Solving the Definite Integral of an Absolute Value Function

We need to evaluate the definite integral of the function \(|2x - 1|\) over the interval from \(x=2\) to \(x=3\).

The function we are integrating is \(f(x) = |2x - 1|\). The absolute value function \(|u|\) is defined as:

• \(|u| = u\) if \(u \ge 0\)
• \(|u| = -u\) if \(u < 0\)

In our case, \(u = 2x - 1\). We need to determine the sign of \(2x - 1\) within the interval of integration [2, 3].

Let's find where \(2x - 1\) changes sign by setting \(2x - 1 = 0\):

\(2x - 1 = 0\)

\(2x = 1\)

\(x = \frac{1}{2}\)

The critical point is \(x = \frac{1}{2}\). Our interval of integration is [2, 3]. This interval is entirely to the right of the critical point \(x = \frac{1}{2}\).

Let's pick a test value within the interval [2, 3], for example, \(x=2.5\):

\(2(2.5) - 1 = 5 - 1 = 4\)

Since \(4 > 0\), we can conclude that for all \(x\) in the interval [2, 3], the expression \(2x - 1\) is positive.

Therefore, within the interval [2, 3], the absolute value function simplifies to:

\(|2x - 1| = 2x - 1\) for \(x \in [2, 3]\)

Now, we can rewrite the definite integral without the absolute value:

\(\int_{2}^{3} |2x - 1| \,dx = \int_{2}^{3} (2x - 1) \,dx\)

Next, we evaluate this standard definite integral. First, find the antiderivative of \(2x - 1\). Using the power rule for integration (\(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\)) and the constant rule (\(\int c \,dx = cx + C\)):

\(\int (2x - 1) \,dx = \int 2x \,dx - \int 1 \,dx\)

\(= 2 \int x^1 \,dx - \int 1 \,dx\)

\(= 2 \left(\frac{x^{1+1}}{1+1}\right) - x + C\)

\(= 2 \left(\frac{x^2}{2}\right) - x + C\)

\(= x^2 - x + C\)

The antiderivative is \(F(x) = x^2 - x\). Now, apply the Fundamental Theorem of Calculus to evaluate the definite integral:

\(\int_{a}^{b} f(x) \,dx = F(b) - F(a)\)

Here, \(a=2\), \(b=3\), and \(F(x) = x^2 - x\).

\(\int_{2}^{3} (2x - 1) \,dx = F(3) - F(2)\)

Calculate \(F(3)\):

\(F(3) = (3)^2 - (3) = 9 - 3 = 6\)

Calculate \(F(2)\):

\(F(2) = (2)^2 - (2) = 4 - 2 = 2\)

Now, subtract \(F(2)\) from \(F(3)\):

\(F(3) - F(2) = 6 - 2 = 4\)

So, the value of the definite integral \(\int_{2}^{3} |2x - 1| \,dx\) is 4.

Revision Table: Definite Integral Steps

Additional Information on Absolute Value Integrals

When integrating absolute value functions, the key step is to understand how the absolute value affects the function within the integration interval. If the expression inside the absolute value changes sign within the interval, you must split the integral into multiple integrals at the points where the expression is zero.

For example, to evaluate \(\int_{0}^{2} |2x - 1| \,dx\):

• The expression \(2x - 1\) is zero at \(x = \frac{1}{2}\).
• The interval [0, 2] contains the critical point \(x = \frac{1}{2}\).
• For \(0 \le x < \frac{1}{2}\), \(2x - 1 < 0\), so \(|2x - 1| = -(2x - 1) = 1 - 2x\).
• For \(\frac{1}{2} \le x \le 2\), \(2x - 1 \ge 0\), so \(|2x - 1| = 2x - 1\).

So, the integral would be split:

\(\int_{0}^{2} |2x - 1| \,dx = \int_{0}^{1/2} (1 - 2x) \,dx + \int_{1/2}^{2} (2x - 1) \,dx\)

You would then evaluate each definite integral separately and sum the results. In the problem solved here, the interval [2, 3] did not contain the point where \(2x - 1\) is zero, simplifying the process to a single integral.

Understanding the graph of the function can also be helpful. The graph of \(y = |2x - 1|\) is a V-shape with its vertex at \((\frac{1}{2}, 0)\). The definite integral represents the area under this curve above the x-axis between the integration limits.