Question

Evaluate the integral \[ I = \int_1^5 \left( |x - 3| + |1 - x| \right) \, dx \]

• A. \( 4 \)
• B. \( 8 \)
• C. \( 12 \)
• D. \( 24 \)

Hint:

When dealing with absolute value functions, split the integral at points where the expression inside the absolute value changes sign.

Answer 1

The Correct Option is C

Step 1: Split the integral based on the absolute value We are asked to evaluate: \[ I = \int_1^5 \left( |x - 3| + |1 - x| \right) \, dx \] First, break the absolute values into pieces based on the points where the expressions inside the absolute values change sign: - \( |x - 3| \) changes sign at \( x = 3 \), - \( |1 - x| \) changes sign at \( x = 1 \). Thus, we split the integral into two parts: \[ I = \int_1^3 \left( (3 - x) + (1 - x) \right) \, dx + \int_3^5 \left( (x - 3) + (x - 1) \right) \, dx \] Step 2: Evaluate the integrals Now evaluate each part: For \( \int_1^3 \left( (3 - x) + (1 - x) \right) \, dx \): \[ I_1 = \int_1^3 (4 - 2x) \, dx = \left[ 4x - x^2 \right]_1^3 = (12 - 9) - (4 - 1) = 3 - 3 = 0 \] For \( \int_3^5 \left( (x - 3) + (x - 1) \right) \, dx \): \[ I_2 = \int_3^5 (2x - 4) \, dx = \left[ x^2 - 4x \right]_3^5 = (25 - 20) - (9 - 12) = 5 + 3 = 8 \] Thus, the value of the integral is: \[ I = 0 + 8 = 12 \]

Answer 2

Evaluate the integral:
\[ I = \int_1^5 \left( |x - 3| + |1 - x| \right) \, dx \]

Step 1: Break the integral at points where the absolute value expressions change their behavior:
Points: \( x = 1 \) and \( x = 3 \)

Step 2: For \( |x - 3| \):
\[ |x - 3| = \begin{cases} 3 - x, & x \leq 3 \\ x - 3, & x \geq 3 \end{cases} \]

Step 3: For \( |1 - x| \):
\[ |1 - x| = \begin{cases} 1 - x, & x \leq 1 \\ x - 1, & x \geq 1 \end{cases} \] Since \( x \geq 1 \) on the interval of integration, \( |1 - x| = x - 1 \) for all \( x \in [1,5] \).

Step 4: Split the integral:
\[ I = \int_1^3 \left( (3 - x) + (x - 1) \right) dx + \int_3^5 \left( (x - 3) + (x - 1) \right) dx \]

Step 5: Simplify integrands:
\[ \int_1^3 (3 - x + x - 1) dx = \int_1^3 (2) dx = 2 \times (3 - 1) = 4 \]
\[ \int_3^5 (x - 3 + x - 1) dx = \int_3^5 (2x - 4) dx \]

Step 6: Calculate second integral:
\[ \int_3^5 (2x - 4) dx = \left[ x^2 - 4x \right]_3^5 = (25 - 20) - (9 - 12) = 5 - (-3) = 8 \]

Step 7: Sum both integrals:
\[ I = 4 + 8 = 12 \]

Therefore, the value of the integral is:
\[ \boxed{12} \]