Question

\[ \int\limits_0^1 x\left|x-\frac{1}{2}\right| \mathrm{d} x= \]

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{12}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{1}{16}\)

Hint:

Split the interval at x = 1/2 where the expression inside the modulus changes sign.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The modulus \(\left|x-\frac{1}{2}\right|\) equals \(\frac{1}{2}-x\) for \(x\lt\frac{1}{2}\) and \(x-\frac{1}{2}\) for \(x\gt\frac{1}{2}\). So the integral is split at \(x=\frac{1}{2}\).

Step 2: Set up
\(I=\int_0^{1/2}x\left(\tfrac{1}{2}-x\right)dx+\int_{1/2}^{1}x\left(x-\tfrac{1}{2}\right)dx\)

Step 3: First part
\(\left[\dfrac{x^2}{4}-\dfrac{x^3}{3}\right]_0^{1/2}=\dfrac{1}{16}-\dfrac{1}{24}=\dfrac{1}{48}\)

Step 4: Second part
\(\left[\dfrac{x^3}{3}-\dfrac{x^2}{4}\right]_{1/2}^{1}=\dfrac{1}{12}-\left(\dfrac{1}{24}-\dfrac{1}{16}\right)=\dfrac{1}{12}+\dfrac{1}{48}=\dfrac{5}{48}\)

Step 5: Calculation
\(I=\dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)

Final Answer:
The value of the integral is \(\frac{1}{8}\). \[ \boxed{\text{(C) }\dfrac{1}{8}} \]