Find the values of c guaranteed by the mean value theorem for integrals.

f(x)= x^3

[0,3]

Is 27/4 right?

You should know immediately that 27/4 is not correct, since it does not lie in the interval [0,3]!

You want c such that f'(c) = (f(3)-f(0))/(3-0)

3c^2 = 27/3
c = √3

To find the values of c guaranteed by the Mean Value Theorem for Integrals for the function f(x) = x^3 over the interval [0,3], we need to follow these steps:

1. Evaluate the definite integral of the function f(x) over the interval [0,3]:
∫[0,3] x^3 dx

To integrate x^3, add 1 to the exponent and divide by the new exponent:
∫[0,3] x^3 dx = (1/4)x^4 | [0,3]
= (1/4)(3)^4 - (1/4)(0)^4
= (1/4)(81 - 0)
= 81/4
= 20.25

2. Next, find the average value of the function f(x) over the interval [0,3] by dividing the value of the definite integral by the width of the interval:
Average value = (1/(b-a)) * ∫[a,b] f(x) dx

In this case, a = 0 and b = 3. So, the average value is:
Average value = (1/(3-0)) * (81/4) = (4/3) * (81/4) = 3

3. According to the Mean Value Theorem for Integrals, there exists a value c in the interval [0,3] such that f(c) is equal to the average value of the function f(x) over the interval [0,3].

In this case, the average value is 3. Therefore, there exists a value c in [0,3] such that f(c) = 3.

Let's solve for c:
f(c) = 3
c^3 = 3
c = ∛3

Therefore, the value of c that satisfies the Mean Value Theorem for Integrals for the function f(x) = x^3 over the interval [0,3] is c = ∛3 (the cube root of 3).