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Integral Involving Irrational values

  1. Jan 26, 2005 #1
    Hello all

    If we want to calculate the definite integral [tex] \int^b_a x^{\alpha} [/tex] for any irrational value of [tex] \alpha [/tex] where [tex] 0<a<b [/tex] do we use the Mean Value Theorem? Would [tex] \alpha [/tex] be represented as a limit of a sequence of rational numbers [tex] \alpha = \lim_{x\rightarrow \infty} \alpha_n [/tex] and [tex] \alpha [/tex] is not equal to -1. Hence [tex] x^{\alpha} = \lim_{x\rightarrow \infty} x^{\alpha_n} [/tex] So we can always find a number such that [tex] |x^{\alpha} - x^{\alpha_n}| < \epsilon [/tex] (how do we prove this)?.

    Now [tex] f(x) = x^\alpha [/tex] and [tex] g(x) = x^{\alpha_n} [/tex]. Now applying the Mean Value Theorem for Integral Calculus we get:

    [tex] -\epsilon(b-a) + \int^b_a x^{\alpha_n} \ dx < \int^b_a x^{\alpha} \ dx < \int^b_a x^{\alpha_n} + \epsilon(b-a) [/tex]

    We know that [tex] \int^b_a x^{\alpha} \ dx = \frac {1}{\alpha +1}(b^{\alpha +1} - a^{\alpha+1}) [/tex].

    [tex] -\epsilon(b-a) + \frac{1}{\alpha_n +1}(b^{\alpha_n+1} - a^{\alpha_n+1}) < \int^b_a x^\alpha \ dx < \frac{1}{\alpha_n +1}(b^{\alpha_n+1} - a^{\alpha_n+1}) + \epsilon(b-a) [/tex]. From here how do we receive

    [tex] \int^b_a \ dx = \frac{1}{\alpha +1}(b^{\alpha+1} - a^{\alpha +1})[/tex]?

    Last edited: Jan 26, 2005
  2. jcsd
  3. Jan 26, 2005 #2
    [tex] x^\alpha = e^{\alpha lnx}[/tex]
  4. Jan 26, 2005 #3


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    We could use the FTAC, because we know the derivative of x&alpha;+1.
    Last edited: Jan 26, 2005
  5. Jan 26, 2005 #4


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    If y= xα then ln(y)= α ln(x) so [itex]\frac{1}{y}\frac{dy}{dx}= \alpha \frac{1}{x}[/itex]. That is, [itex]\frac{dy}{dx}= \alpha\frac{1}{x}y= \alpha\frac{1}{x}x^{\alpha}= \alpha x^{\alpha-1}[/itex]

    From that, it follows that [itex]\frac{1}{\alpha+1}x^{\alpha+1}[/itex] is an anti-derivative of [itex]x^{\alpha;}[/itex].

    [itex] \int^b_a x^{\alpha}= \frac{1}{\alpha+1}x^{\alpha+1}+ C[/itex].
    Last edited: Jan 26, 2005
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