Lời giải

\(I = \int_{-1}^{\frac{1}{2}} f(1 - 2x) \, dx + \int_{\frac{1}{2}}^1 f(2x - 1) \, dx\)

•  Đặt \(t = 1 - 2x \Rightarrow dt = -2 \, dx \)  

\(\begin{cases}
x = -1 \Rightarrow t = 3 \\
x = \frac{1}{2} \Rightarrow t = 0
\end{cases}\)

\(\int_{-1}^{\frac{1}{2}} f(1 - 2x) \, dx = \int_3^0 f(t)  \left(-\frac{1}{2}\right) \, dt = \frac{1}{2} \int_0^3 f(t) \, dt  = 3\)

•   Đặt \(t = 2x - 1 \Rightarrow dt = 2 \, dx\)  

\(\begin{cases}
x = \frac{1}{2} \Rightarrow t = 0 \\
x = 1 \Rightarrow t = 1
\end{cases}\)

\(\int_{\frac{1}{2}}^1 f(2x - 1) \, dx = \int_0^1 f(t) \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 f(t) \, dt = 1\)

\(I = 3 + 1 = 4\)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)