A function f has average value on [0,2] equal to (1/2) ∫_0^2 f(x) dx. What is the average value of f(x) = x^2 on [0,2]?

A function f has average value on [0,2] equal to (1/2) ∫_0^2 f(x) dx. What is the average value of f(x) = x^2 on [0,2]?

• A. 4/3
• B. 2/3
• C. 1
• D. 8/3

Answer: (A)

The quantity being tested is how to find the average value of a function over an interval. The average value on [a,b] is (1/(b−a)) times the integral from a to b of f(x) dx. Here the interval is [0,2], so the average value of f(x) = x^2 is (1/2) ∫_0^2 x^2 dx. Evaluate the integral: ∫_0^2 x^2 dx = [x^3/3]_0^2 = 8/3. Multiply by 1/2 to get (1/2) * (8/3) = 4/3. So the average value is 4/3.

The quantity being tested is how to find the average value of a function over an interval. The average value on [a,b] is (1/(b−a)) times the integral from a to b of f(x) dx. Here the interval is [0,2], so the average value of f(x) = x^2 is (1/2) ∫_0^2 x^2 dx. Evaluate the integral: ∫_0^2 x^2 dx = [x^3/3]_0^2 = 8/3. Multiply by 1/2 to get (1/2) * (8/3) = 4/3. So the average value is 4/3.