Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. (Enter your answers as a comma-separated list.)

Question

f(x) = x³, [0, 3]

Answer 1

(1/(3-0))•₀∫³ x³ dx = (1/3)(1/4)x⁴ evaluated between 0 and 3 = (1/12)(3)⁴ - (1/12)(0)⁴ = 81/12 = 27/4

To find the "c" value where f(c) = 27/4:

27/4 = x³

x = cuberoot(27/4) = 3/cuberoot(4) = 3•cuberoot(2)/cuberoot(8) = 3cuberoot(2)/2

So c = 3cuberoot(2)/2

Another way to say this is that 3cuberoot(2)/2 is the average value of f(x) = x³ between 0 and 3