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

Algebra
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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
oh, also: \[0\le b \le1/2\]
Thx, I'll take a look.

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also \[\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\] is always true
Ok, I get the first part. What happens when the integrals are calculated?
\(\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\) true if \(a \le c \le b\)
actually it is always true, so long as the integral exists

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