What is the average value of f on [0, π] when f(x) = sin x?

What is the average value of f on [0, π] when f(x) = sin x?

• A. 2/π
• B. π/2
• C. 0
• D. 1

Answer: (A)

The average value of a function on an interval is found by taking the integral over the interval and dividing by its length. For f(x) = sin x on [0, π], this is (1/π) ∫_0^π sin x dx. The integral ∫_0^π sin x dx equals [-cos x]_0^π = (-cos π) - (-cos 0) = 1 - (-1) = 2. Therefore, the average value is 2/π, which is about 0.637 and lies between 0 and 1 as sin x stays between 0 and 1 on that interval. The other numbers would not be possible given the function’s range.

The average value of a function on an interval is found by taking the integral over the interval and dividing by its length. For f(x) = sin x on [0, π], this is (1/π) ∫_0^π sin x dx. The integral ∫_0^π sin x dx equals [-cos x]_0^π = (-cos π) - (-cos 0) = 1 - (-1) = 2. Therefore, the average value is 2/π, which is about 0.637 and lies between 0 and 1 as sin x stays between 0 and 1 on that interval. The other numbers would not be possible given the function’s range.