`int_(0)^(a)f(x)dx=`

Options

A. `int_(0)^(a)f(a+x)dx`
B. `int_(0)^(a)f(2a+x)dx`
C. `int_(0)^(a)f(x-a)dx`
D. `int_(0)^(a)f(a-x)dx`

Answer

To solve the integral \( I = \int_{0}^{a} f(x) \, dx \) and prove the property, we will follow the steps outlined in the video transcript.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_{0}^{a} f(x) \, dx \]

2. **Substitute \( x \) with \( a - t \)**:
We will perform a substitution where \( x = a - t \). Then, the differential \( dx \) becomes:
\[ dx = -dt \]

3. **Change the Limits of Integration**:
When \( x = 0 \):
\[ t = a \]
When \( x = a \):
\[ t = 0 \]
Thus, the limits change from \( x = 0 \) to \( x = a \) into \( t = a \) to \( t = 0 \).

4. **Rewrite the Integral**:
Substitute \( x \) and \( dx \) in the integral:
\[ I = \int_{a}^{0} f(a - t)(-dt) = \int_{0}^{a} f(a - t) \, dt \]

5. **Rename the Variable**:
Since the variable of integration is a dummy variable, we can replace \( t \) with \( x \):
\[ I = \int_{0}^{a} f(a - x) \, dx \]

6. **Conclusion**:
We have shown that:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
This indicates a property of definite integrals.

### Final Result:
The integral \( \int_{0}^{a} f(x) \, dx \) is equal to \( \int_{0}^{a} f(a - x) \, dx \).