Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.

Question

(Round your answer to four decimal places. Enter your answers as a comma-separated list.)

f(x) = x2

9

, [0, 3] (its supposed to be a fraction)

Answer 1

1/(3-0) • ∫₀³ (x²/9)dx = (1/3)(3³/27 - 0³/27) = (1/3)(27/27) = 1/3

f(c) = 1/3

x²/9 = 1/3

x² = 3

x = √3

Therefore, c=√3, which means that the average value of f(x) = x²/9 between 0 and 3 is √3.