integral of sqrt[abs[x]]
Answer: [tex]\int_0^x \sqrt{|X|}dX = \frac{2}{3}x\sqrt{|x|}[/tex]
Proof: Consider that for [tex]x\geq 0,[/tex] we have
[tex]\int_0^x \sqrt{|X|}dX =\int_0^x \sqrt{X}dX = \frac{2}{3}x\sqrt{x},\mbox{ for }x\geq 0[/tex].
Also, if [tex]x\leq 0,[/tex], set [tex]t=-x[/tex] so that [tex]t\geq 0,[/tex] and we have
[tex]\int_0^x \sqrt{|X|}dX =\int_0^{-t} \sqrt{|X|}dX[/tex]
now let [tex]u=-X[/tex] so that [tex]du=-dX[/tex] and [tex]0\leq X\leq -t[/tex] becomes [tex]0\leq u\leq t[/tex] and the integral becomes
[tex]\int_0^{-t} \sqrt{|X|}dX = -\int_0^{t} \sqrt{|-u|}du = -\int_0^{t} \sqrt{u}du = -\frac{2}{3}t\sqrt{t}= \frac{2}{3}x\sqrt{-x},\mbox{ for }x\leq 0[/tex]
putting these togeather we have
[tex]\int_0^x \sqrt{|X|}dX =\left\{\begin{array}{cc}\frac{2}{3}x\sqrt{-x}, & \mbox{ if } x\leq 0\\ \frac{2}{3}x\sqrt{x},&\mbox{ if }<br />
x\geq 0\end{array}\right. =\frac{2}{3}x\sqrt{|x|}[/tex]