Question

$\int_{a}^{b} (|x-a| + |x-b|) dx = , (0 < a < b)$

• A. $(b-a)^2$
• B. $(b-a)$
• C. $(b+a)$
• D. $(b+a)^2$

Hint:

Between $a$ and $b$, the sum of distances $|x-a| + |x-b|$ is always constant and equal to the distance between the endpoints, $(b-a)$.

Answer 1

The Correct Option is A

Step 1: Concept
For $x$ in the interval $[a, b]$, $|x-a| = x-a$ because $x \ge a$, and $|x-b| = b-x$ because $x \le b$.

Step 2: Meaning
The integrand simplifies to $(x-a) + (b-x)$.

Step 3: Analysis
$(x-a) + (b-x) = b-a$. Since $a$ and $b$ are constants, $b-a$ is a constant.

Step 4: Conclusion
$\int_{a}^{b} (b-a) dx = (b-a) [x]_{a}^{b} = (b-a)(b-a) = (b-a)^2$. Final Answer: (A)