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In our discussions on derivatives, you learned about the Mean Value Theorem - an important theorem that claims that a function will take on its average value over an interval at least once. The Mean Value Theorem also has an application for integrals that is a consequence of the Mean Value Theorem and the Fundamental Theorem of Calculus.
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Sign up for freeWhat does the Mean Value Theorem for Integrals state about a function that is continuous on a closed interval [a, b]?
What is the geometric interpretation of the Mean Value Theorem for Integrals?
What theorem is the Mean Value Theorem for Integrals a consequence of?
In Example 1, what was the average value f(x) took on over the interval [1, 4]?
In Example 2, for the function f(x) = x + sin(2x), what was the average value over the interval [0, 2π]?
What does the Mean Value Theorem for Integrals state about a function that is continuous on a closed interval [a, b]?
What is the geometric interpretation of the Mean Value Theorem for Integrals?
What theorem is the Mean Value Theorem for Integrals a consequence of?
In Example 1, what was the average value f(x) took on over the interval [1, 4]?
In Example 2, for the function f(x) = x + sin(2x), what was the average value over the interval [0, 2π]?
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Team Mean Value Theorem for Integrals Teachers
The Mean Value Theorem for integrals states that if a function f is continuous on the closed interval [a, b], then there is a number c such that
Clearly, the left-hand side of the equation is the area under the curve of f on the interval (a, b). The right-hand side can be thought of as the area of a rectangle. So, the theorem states that the area under the curve is equal to the area of a rectangle with a width of the interval (b - a) and a height equal to the average value of the function f. Rearranging this equation to solve for f(c), the average value, we find
Let us visualize the Mean Value Theorem for integrals geometrically.
The area under the curve of a function f on the interval [a, b] is equal to a rectangle with a width of b - a and a height of the average value of f, f(c) - Vaia Original
Consider the definition of an antiderivative where
By the Fundamental Theorem of Calculus
and
Because F is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), we can apply the Mean Value Theorem, which says there is a number c such that and
Using the outcomes of the Fundamental Theorem of Calculus
For the function over the interval [1, 4], find the value c (the x-value where f(x) takes on its average value).
Since f(x) is a polynomial, we know it is continuous over the interval [1, 4].
So, the average value that f(x) takes on is 14.5.
In Step 2, we found that the area under the curve is . To find the area of the rectangle, we multiply the width by the height.
Thus, the Mean Value Theorem for integrals holds.
Since and we want to find c, we can set f(x) equal to 14.5.
To solve for x, we apply the quadratic formula.
Since is outside of the interval, .
For the function , find the x-value where f(x) takes on the average value over the interval
The function sin(x) is continuous everywhere.
Use your knowledge of the unit circle to solve the trigonometric equations! Remember, is just a multiple of .
So, the average value that f(x) takes on is .
In Step 2, we found that the area under the curve is units2. To find the area of the rectangle, we multiply the width by the height.
units2
Thus, the Mean Value Theorem for integrals holds.
Since and we want to find c, we can set f(x) equal to .
Solving this equation graphically, we find that .
As a reminder
Geometrically speaking, the area under the curve is equal to the area of a rectangle with a width of b - a and a height of the average value of f(x), f(c)
The Mean Value Theorem for integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus
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What is the mean value theorem for integrals?
The Mean Value Theorem for integrals states that if a function f is continuous on the closed interval [a, b], then the area under the curve is equal to the are of a rectangle with width b - a and height equal to the average value of the function f.
How do you use the Mean Value Theorem for integration?
To use the Mean Value Theorem, integrate the function f over the given interval (a, b). Multiply the area under the curve by 1/(b-a) to find the average value over the given interval.
How do you find the values of C that satisfy the Mean Value Theorem for Integrals?
To find the value c, apply the Mean Value Theorem for integrals to find the function value at c. Then, set f(x) equal to f(c) and solve for x.
What is an example for the mean value theorem for integral?
A simple example of the Mean Value Theorem for integrals is the function f(x)=x over the interval [0, 1] has an average value of 1/2 at x = 1/2. This means that the area under the of f(x) over the interval [0, 1] is equal to the area of a rectangle with a width of 1 and a height of 1/2.
What is the formula and equation for finding the mean value theorem for integrals?
The formula for the Mean Value Theorem for integrals says that the definite integral from a to b is equal to f(c)(b - a) for some c value.
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