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Mean Value Theorem for Definite Integrals

  1. Jul 2, 2015 #1
    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]##.
    Last edited: Jul 2, 2015
  2. jcsd
  3. Jul 3, 2015 #2
    From what I recall, the proof uses the intermediate value theorem, which requires continuity of f
  4. Jul 3, 2015 #3
    I looked it up on wikipedia and got the same answer. But it doesn't explain why IVM requires continuity of f
  5. Jul 3, 2015 #4
    I assume you meant IVT? (Or did you mean MVT?)

    The IVT requires continuity of f because other wise it would be possible for the image of f, to be "split" (actual term is separated or disconnected) into two separate, distinct parts and completely miss an intermediate point. It comes down to a concept called connectedness. Suppose you have a map f from [a,b] = X to R. Let's make things simple and suppose that f(a) < 0 and f(b) > 0 and you want to prove there's a c in (a,b) so that f(c) = 0. Now consider the set f(X). You wish to prove that 0 is in this set. However, if f is not continuous then it may very well be possible to "split" (disconnect) f(X) into two pieces. One piece where all its points are negative and another piece where all its points are positive. This would cause f to miss 0. But if f is continuous then it can be proven that since X cannot be "split" (disconnected), then f(X) also cannot be "split". Thus if it contains negative AND positive values, it must have the middle piece as well, 0.

    You can visualize this. Imagine the graph of f. Suppose that graph dips below the x axis then rises above the x axis and suppose f is continuous. It HAS to cross the x axis somewhere, right?

    I used some imprecise language to explain this because I don't think you have learned about connectedness as of yet, but it should give you the gist of things.

    If you really meant MVT, then apply my explanation of the IVT to the proof of the MVT.
  6. Jul 3, 2015 #5
    What you said is too complicated for a high school student's point of view. Doesn't it have a simpler explanation as to why ##f## has to be continuous?
  7. Jul 3, 2015 #6
    For the IVT, think of it like this: in high school, students are usually taught that a function f is continuous if its graph can be drawn without lifting up the pen. So suppose f(a) < f(b) for some two values, then intuitively, f must take ALL values between f(a) and f(b) because the pen cannot be lifted up, right?

    For the MVT, think of it like this. The integral of f from a to b measures the average value of the function then multiplies it by (b-a). How do we know this? Well, split the interval [a,b] into n equal parts. The length of each part is h=(b-a)/n. So int(f, a to b) is approximately equal to [(b-a)/n]*[f(a +h) + f(a + 2h) + f(a + 3h) + ... + f(a + nh)] =(b-a) * [f(a+h) + f(a+2h) + ...+ f(a+nh)]/n. Now the [f(a+h) + f(a+2h) + ...+ f(a+nh)]/n part is just the average of those n points right? So let h approach 0 (which is the same as letting n approach infinity) Then the integral gives you the average of the function f and multiplies it by the length of the interval (b-a). Now why is continuity required? Well you know the average of f must be between m and M, where m is f's min value and f's max value (It must take on a min and max by the EVT, which also requires continuity). So if the integral gives you an average value of f and multiplies it by (b-a), then this average value must be between m*(b-a) and M*(b-a), right? So to ensure that there is a number in the interval [a,b], say c, so that f(c) = average value of f, then we require continuity (to use the IVT).

    If you didn't want to read all that: Basically, the integral of f from a to b gives you f's average value and multiplies it by (b-a). But you need continuity to make sure that there is a number in the interval of integration where f takes on its average value.
    Last edited: Jul 3, 2015
  8. Jul 3, 2015 #7
    And why is that? Why do we need continuity?
  9. Jul 3, 2015 #8
    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. Then what? How's that useful for us? The MVT is useful because it guarantees the existence of such a number c in [a,b]. If f weren't continuous then all we could say is "this number c may or may not exist, we don't know", in which case a discussion about the relationship between the integral of f and its average value is pointless.
  10. Jul 3, 2015 #9
    How is this true? Why is this true?
  11. Jul 3, 2015 #10
    Well the average value of f must be between f(m) and f(M) (f(m) and f(M) being the min and max of f), right? Why should there be a c between m and M so that f(c) = average of f? We need the IVT to guarantee this is true, which requires continuity. Why does the IVT require continuity? Look at the graph and lifting up pen explanation I posted before.

    Take for example the function f from [0,1] to R defined by f(x) = 0 if x is in [0, 0.5] and f(x) = 2 if x is in (0.5,1]. f is not continuous (it's discontinuous at x = 0.5). Its average value is 1. But since f is not continuous, there is no number c in the interval [0,1] where f(c) = 1.
  12. Jul 3, 2015 #11
    That makes a lot of sense with the example. Without the example its pretty hard to understand why it has to be continuous for an average number to exist.
  13. Jul 3, 2015 #12
    You have likely only defined the integral for continuous functions to begin with.

    Anyway, take the function ##f:[-1,1]\rightarrow \mathbb{R}## defined by ##f(x) = 0## for ##x\leq 0## and ##f(x) = 1## for ##x>0##. What does the mean value theorem say in this case? Is it satisfied for this function?
  14. Jul 3, 2015 #13


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    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.
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