To solve the problem, we start with the equation given:
\[ f(k - x) + f(x) = \sin x \]
We need to find the value of the integral:
\[ I = \int_0^k f(x) \, dx \]
### Step 1: Substitute \( x \) with \( k - x \)
We can substitute \( x \) with \( k - x \) in the original equation:
\[ f(k - (k - x)) + f(k - x) = \sin(k - x) \]
This simplifies to:
\[ f(x) + f(k - x) = \sin(k - x) \]
### Step 2: Write the two equations
Now we have two equations:
1. \( f(k - x) + f(x) = \sin x \) (Equation 1)
2. \( f(x) + f(k - x) = \sin(k - x) \) (Equation 2)
### Step 3: Combine the equations
From both equations, we can see that:
\[ \sin(k - x) = \sin x \]
This implies that:
\[ f(k - x) + f(x) = \sin x \]
\[ f(k - x) + f(x) = \sin(k - x) \]
### Step 4: Set up the integral
Now, we can express the integral \( I \):
\[ I = \int_0^k f(x) \, dx \]
### Step 5: Change of variable in the integral
Let's change the variable in the integral by substituting \( x \) with \( k - x \):
When \( x = 0 \), \( k - x = k \) and when \( x = k \), \( k - x = 0 \). Thus, we have:
\[ I = \int_k^0 f(k - x) (-dx) = \int_0^k f(k - x) \, dx \]
### Step 6: Use the original equation
Using the original equation \( f(k - x) = \sin x - f(x) \):
\[ I = \int_0^k (\sin x - f(x)) \, dx \]
### Step 7: Split the integral
Now we can split the integral:
\[ I = \int_0^k \sin x \, dx - \int_0^k f(x) \, dx \]
### Step 8: Substitute back for \( I \)
This gives us:
\[ I = \int_0^k \sin x \, dx - I \]
### Step 9: Solve for \( I \)
Now, we can solve for \( I \):
\[ 2I = \int_0^k \sin x \, dx \]
### Step 10: Calculate the integral
The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x \]
Thus,
\[ \int_0^k \sin x \, dx = [-\cos x]_0^k = -\cos k + \cos 0 = 1 - \cos k \]
### Step 11: Final expression for \( I \)
Now substituting back:
\[ 2I = 1 - \cos k \]
So,
\[ I = \frac{1 - \cos k}{2} \]
### Step 12: Final result
Using the identity \( 1 - \cos k = 2 \sin^2\left(\frac{k}{2}\right) \):
\[ I = \sin^2\left(\frac{k}{2}\right) \]
Thus, the value of the integral \( I \) is:
\[ I = \sin^2\left(\frac{k}{2}\right) \]
