Let `g(x)=int_(0)^(x)f(t)dt`, where `f` is such that `1/2lef(t)le1`, for `tepsilon[0,1]` and `0lef(t)le1/2`, for `tepsilon[1,2]`. Then prove that `1/2leg(2)le3/2`.

Explanation

To solve the problem, we need to prove that \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \), where \( g(x) = \int_0^x f(t) dt \) and the function \( f(t) \) is defined piecewise over the intervals \( [0, 1] \) and \( [1, 2] \).

### Step-by-Step Solution:

1. **Define the Function \( g(x) \)**:
\[ g(x) = \int_0^x f(t) dt \]
...