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 2 years ago
Please explain this result:
\[\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx=\sqrt{1/2b}\]
\[0\le b \le1/2\]
Shouldn't it be: \(=\sqrt{1/2b}\)
 2 years ago
Please explain this result: \[\frac{1}{2}\int_{b1/2}^{0}\frac{1}{\sqrt{x}}dx=\sqrt{1/2b}\] \[0\le b \le1/2\] Shouldn't it be: \(=\sqrt{1/2b}\)

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joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1Because of the absolute value, and the fact that b is less than 1/2 (making the bottom limit of the integrand negative), it follows that:\[\int\limits_{b\frac{1}{2}}^0\frac{1}{\sqrt{x}}dx=\int\limits_{0}^{\frac{1}{2}b}\frac{1}{\sqrt{x}}dx\] Another way to think about it is that the you are taking the integral of is above the x axis, and area under the curve above the x axis is always positive, not negative.

Accipiter46
 2 years ago
Best ResponseYou've already chosen the best response.0Awesome, thanks! :)
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