Let f(x) be an odd continuous function which is periodic with period 2. if `g(x)=underset(0)overset(x)intf(t)dt`, then

Options

A. `g (x)` is an odd function
B. `g(n)=0 "for all" n in N`
C. `g(2n)=0 "for all" n in N`
D. `g(x)` is non-peridic

Answer

To solve the problem, we will analyze the function \( g(x) = \int_0^x f(t) \, dt \) where \( f(x) \) is an odd continuous function that is periodic with period 2.

### Step-by-Step Solution:

1. **Understanding the Function**:
Since \( f(x) \) is an odd function, we have:
\[ f(-t) = -f(t) \]
for all \( t \). This property will be useful when evaluating the integral.

2. **Evaluating \( g(x + 2) \)**:
We can express \( g(x + 2) \) as follows:
\[ g(x + 2) = \int_0^{x + 2} f(t) \, dt \]
We can split this integral into two parts:
\[ g(x + 2) = \int_0^2 f(t) \, dt + \int_2^{x + 2} f(t) \, dt \]

3. **Changing the Variable in the Second Integral**:
For the second integral, we can change the variable by letting \( u = t - 2 \). Then \( dt = du \) and when \( t = 2 \), \( u = 0 \), and when \( t = x + 2 \), \( u = x \):
\[ \int_2^{x + 2} f(t) \, dt = \int_0^x f(u + 2) \, du \]
Since \( f(t) \) is periodic with period 2, we have \( f(u + 2) = f(u) \):
\[ \int_2^{x + 2} f(t) \, dt = \int_0^x f(u) \, du = g(x) \]

4. **Combining the Results**:
Now we can substitute back into our expression for \( g(x + 2) \):
\[ g(x + 2) = \int_0^2 f(t) \, dt + g(x) \]

5. **Evaluating \( \int_0^2 f(t) \, dt \)**:
Since \( f(t) \) is an odd function, we can evaluate the integral over one period:
\[ \int_{-1}^{1} f(t) \, dt = 0 \]
Therefore, the integral from 0 to 2 can be expressed as:
\[ \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt \]
By changing the variable in the second integral (let \( u = t - 2 \)):
\[ \int_1^2 f(t) \, dt = \int_{-1}^0 f(u + 2) \, du = \int_{-1}^0 f(u) \, du = -\int_0^1 f(t) \, dt \]
Thus, we find:
\[ \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt - \int_0^1 f(t) \, dt = 0 \]

6. **Final Result**:
Substituting back, we have:
\[ g(x + 2) = 0 + g(x) = g(x) \]
This shows that \( g(x) \) is periodic with period 2.

### Conclusion:
Thus, the final result is that \( g(x) \) is periodic with period 2, and we can conclude that:
\[ g(x + 2) = g(x) \quad \text{for all } x \]