Question

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: \(=-\sqrt{1/2-b}\)

Answer 1

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.

Answer 2

Awesome, thanks! :-)