If f(x) is an integrable function over every interval on the real line such that f(t+x)=f(x) for every x and real t, then
` int_(a)^(a+t) f(x)dx` is equal to

Options

A. `overset(a)underset(0)int f(x)dx`
B. `overset(t)underset(0)int f(x)dx`
C. `overset(t)underset(a)int f(x)dx`
D. none of these

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{a}^{a+t} f(x) \, dx \) given the property \( f(t+x) = f(x) \) for every \( x \) and real \( t \).

### Step-by-Step Solution:

1. **Define the Integral**:
Let \( I = \int_{a}^{a+t} f(x) \, dx \).

2. **Split the Integral**:
We can split the integral into two parts:
\[ I = \int_{a}^{0} f(x) \, dx + \int_{0}^{t} f(x) \, dx + \int_{t}^{a+t} f(x) \, dx \]

3. **Change of Variable**:
For the integral \( \int_{t}^{a+t} f(x) \, dx \), we will perform a change of variable. Let \( x = y + t \), then \( dx = dy \).
- When \( x = t \), \( y = 0 \).
- When \( x = a+t \), \( y = a \).
Thus, we have:
\[ \int_{t}^{a+t} f(x) \, dx = \int_{0}^{a} f(y+t) \, dy \]

4. **Use the Given Property**:
From the property \( f(t+x) = f(x) \), we can replace \( f(y+t) \) with \( f(y) \):
\[ \int_{0}^{a} f(y+t) \, dy = \int_{0}^{a} f(y) \, dy \]

5. **Combine the Integrals**:
Now substituting back into our expression for \( I \):
\[ I = \int_{a}^{0} f(x) \, dx + \int_{0}^{t} f(x) \, dx + \int_{0}^{a} f(x) \, dx \]

6. **Reversing the Limits**:
The integral \( \int_{a}^{0} f(x) \, dx \) can be rewritten by reversing the limits:
\[ \int_{a}^{0} f(x) \, dx = -\int_{0}^{a} f(x) \, dx \]

7. **Final Expression**:
Substituting this back into our expression for \( I \):
\[ I = -\int_{0}^{a} f(x) \, dx + \int_{0}^{t} f(x) \, dx + \int_{0}^{a} f(x) \, dx \]
The \( -\int_{0}^{a} f(x) \, dx \) and \( +\int_{0}^{a} f(x) \, dx \) cancel each other out:
\[ I = \int_{0}^{t} f(x) \, dx \]

### Conclusion:
Thus, the value of the integral \( \int_{a}^{a+t} f(x) \, dx \) is:
\[ \int_{0}^{t} f(x) \, dx \]