## Accipiter46 Group Title 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: $$=-\sqrt{1/2-b}$$ one year ago one year ago

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1. joemath314159 Group Title

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.

2. Accipiter46 Group Title

Awesome, thanks! :-)