Find the values of C guaranteed by the mean value theorem for integrals.
f(x)=x^3
[0,3]
well the integral is
x^4/4 + C
Now in between x = 0 and x = 3 the slope of the integral must be
f(0) = 0
f(3) = 27
so the integral must be between
0 + 0(3) = 0
or
0 + 27*3 = 81
In fact the integral is 3^4/4 = 81/4 or about 20
To find the values of C guaranteed by the mean value theorem for integrals for the given function f(x) = x^3 on the interval [0,3], we can follow these steps:
1. Apply the mean value theorem for integrals, which states that if a function f(x) is continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that the definite integral of f(x) from a to b is equal to f(c) multiplied by the length of the interval (b - a).
2. Calculate the definite integral of f(x) from 0 to 3. To do this, we need to find the antiderivative of f(x) first. The antiderivative of x^3 is (1/4)x^4. Then, calculate the definite integral using the fundamental theorem of calculus: ∫[0,3] x^3 dx = [(1/4)x^4] evaluated from 0 to 3.
∫[0,3] x^3 dx = [(1/4)(3)^4] - [(1/4)(0)^4]
= (1/4)(81) - (1/4)(0)
= 81/4.
3. Calculate the length of the interval [0,3]. The length of an interval [a, b] is given by (b - a). In this case, the length of [0,3] is (3 - 0) = 3.
4. Now, set up the equation from the mean value theorem for integrals: f(c) * (b - a) = ∫[a,b] f(x) dx.
For our given interval [a, b] = [0, 3], we have:
f(c) * (3 - 0) = 81/4.
5. Solve the equation for f(c):
f(c) * 3 = 81/4.
6. Divide both sides of the equation by 3:
f(c) = 81/4 * (1/3).
7. Simplify the right side of the equation:
f(c) = 27/4.
So, the value of C guaranteed by the mean value theorem for integrals for the function f(x) = x^3 on the interval [0,3] is C = 27/4.