Problem 8 - Calculus and its applications

Question

Find the area represented by each definite integral. $$ \int_{-1}^{1}|3 x-2| d x $$

Step-by-Step Solution

Answer

The area under the curve is \(\frac{1}{3}\).

Step 1: Understanding the Problem

We need to find the area under the curve of the function \(|3x - 2|\) from \(x = -1\) to \(x = 1\). This will require evaluating the definite integral of \(|3x - 2|\) over the interval.

Step 2: Identify Critical Points

The function \(3x - 2 = 0\) gives us the point where the expression inside the absolute value changes its sign. Solving \(3x - 2 = 0\) yields \(x = \frac{2}{3}\). Thus, we have three regions to consider: from \(-1\) to \(\frac{2}{3}\) and from \(\frac{2}{3}\) to \(1\).

Step 3: Absolute Value Expression

For \(x < \frac{2}{3}\), \(3x - 2\) is negative, thus \(|3x - 2| = -(3x - 2) = -3x + 2\). For \(x > \frac{2}{3}\), \(3x - 2\) is positive, so \(|3x - 2| = 3x - 2\).

Step 4: Integrate Over Subintervals

First, integrate from \(-1\) to \(\frac{2}{3}\): \[ \int_{-1}^{\frac{2}{3}} (-3x + 2) \, dx \] Second, integrate from \(\frac{2}{3}\) to \(1\): \[ \int_{\frac{2}{3}}^{1} (3x - 2) \, dx \]

Step 5: Calculate the First Integral

Calculate \( \int_{-1}^{\frac{2}{3}} (-3x + 2) \, dx \). Find the antiderivative: \(-\frac{3}{2}x^2 + 2x\). Evaluate it from \(-1\) to \(\frac{2}{3}\): \[ \left(-\frac{3}{2}(\frac{2}{3})^2 + 2(\frac{2}{3})\right) - \left(-\frac{3}{2}(-1)^2 + 2(-1)\right) = -\frac{2}{3} + \frac{4}{3} + \frac{3}{2} - 2 \Rightarrow \frac{1}{6}.\]

Step 6: Calculate the Second Integral

Calculate \( \int_{\frac{2}{3}}^{1} (3x - 2) \, dx \). Antiderivative is \(\frac{3}{2}x^2 - 2x\). Evaluate it from \(\frac{2}{3}\) to \(1\): \[ \left(\frac{3}{2}(1)^2 - 2(1)\right) - \left(\frac{3}{2}(\frac{2}{3})^2 - 2(\frac{2}{3})\right). = -\frac{1}{2} + \frac{2}{3} - \frac{2}{3} + \frac{1}{2} \Rightarrow \frac{1}{6}.\]

Step 7: Add the Integrals' Results

Add the absolute areas calculated from each integral: \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}.\) This is the total area under the curve \(|3x - 2|\) from \(-1\) to \(1\).

Key Concepts

Absolute Value

The concept of absolute value is crucial when dealing with definite integrals like the one in this exercise. Absolute value, denoted as \(|x|\), represents the non-negative value of a number \(x\) without regard to its sign. In simpler terms, it measures the distance from zero on a number line, always resulting in a non-negative number.

When we encounter an absolute value function within an integral, such as \(|3x - 2|\), it implies that the expression inside it may change its sign over the interval. Thus, we must split the integral into parts where the expression's sign is consistent. In the example \(3x - 2\), we identify the critical point where the sign changes by setting the expression to zero and solving for \(x\). This gives us \(x = \frac{2}{3}\). Therefore, to compute the definite integral, we must evaluate two separate integrals corresponding to when \(3x - 2\) is negative and when it is positive.

This step ensures that the integral forms a true representation of the "area under the curve" regardless of the algebraic sign. It's a methodical way to handle integrals involving absolute value functions.

Area Under the Curve

Calculating the area under the curve of a function is a fundamental concept in calculus. Essentially, when we compute a definite integral over an interval, we are finding the net area between the curve and the x-axis. Positive areas are those where the function is above the x-axis, while negative areas are where the function is below.

In the exercise, given the function \(|3x - 2|\), the challenge is to find this area from \(x = -1\) to \(x = 1\). Due to the nature of absolute values, our function does not dip below the x-axis. Thus, all parts of the curve contribute positively to the total area.

By dividing the evaluation into manageable parts, based on sign and critical points, we ensure that all segments contribute correctly to the final integral value. This is reflected in adding together the results of both sub-interval integrals, resulting in the total area under the curve, which in this case, is \(\frac{1}{3}\).

This step-wise approach helps in breaking down complex problems and providing a clear picture of the area under a curve, especially when dealing with absolute values.

Antiderivative

An antiderivative, often called an indefinite integral, is a function that reverses differentiation. In simpler terms, it is a function whose derivative yields the original function. Finding an antiderivative is a crucial part of evaluating definite integrals.

In this exercise, after determining where the absolute value function \(|3x - 2|\) changes its sign, we calculate the antiderivatives for each part. For \(-3x + 2\), the antiderivative is \(-\frac{3}{2}x^2 + 2x\), and for \(3x - 2\), it is \(\frac{3}{2}x^2 - 2x\). These expressions are derived from basic rules of integration, where power and constant rules are applied.

Once obtained, these antiderivatives are evaluated over their respective bounds to find the definite integrals. Substituting the limits into the antiderivative gives the net area for each section. Finally, we sum these results to find the total area. This methodically calculated process illustrates the power of antiderivatives in solving definite integrals.