`int_0^a f(a-x)` dx=

Options

A. `int_0^(2a) f(x) dx`
B. `int_(-a)^a f(x) dx`
C. `int_0^a f(x) dx`
D. `int_a^0 f(x) dx`.

Answer

To solve the integral \( \int_0^a f(a-x) \, dx \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_0^a f(a-x) \, dx \]

### Step 2: Make a Substitution
We will use the substitution \( b = a - x \). Then, differentiating both sides gives us:
\[ dx = -db \]

### Step 3: Change the Limits of Integration
Now, we need to change the limits of integration according to our substitution:
- When \( x = 0 \), \( b = a - 0 = a \)
- When \( x = a \), \( b = a - a = 0 \)

Thus, the integral becomes:
\[ I = \int_a^0 f(b) (-db) = \int_0^a f(b) \, db \]

### Step 4: Simplify the Integral
Since the limits have changed, we can drop the negative sign:
\[ I = \int_0^a f(b) \, db \]

### Step 5: Rename the Variable
We can rename the variable \( b \) back to \( x \) (since the variable of integration is just a dummy variable):
\[ I = \int_0^a f(x) \, dx \]

### Conclusion
Thus, we have shown that:
\[ \int_0^a f(a-x) \, dx = \int_0^a f(x) \, dx \]

### Final Answer
The value of the integral \( \int_0^a f(a-x) \, dx \) is:
\[ \int_0^a f(x) \, dx \]

---