Find the Value(s) of C guaranteed by the Mean Value Theorem for Integrals for f(x) =x^3 in the interval [0,3]
To find the value(s) of C guaranteed by the Mean Value Theorem for Integrals for the function f(x) = x^3 in the interval [0,3], we need to follow these steps:
Step 1: Calculate the definite integral of f(x) over the interval [0,3].
The definite integral of f(x) over [a,b] can be represented as ∫[a,b] f(x) dx. In this case, we have ∫[0,3] x^3 dx.
To find the antiderivative of x^3, we use the power rule for integration.
∫ x^n dx = (1/(n+1)) * x^(n+1) + C,
where C is the constant of integration.
Applying the power rule, we integrate x^3 as follows:
∫ x^3 dx = (1/(3+1)) * x^(3+1) + C = (1/4) * x^4 + C.
Step 2: Evaluate the integral at the endpoints of the interval [0,3].
To find the value of the definite integral at the endpoints, substitute the values of a and b into the antiderivative of f(x):
(1/4) * x^4 + C evaluated from x = 0 to x = 3:
[(1/4) * 3^4 + C] - [(1/4) * 0^4 + C]
= (1/4) * 81 + C - (1/4) * 0 + C
= (1/4) * 81 + C - 0 + C
= (1/4) * 81 + 2C.
Step 3: Apply the Mean Value Theorem for Integrals.
According to the Mean Value Theorem for Integrals, if a function f(x) is continuous over the closed interval [a, b], and differentiable over the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that the integral of f(x) over [a, b] is equal to f(c) times the length of the interval (b-a).
In this case, the length of the interval [0,3] is 3 - 0 = 3.
Therefore, based on the Mean Value Theorem for Integrals, we have:
(1/4) * 81 + 2C = f(c) * 3.
Simplifying this equation gives us:
(1/4) * 81 + 2C = 3f(c).
Step 4: Solve the equation to find the value(s) of C.
To find the value(s) of C, we need to find the value(s) of c in the interval (0,3).
Since f(x) = x^3, f(c) = c^3.
Substituting these values into the equation, we have:
(1/4) * 81 + 2C = 3c^3.
Simplifying this equation gives us:
81/4 + 2C = 3c^3.
To find the value(s) of c, we can use numerical methods or approximation techniques. In this case, there is no simple algebraic solution.
To find the value(s) of C guaranteed by the Mean Value Theorem for Integrals for the function f(x) = x^3 in the interval [0,3], we need to follow these steps:
Step 1: Calculate the definite integral of f(x) over the interval [0,3]. The definite integral of f(x) is denoted as ∫[0,3] x^3 dx and represents the area under the curve f(x) between x=0 and x=3.
Step 2: Use the Mean Value Theorem for Integrals, which states that there exists at least one value c in the interval [0,3] such that the integral of f(x) over the interval [0,3] is equal to f(c) multiplied by the length of the interval [0,3]. Mathematically, this can be expressed as:
∫[0,3] x^3 dx = f(c) * (3 - 0)
Step 3: Simplify the equation from step 2 and solve for f(c). Since f(x) = x^3, we have:
∫[0,3] x^3 dx = f(c) * 3
∫[0,3] x^3 dx = c^3 * 3
Step 4: Calculate the definite integral of x^3 over the interval [0,3]. The integral of x^3 can be found by applying the power rule for integration:
∫[0,3] x^3 dx = [1/4 * x^4] evaluated from x=0 to x=3
∫[0,3] x^3 dx = [1/4 * 3^4] - [1/4 * 0^4]
∫[0,3] x^3 dx = 81/4 - 0
∫[0,3] x^3 dx = 81/4
Step 5: Equate the result from step 4 to the equation in step 3:
81/4 = 3c^3
Step 6: Solve the equation from step 5 for c. Divide both sides of the equation by 3:
c^3 = (81/4) / 3
c^3 = 81/12
c^3 = 27/4
Step 7: Take the cube root of both sides to solve for c:
c = ∛(27/4)
c = ∛27 / ∛4
c = 3 / ∛4
c = 3 / (4^(1/3))
c = 3 / 1.587401...
So, the value of C guaranteed by the Mean Value Theorem for Integrals for f(x) = x^3 in the interval [0,3] is approximately 1.891.
f(3) = 27
f(0) = 0
so, the slope of the secant line is 27/3 = 9
So, you need
3c^2 = 9
c = √3