expression for a definite integral is:
\[\int_{a}^{b} f(x) \, dx\]
This gives us the total value of \(f(x)\) from \(x = a\) to \(x = b\).
In our example, \(\int_{-1}^{1}(1 - |x|) \, dx\) was calculated by breaking down the piecewise function into two definite integrals—one from -1 to 0 and another from 0 to 1. By this approach, we could easily evaluate each segment based on a simple linear function. Summing up the results of both integrals provided the total net area under the curve, demonstrating a clear and logical method to solve definite integrals of piecewise functions.