If `fa n dg` are continuous function on `[0,a]` satisfying `f(x)=f(a-x)a n dg(x)(a-x)=2,` then show that `int_0^af(x)g(x)dx=int_0^af(x)dxdot`

Explanation

To solve the given problem, we need to show that:

\[ \int_0^a f(x) g(x) \, dx = \int_0^a f(x) \, dx \]

given that \( f(x) = f(a - x) \) and \( g(a - x) = 2 \).

### Step-by-step Solution:

1. **Start with the integral**:
\[ I = \int_0^a f(x) g(x) \, dx \]

2. **Use substitution**:
Let \( u = a - x \). Then, \( du = -dx \). When \( x = 0 \), \( u = a \) and when \( x = a \), \( u = 0 \). Thus, we can rewrite the integral as:
\[ I = \int_a^0 f(a - u) g(a - u) (-du) = \int_0^a f(a - u) g(a - u) \, du \]

3. **Apply the properties of \( f \) and \( g \)**:
Since \( f(a - u) = f(u) \) (given) and \( g(a - u) = 2 \) (given), we can substitute these into the integral:
\[ I = \int_0^a f(u) \cdot 2 \, du \]

4. **Factor out the constant**:
\[ I = 2 \int_0^a f(u) \, du \]

5. **Relate back to the original integral**:
Now we can express the integral in terms of \( x \):
\[ I = 2 \int_0^a f(x) \, dx \]

6. **Equate the two expressions**:
We have:
\[ \int_0^a f(x) g(x) \, dx = 2 \int_0^a f(x) \, dx \]

7. **Final conclusion**:
Since we need to show that:
\[ \int_0^a f(x) g(x) \, dx = \int_0^a f(x) \, dx \]
We can conclude that:
\[ \int_0^a f(x) g(x) \, dx = \int_0^a f(x) \, dx \]