Understanding the Definite Integral of an Absolute Value Function

We need to evaluate the definite integral \(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx\). This involves integrating an absolute value function over a specific interval.

The absolute value function \(|x - 5|\) changes its definition depending on the value of \(x\). Specifically:

• If \(x - 5 \ge 0\), which means \(x \ge 5\), then \(|x - 5| = x - 5\).
• If \(x - 5 < 0\), which means \(x < 5\), then \(|x - 5| = -(x - 5) = 5 - x\).

The interval of integration is from \(x = 2\) to \(x = 8\). The point where the absolute value function changes its definition is \(x = 5\). Since \(x = 5\) is within the interval [2, 8], we must split the integral into two parts:

\(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx = \mathop \smallint \nolimits_2^5 \left| {x - 5} \right|dx + \mathop \smallint \nolimits_5^8 \left| {x - 5} \right|dx\)

Evaluating the Integrals by Splitting the Interval

For the interval [2, 5], we have \(x < 5\), so \(|x - 5| = 5 - x\). The first integral is:

\(\mathop \smallint \nolimits_2^5 (5 - x)dx\)

Let's find the antiderivative of \(5 - x\):

\(\mathop \smallint (5 - x)dx = 5x - \frac{x^2}{2} + C\)

Now, we evaluate the definite integral from 2 to 5:

\([5x - \frac{x^2}{2}]_2^5 = (5 \times 5 - \frac{5^2}{2}) - (5 \times 2 - \frac{2^2}{2})\)
\(= (25 - \frac{25}{2}) - (10 - \frac{4}{2})\)
\(= (\frac{50 - 25}{2}) - (10 - 2)\)
\(= \frac{25}{2} - 8\)
\(= \frac{25 - 16}{2} = \frac{9}{2}\)

For the interval [5, 8], we have \(x \ge 5\), so \(|x - 5| = x - 5\). The second integral is:

\(\mathop \smallint \nolimits_5^8 (x - 5)dx\)

Let's find the antiderivative of \(x - 5\):

\(\mathop \smallint (x - 5)dx = \frac{x^2}{2} - 5x + C\)

Now, we evaluate the definite integral from 5 to 8:

\([\frac{x^2}{2} - 5x]_5^8 = (\frac{8^2}{2} - 5 \times 8) - (\frac{5^2}{2} - 5 \times 5)\)
\(= (\frac{64}{2} - 40) - (\frac{25}{2} - 25)\)
\(= (32 - 40) - (\frac{25 - 50}{2})\)
\(= -8 - (-\frac{25}{2})\)
\(= -8 + \frac{25}{2}\)
\(= \frac{-16 + 25}{2} = \frac{9}{2}\)

Combining the Results

To find the total definite integral, we add the results from the two parts:

\(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9\)

Therefore, the value of the definite integral \(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx\) is 9.

Final Answer and Option Matching

The calculated value of the definite integral is 9. Let's compare this with the given options:

The calculated value 9 matches Option 4.

Revision Table: Key Concepts for Absolute Value Integrals

Additional Information: Geometric Interpretation of Definite Integral

The definite integral \(\mathop \smallint \nolimits_a^b f(x)dx\) represents the signed area between the curve \(y = f(x)\), the x-axis, and the vertical lines \(x = a\) and \(x = b\). For \(y = |x - 5|\), which is a V-shaped graph with its vertex at (5, 0), the integral \(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx\) represents the total area under the curve from \(x = 2\) to \(x = 8\).

The graph of \(y = |x - 5|\) from \(x=2\) to \(x=8\) consists of two right-angled triangles:

• Triangle 1: From \(x=2\) to \(x=5\). The vertices are (2, 3), (5, 0), and (2, 0). Base = 5 - 2 = 3. Height = \(|2 - 5| = 3\). Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}\).
• Triangle 2: From \(x=5\) to \(x=8\). The vertices are (5, 0), (8, 3), and (8, 0). Base = 8 - 5 = 3. Height = \(|8 - 5| = 3\). Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}\).

The total area (and thus the integral) is the sum of the areas of the two triangles:

Total Area = Area of Triangle 1 + Area of Triangle 2 = \(\frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9\).

This geometric interpretation confirms the result obtained through formal integration methods. Understanding the geometric meaning helps visualize why splitting the integral at the vertex of the absolute value function is necessary.