Problem 48 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral by the most convenient method. Explain your approach. $$ \int_{0}^{3}|2 x-3| d x $$

Step-by-Step Solution

Answer

The result of the definite integral \( \int_{0}^{3}|2 x-3| d x \) is 1.5.

Step 1: Identify the function inside the modulus

Look inside the modulus |2x-3|, the function is 2x-3. We need to establish the point where this function changes sign. This can be done by setting the function to zero and solving for x. \n Solve \(2x-3=0\). This simplifies to \(x=3/2\), which is the inflection point

Step 2: Split the Integral

We split the integral into two from 0 to \(3/2\) and \(3/2\) to 3. Mathematically, \(\int_{0}^{3}|2 x-3| d x = \int_{0}^{1.5}-(2x-3) dx+\int_{1.5}^{3}(2x-3) dx\), because for x< \(3/2\), 2x-3 is negative and for x>\(3/2\), 2x-3 is positive.

Step 3: Evaluate the Integrals

Now evaluate both integrals separately. \nFor the first integral: \(\int_{0}^{1.5}-(2x-3) dx = -(x^2 -3x)|_{0}^{1.5} = -(1.5^2 -3*1.5 - (0^2 - 3*0)) = -0.75\). \nFor the second integral: \(\int_{1.5}^{3}(2x-3) dx = (x^2 - 3x)|_{1.5}^{3} = (3^2 - 3*3) - (1.5^2 - 1.5*3) = 0.75\)

Step 4: Sum the results

To conclude, sum the absolute values of the results of the two integrals: \(-0.75 + 0.75 = 1.5\)

Key Concepts

Modulus Function

Understanding the modulus function is crucial when manipulating expressions within definite integrals. In calculus, the modulus function, often represented as 'abs' or with vertical bars (| |), gives the absolute value of a number or expression, meaning it represents the distance from zero without considering direction. For instance, both |-5| and |5| equal 5, because both -5 and 5 are five units away from zero on the number line.

The integral evaluation process involving a modulus function usually requires additional steps. When the modulus function encloses an expression, like |2x - 3|, the expression inside can be positive or negative depending on the value of 'x'. Therefore, to integrate such a function, one must find the point(s) at which the expression changes sign. In the example of \(2x - 3\), the sign changes when x equals \(\frac{3}{2}\), also called the critical point. It's this critical point that often dictates how the integral should be split for evaluation.

Integral Evaluation

The process of integral evaluation involves finding the area under the curve of a function on a specified interval. In the context of a modulus function, as with the example \(\int_{0}^{3}|2x - 3|dx\), the area of interest can be above and below the x-axis since the modulus denotes absolute value.

Considering the critical point, the integral is assessed over intervals where the function inside the modulus maintains a consistent sign. This way, the absolute values of the resulting areas—whether above or below the x-axis—are added together to yield the total area under the original curve, without sign considerations. The step-by-step solution illustrates this as it evaluates the areas from 0 to \(\frac{3}{2}\) and from \(\frac{3}{2}\) to 3 separately. Integral evaluation usually ends with applying the limits of integration to the antiderivative of the function and subtracting to find the net result.

Splitting the Integral

When an integral contains a modulus function with an expression that can take both positive and negative values, splitting the integral becomes an effective strategy.

The technique involves dividing the integral at the points where the enclosed expression changes sign—the critical points. From here, separate integrals are set up reflecting the respective positive and negative intervals of the expression within the modulus. In our example, the integral is split at \(\frac{3}{2}\), resulting in two integrals: \(\int_{0}^{\frac{3}{2}}-(2x-3)dx\) and \(\int_{\frac{3}{2}}^{3}(2x-3)dx\).

This approach allows the accurate computation of areas even when a function crosses the x-axis. Within each divided range, the modulus function can be dispensed with, treating the expression as either strictly positive or strictly negative for the sake of integration. After evaluating each integral, the absolute areas they represent are added together to provide the total area associated with the original integral.