Question

Which of the following functions is/are Riemann integrable on [0, 1] ?

• A. \(f(x)=\int\limits^x_0|\frac{1}{2}-t|dt\)
• B. \(f(x)=\begin{cases}  x \sin(1/x) & \text{if }x \ne0 \\     0 & \text{if }x=0 \end{cases}\)
• C. \(f(x)=\begin{cases} 1 & \text{if }x \in Q ∩[0,1] \\     -1 & \text{otherwise} \end{cases}\)
• D. \(f(x)=\begin{cases} x & \text{if }x \in [0,1) \\     0 & \text{if } x=1\end{cases}\)

Answer 1

The Correct Option is A, B, D

To determine which functions are Riemann integrable on the interval \([0, 1]\), we will analyze each option based on the criteria for Riemann integrability. A function is Riemann integrable on a closed interval if it is bounded and its set of discontinuities has Lebesgue measure zero. 

1. \(f(x) = \int\limits^x_0 |\frac{1}{2}-t| \, dt\)
2. \(f(x) = \begin{cases} x \sin(1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}\)
3. \(f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \cap [0,1] \\ -1 & \text{otherwise} \end{cases}\)
4. \(f(x) = \begin{cases} x & \text{if } x \in [0,1) \\ 0 & \text{if } x=1 \end{cases}\)

Thus, the functions \( \int\limits^x_0 |\frac{1}{2} - t| \, dt \), the piecewise function \( x \sin(1/x) \), and the function \( f(x) = x \) for \( x \in [0,1) \) and \( f(1) = 0 \) are Riemann integrable on the interval \([0, 1]\).