anonymous
  • anonymous
Please explain this result: \[\frac{1}{2}\int_{b-1/2}^{0}\frac{1}{\sqrt{|x|}}dx=\sqrt{1/2-b}\] \[0\le b \le1/2\] Shouldn't it be: \(=-\sqrt{1/2-b}\)
Algebra
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Because 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.
anonymous
  • anonymous
Awesome, thanks! :-)

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