I have so far written the following proof in terms I understand for the Mean Value Theorem for Integrals:'If f is continuous on an interval I = [ a , b ], then for some c ∈ I we have: ∫ a b f ( x ) ⋅ d x = f ( c ) ( b − a )Let m and M denote the minimum and maximum values of f on I. Then: m ≤ f ( x ) ≤ M for all x ∈ [ a , b ]. Integrating the inequalities we should find that: ∫ a b m ≤ ∫ a b f ( x ) ≤ ∫ a b M m ( b − a ) ≤ ∫ a b f ( x ) ≤ M ( b − a ) m ≤ 1 b − a ∫ a b f ( x ) ≤ MHowever, the book I am using now says: 'label 1 b − a ∫ a b f ( x )as A ( f ). But now the intermediate value theorem says A ( f ) = f ( c ) for some c ∈ I. I am confused how the intermediate value theorem shows this and was wondering if anyone could help me understand or clarify it?

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I have so far written the following proof in terms I understand for the Mean Value Theorem for Integrals:&#039;If   f is continuous on an interval   I  =  [  a  ,  b  ], then for some   c  ∈  I we have:      ∫          a              b        f  (  x  )  ⋅  d  x  =  f  (  c  )  (  b  −  a  )Let   m and   M denote the minimum and maximum values of   f on   I. Then:   m  ≤  f  (  x  )  ≤  M for all   x  ∈  [  a  ,  b  ]. Integrating the inequalities we should find that:      ∫          a              b        m  ≤      ∫          a              b        f  (  x  )  ≤      ∫          a              b        M  m  (  b  −  a  )  ≤      ∫          a              b        f  (  x  )  ≤  M  (  b  −  a  )  m  ≤      1          b      −      a            ∫          a              b        f  (  x  )  ≤  MHowever, the book I am using now says: &#039;label      1          b      −      a            ∫          a              b        f  (  x  )as   A  (  f  ). But now the intermediate value theorem says   A  (  f  )  =  f  (  c  ) for some   c  ∈  I. I am confused how the intermediate value theorem shows this and was wondering if anyone could help me understand or clarify it?

Rowan Dyer · Accepted Answer

There exists a point   x  ∈  [  a  ,  b  ] satisfying   f  (  x  )  =  m. There exists a point   y  ∈  [  a  ,  b  ] satisfying   f  (  y  )  =  M. For every value   λ  ∈  (  m  ,  M  ) there exists a point   c in between   x and   y satisfying   f  (  c  )  =  λ.Take   λ  =            1              b        −        a                    ∫      a      b        f    (    x    )        d    x    .

nabakhi72 · Accepted Answer

Assuming   a  &amp;lt;  b  . Let   F  (  y  )  =      ∫    a    y    f  (  x  )  d  x  . Since   f is continuous, the Fundamental Theorem of Calculus implies that   F is differentiable and that       F    ′    =  f  . And (obviously)   F  (  a  )  =  0. So      1          b      −      a            ∫    a    b    f  (  x  )  d  x  =            F      (      b      )              b      −      a        =            F      (      b      )      −      F      (      a      )              b      −      a        .By the Intermediate Value Theorem for derivatives there exists   c  ∈  (  a  ,  b  ) such that            F      (      b      )      −      F      (      a      )              b      −      a        =      F    ′    (  c  )  .And       F    ′    =  f so       F    ′    (  c  )  =  f  (  c  )  .I think the book should have been clearer about this.Your approach using   m and   M is also perfectly valid.

I have so far written the following proof in terms I understand for the Mean Value Theorem for Integrals: 'If f is continuous on an interval I=[a,b], then for some c in I we have: int^_a(f(x)*dx=f(c)(b−a))

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