What is the average value of the function f(x) = 3x^2 - 2x on the interval [0, 2]?
Answer and explanation
Correct answer: B. 2
- Step 1: The average value of a continuous function on [a, b] is: f_avg = [1 / (b - a)] * integral from a to b of f(x) dx.
- Step 2: Here a = 0 and b = 2, so 1 / (b - a) = 1/2.
- Step 3: Compute the integral: integral from 0 to 2 of (3x^2 - 2x) dx = [x^3 - x^2] from 0 to 2 = (8 - 4) - 0 = 4.
- Step 4: Multiply by 1/2: f_avg = (1/2) * 4 = 2.
Answer
2.
Why not the others: '4' is the total definite integral before dividing by the interval length (b - a = 2); '3' makes an arithmetic error; '6' calculates (f(0) + f(2)) / 2 without integrating.
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