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andyrk
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In the MVT for Integrals: ##f(c)(b-a)=\int_a^bf(x)dx##, why does ##f(x)## have to be continuous in ##[a,b]##.
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andyrk said:I looked it up on wikipedia and got the same answer. But it doesn't explain why IVM requires continuity of f
And why is that? Why do we need continuity?JonnyG said:But you need continuity to make sure that there is a number in the interval of integration where f takes on its average value.
andyrk said:And why is that? Why do we need continuity?
How is this true? Why is this true?JonnyG said:If f weren't continuous then it's possible that there is NO number c in [a,b] so that f(c) =its average value
andyrk said:How is this true? Why is this true?
I assume that there is a point c a<c<b satisfying the equation. If f(x) is discontinuous, say = -1 for half the interval and +1 for the other half, then the integral = 0, but there is no point where f(c)=0.andyrk said:In the MVT for Integrals: ##f(c)(b-a)=\int_a^bf(x)dx##, why does ##f(x)## have to be continuous in ##[a,b]##.
The Mean Value Theorem for Definite Integrals is a theorem in calculus that states that at some point within a given interval, the average value of a function over that interval is equal to the value of the function at that point.
The Mean Value Theorem for Definite Integrals deals with the average value of a function over an interval, while the Mean Value Theorem for Derivatives deals with the instantaneous rate of change of a function at a specific point.
No, the Mean Value Theorem for Definite Integrals can only be applied to continuous functions on a closed interval. This means that the function has no breaks or jumps in its graph and is defined for all values within the interval.
The Mean Value Theorem for Definite Integrals is significant because it allows us to find the average value of a function over a given interval, which can be useful in many real-world applications such as calculating average temperature or average velocity.
The Mean Value Theorem for Definite Integrals can be used to prove that a definite integral exists for a given function on a closed interval. It can also be used to simplify the calculation of definite integrals by finding the average value of the function and multiplying it by the length of the interval.