Answer:
To prove the given statement, we will use a simple variable substitution. Let's start with the left side of the equation:
∫[a to b] f(x + c) dx
Now, let's make a substitution:
Let u = x + c
Then, du = dx (since c is a constant)
Now, we need to change the limits of integration:
When x = a, u = a + c
When x = b, u = b + c
So, the new limits of integration become a + c and b + c:
∫[a to b] f(x + c) dx = ∫[a + c to b + c] f(u) du
Now, this is in the same form as the right side of the equation:
∫[a + c to b + c] f(u) du = ∫[a - c to b - c] f(u) du
Since the variable of integration is a dummy variable (we can use any variable name), we can change u back to x:
∫[a + c to b + c] f(u) du = ∫[a - c to b - c] f(x) dx
So, we have shown that:
∫[a to b] f(x + c) dx = ∫[a - c to b - c] f(x) dx
This proves the given statement.
Explanation: