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Integral = 0, why?

  1. Jun 12, 2007 #1
    Why does
    [tex] \int^{1\sqrt3}_{1\sqrt3} ^5{\sqrt{x^2 + arccos{x}}}dx [/tex]

    = 0
  2. jcsd
  3. Jun 12, 2007 #2
    because by this you mean find the area enclosed between the curve and the x axis between squareroot3 and squareroot3. its the integral of it at squareroot3 minus the integral of it at squareroot3. so if we let the integral function be g(x) your expression says "g(squareroot3)-g(squareroot3)"
  4. Jun 12, 2007 #3
    And just what is [tex]\arccos {\sqrt{3}} [/tex]???
  5. Jun 12, 2007 #4


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    triggy, itsjustme's point is that
    [tex]\int_a^a f(x)dx= F(a)- F(a)= 0[/tex]
    for any integrable function f (F is, of course, an anti-derivative of f).

    Werg22's point is that, since [itex]\sqrt{3}[/itex] is larger than 1, the problem makes no sense as a real valued integral!
  6. Jun 12, 2007 #5

    George Jones

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    My guess is that Triggy made a typo, i.e., Triggy meant

  7. Jun 14, 2007 #6
    either that or arctan instead of arccos but i highly doubt it seeing it is 1squareroot3 not just squareroot3
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