Question

Evaluate the integral: \[ \int_{3}^{5} |x-4|\,dx \]

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For integrals involving modulus: \[ \int |f(x)|dx \] first find the points where \(f(x)=0\) and split the interval there. Then remove the modulus by considering the sign of \(f(x)\) in each interval.

Answer 1

The Correct Option is A

Concept: The absolute value function is defined as \[ |x-a|= \begin{cases} a-x, & x x-a, & x\ge a \end{cases} \] Thus, the interval must be split at the point where the expression inside the modulus becomes zero.
Step 1: {Find where the expression inside modulus becomes zero.} \[ x-4=0 \] \[ x=4 \] Since \(4\) lies between \(3\) and \(5\), split the integral at \(x=4\). 
Step 2: {Rewrite the integral.} For \(3 \le x < 4\), \[ |x-4| = 4-x \] For \(4 \le x \le 5\), \[ |x-4| = x-4 \] Thus, \[ \int_{3}^{5} |x-4|dx = \int_{3}^{4} (4-x)dx + \int_{4}^{5} (x-4)dx \] 
Step 3: {Evaluate both integrals.} First integral: \[ \int_{3}^{4} (4-x)dx = \left[4x-\frac{x^2}{2}\right]_{3}^{4} \] \[ = \frac{1}{2} \] Second integral: \[ \int_{4}^{5} (x-4)dx = \left[\frac{x^2}{2}-4x\right]_{4}^{5} \] \[ = \frac{1}{2} \] 
Step 4: {Add the results.} \[ \frac{1}{2}+\frac{1}{2}=1 \] Thus, \[ \int_{3}^{5}|x-4|dx=1 \] Hence, the correct option is **(A)**.