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Accipiter46

  • 2 years ago

Please explain this to me: \[\frac{1}{2}\int_{b-1/2}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=\] \[\frac{1}{2}\int_{b-1/2}^{0}\frac{1}{\sqrt{|x|}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=\] \[\sqrt{1/2-b}+\sqrt{1/2+b}\]

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  1. BluFoot
    • 2 years ago
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    This is the fundamental theorem of calculus. Check out this video, there's an example similar to yours at 6:00 http://www.youtube.com/watch?v=PGmVvIglZx8

  2. Accipiter46
    • 2 years ago
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    oh, also: \[0\le b \le1/2\]

  3. Accipiter46
    • 2 years ago
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    Thx, I'll take a look.

  4. satellite73
    • 2 years ago
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    also \[\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\] is always true

  5. Accipiter46
    • 2 years ago
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    Ok, I get the first part. What happens when the integrals are calculated?

  6. hartnn
    • 2 years ago
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    \(\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\) true if \(a \le c \le b\)

  7. satellite73
    • 2 years ago
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    actually it is always true, so long as the integral exists

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