Question

Evaluate:
\[ I = \int_2^4 \left( |x - 2| + |x - 3| + |x - 4| \right) dx \]

Answer 1

We need to handle the absolute value expressions carefully. Let’s analyze where each expression inside the absolute value changes sign:

• \( |x - 2| \): changes at \( x = 2 \)
• \( |x - 3| \): changes at \( x = 3 \)
• \( |x - 4| \): changes at \( x = 4 \)

So, we split the interval \( [2, 4] \) at \( x = 3 \):

1. Interval \( [2, 3] \):

In this interval:

• \( |x - 2| = x - 2 \)
• \( |x - 3| = 3 - x \)
• \( |x - 4| = 4 - x \)

So the integrand becomes:

\[ (x - 2) + (3 - x) + (4 - x) = 5 - x \]

\[ \int_2^3 (5 - x) \, dx = \left[5x - \frac{x^2}{2} \right]_2^3 \] \[ = \left(15 - \frac{9}{2}\right) - \left(10 - \frac{4}{2}\right) = \left(\frac{30 - 9}{2}\right) - \left(\frac{20 - 4}{2}\right) = \frac{21}{2} - \frac{16}{2} = \frac{5}{2} \]

2. Interval \( [3, 4] \):

In this interval:

• \( |x - 2| = x - 2 \)
• \( |x - 3| = x - 3 \)
• \( |x - 4| = 4 - x \)

So the integrand becomes:

\[ (x - 2) + (x - 3) + (4 - x) = x - 1 \]

\[ \int_3^4 (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_3^4 = \left( \frac{16}{2} - 4 \right) - \left( \frac{9}{2} - 3 \right) = (8 - 4) - \left( \frac{9 - 6}{2} \right) = 4 - \frac{3}{2} = \frac{5}{2} \]

Final Answer:

\[ I = \frac{5}{2} + \frac{5}{2} = \boxed{5} \]