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Math Help - when is this true for integral

  1. #1
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    when is this true for integral



    what is the condition for this statement to be true?
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    Quote Originally Posted by szpengchao View Post


    what is the condition for this statement to be true?
    They may be more general hypothesis for the Fundemental Theorem of Calculus, but I know that this works.

    If g is differentiable on [a,b] and g' integrable on [a,b], then

    \int_{a}^{x}g'(t)dt=g(x)-g(a)

    for each x \in [a,b]
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    cheers

    ok, cheers.
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    prove

    how to prove sqrt(x) is differentiable?
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  5. #5
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    Quote Originally Posted by szpengchao View Post
    how to prove sqrt(x) is differentiable?
    Use the definition of the derivative.


    f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}

    Multilpy the numerator and denominator by the conjugate to get


    f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\left( \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri  ght)=\lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}

    Good luck
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    Quote Originally Posted by TheEmptySet View Post
    Use the definition of the derivative.


    f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}

    Multilpy the numerator and denominator by the conjugate to get


    f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\left( \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri  ght)=\lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}

    Good luck
    I think that this works as long as x \ne 0.

    Here is why: if It is differentiable at zero then this limit must exist

    f'(0)=\lim_{h \to 0^+}\frac{\sqrt{0+h}-\sqrt{0}}{h}=\lim_{h \to 0^+}\frac{\sqrt{h}}{h}=\lim_{h \to 0^+}\frac{1}{\sqrt{h}}=\infty

    So f(x)=\sqrt{x} is only differentable on (0,\infty)
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  7. #7
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    Edit: I misread your other thread.

    Okay so I just saw your other thread.

    I think you want to show that

    \int_{0}^{1}\sqrt{x}dx is Riemann integrabole on [0,1]

    Here is one method let \epsilon > 0

    f(x)=\sqrt{x} is continuous on [0,1] so it is uniformly continuous on [0,1]

    Since f is unifomly continous on [0,1] there exists a \delta such that |f(x)-f(y)|< \epsilon

    Now let P=\{x_0,x_1,...x_n\} be any partition of [0,1] such that |x_{j}-x_{j-1}|<\delta. Then by the extreme value theorem there are points x_M,x_m are the max and min on x_j,x_{j-1} for each j.

    Since the mesh of P is less than delta we have |x_M-x_m|< \delta so |f(x_M)-f(x_m) |< \epsilon
    So the upper and lower sums are

    U(f,P)=\sum_{j=1}^{n}f(x_M)(x_j-x_{j-1})
    L(f,P)=\sum_{j=1}^{n}f(x_m)(x_j-x_{j-1})

    U(f,P)-L(f,P)=\sum_{j=1}^{n}(f(x_M)-f(x_m)(x_j-x_{j-1})< \epsilon \sum_{j=1}^{n}(x_j-x_{j-1})=\epsilon(1-0)=\epsilon
    Last edited by TheEmptySet; May 28th 2008 at 02:48 PM. Reason: misread other post doesn't apply
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