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Accipiter46
 2 years ago
Please explain this to me:
\[\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx=\]
\[\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx=\]
\[\sqrt{1/2b}+\sqrt{1/2+b}\]
Accipiter46
 2 years ago
Please explain this to me: \[\frac{1}{2}\int_{b1/2}^{b+1/2}\frac{1}{\sqrt{x}}dx=\] \[\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{x}}dx=\] \[\sqrt{1/2b}+\sqrt{1/2+b}\]

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BluFoot
 2 years ago
Best ResponseYou've already chosen the best response.0This 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

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0oh, also: \[0\le b \le1/2\]

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0Thx, I'll take a look.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0also \[\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\] is always true

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, I get the first part. What happens when the integrals are calculated?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0\(\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx\) true if \(a \le c \le b\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0actually it is always true, so long as the integral exists
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