Problem 19 - University Calculus: Early Transcendentals

Question

Graph the integrands and use known area formulas to evaluate the integrals. $$\int_{-2}^{1}|x| d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{5}{2}\).

Step 1: Identify the Function

The integrand is the absolute value function \(|x|\), which changes its expression depending on whether \(x\) is positive or negative. It can be expressed piecewise as \[ |x| = \begin{cases} -x, & \text{if } x < 0 \ x, & \text{if } x \geq 0. \end{cases} \]

Step 2: Determine the Integration Range

We are integrating from \(x = -2\) to \(x = 1\). Note that there is a change in the function's definition at \(x = 0\). Therefore, we need to split the integral into two parts: from \(-2\) to \(0\) and from \(0\) to \(1\).

Step 3: Split the Integral

Write the integral as the sum of two integrals based on the piecewise definition:\[\int_{-2}^{1}|x| dx = \int_{-2}^{0} -x \, dx + \int_{0}^{1} x \, dx.\]

Step 4: Evaluate the First Integral

Calculate the integral \(\int_{-2}^{0} -x \, dx\):\[\int_{-2}^{0} -x \, dx = \left[ -\frac{x^2}{2} \right]_{-2}^{0}= \left( -\frac{0^2}{2} \right) - \left( -\frac{(-2)^2}{2} \right) = 0 + 2 = 2.\]

Step 5: Evaluate the Second Integral

Calculate the integral \(\int_{0}^{1} x \, dx\):\[\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1}= \left( \frac{1^2}{2} \right) - \left( \frac{0^2}{2} \right) = \frac{1}{2}.\]

Step 6: Sum the Results

Combine the results of the two integrals to find the total area:\[\int_{-2}^{1}|x| dx = 2 + \frac{1}{2} = \frac{5}{2}.\]

Step 7: Graph the Function

Plot \(|x|\) over the interval \([-2, 1]\) to visualize the areas under the curve, confirming the positive area values found from \(-2\) to 0 and 0 to 1.

Key Concepts

Absolute Value Function

The absolute value function, denoted by \(|x|\), is a special type of function that always returns a non-negative value. It evaluates to different expressions depending on whether the input value x is negative or non-negative. This is expressed mathematically as a piecewise function:

• \(|x| = -x\) when \(x < 0\)
• \(|x| = x\) when \(x \geq 0\)

This characteristic makes the absolute value function intriguing as it has a unique, V-shaped graph. Its vertex lies at the origin (0,0), and its two legs extend symmetrically towards positive and negative infinities. Understanding this function is important in calculus, especially when dealing with integrals involving intervals on either side of zero. In such cases, it often requires breaking the integral into sections that align with the piecewise definition of the absolute value function.

Definite Integral

A definite integral is a core concept in calculus used to calculate the accumulation of quantities, like areas under a curve, over a specified interval. It is denoted as \(\int_{a}^{b} f(x) \, dx\), signifying the area from x = a to x = b under the curve of \(f(x)\). This value represents the net signed area, which takes into account whether the area lies above or below the x-axis (positive or negative accumulations).When dealing with piecewise functions like the absolute value function, as in the given exercise, it's vital to split the integral at points where the function's behavior changes. Here, we break the integral into two parts at x = 0 to properly account for the function's different expressions on either side of this point.e.g., computing \(|x|\) from -2 to 1 involves:

• \(\int_{-2}^{0} -x \, dx\)
• \(\int_{0}^{1} x \, dx\)

Each part is integrated separately, and the results are combined to find the total area under the curve.

Area Under a Curve

The concept of the area under a curve is pivotal in understanding definite integrals. When we refer to the area under a curve, we're talking about the region bounded by the curve of the function, the x-axis, and two vertical lines at two points. This region represents the accumulation of space between these boundaries and is calculated using integration.In the exercise analyzed, finding the area under the curve of \(|x|\) from x = -2 to x = 1 gives a practical example of how this works. By splitting the integration into two parts, we calculate the areas separately:

• From x = -2 to x = 0, the area is positive 2, due to the negative slope of \(-x\), but the absolute value ensures a non-negative area.
• From x = 0 to x = 1, the function \(x\) simply slopes upwards, yielding an area of 0.5.

Combining these, we find that the total area is \(\frac{5}{2}\). This illustrates both the precision and utility of using definite integrals to calculate areas, accommodating any complexities introduced by a change in the function's nature across its domain.