Question

Using integration, find the area of the region enclosed by the curve \( y = |x - 6| \), the x-axis, and between \( x = 4 \) and \( x = 8 \).

Hint:

• The graph of \( y = |x-6| \) is V-shaped. You can verify the area geometrically as two triangles of base 2 and height 2: \( 2 \times (\frac{1}{2} \times 2 \times 2) = 4 \).
• Always split integrals at points where modulus functions change sign.

Answer 1

Concept:

• Definition of absolute value: \( |x-6| = \begin{cases} 6-x, & x \\ x-6, & x \ge 6 \end{cases} \).
• Area under curve: \( A = \int_a^b |f(x)| dx \).

Step 1: Split the integral at the turning point
The function \( y = |x-6| \) changes definition at \( x = 6 \). The limits are from 4 to 8.
\( \text{Area} = \int_4^6 (6 - x) dx + \int_6^8 (x - 6) dx \)

Step 2: Evaluate the first integral
\( \int_4^6 (6 - x) dx = [ 6x - \frac{x^2}{2} ]_4^6 \)
\( = (36 - 18) - (24 - 8) = 18 - 16 = 2 \)

Step 3: Evaluate the second integral
\( \int_6^8 (x - 6) dx = [ \frac{x^2}{2} - 6x ]_6^8 \)
\( = (32 - 48) - (18 - 36) = -16 - (-18) = 2 \)

Step 4: Sum the areas
\( \text{Total Area} = 2 + 2 = 4 \text{ sq. units} \).