# Integral (|x|)^0.5dx

1. Dec 12, 2006

### somebody-nobody

I need help solving integral

(|X|)^0.5dx

is it sam integral as (x)^0.5 except that i will need to integrate from 0 to x since it is absolute value of x

Thanks

2. Dec 12, 2006

I suggest you look at what the functions |x|^(1/2) and x^(1/2) look like.

Last edited: Dec 12, 2006
3. Dec 12, 2006

### benorin

integral of sqrt[abs[x]]

Answer: $$\int_0^x \sqrt{|X|}dX = \frac{2}{3}x\sqrt{|x|}$$

Proof: Consider that for $$x\geq 0,$$ we have

$$\int_0^x \sqrt{|X|}dX =\int_0^x \sqrt{X}dX = \frac{2}{3}x\sqrt{x},\mbox{ for }x\geq 0$$.

Also, if $$x\leq 0,$$, set $$t=-x$$ so that $$t\geq 0,$$ and we have

$$\int_0^x \sqrt{|X|}dX =\int_0^{-t} \sqrt{|X|}dX$$

now let $$u=-X$$ so that $$du=-dX$$ and $$0\leq X\leq -t$$ becomes $$0\leq u\leq t$$ and the integral becomes

$$\int_0^{-t} \sqrt{|X|}dX = -\int_0^{t} \sqrt{|-u|}du = -\int_0^{t} \sqrt{u}du = -\frac{2}{3}t\sqrt{t}= \frac{2}{3}x\sqrt{-x},\mbox{ for }x\leq 0$$

putting these togeather we have

$$\int_0^x \sqrt{|X|}dX =\left\{\begin{array}{cc}\frac{2}{3}x\sqrt{-x}, & \mbox{ if } x\leq 0\\ \frac{2}{3}x\sqrt{x},&\mbox{ if } x\geq 0\end{array}\right. =\frac{2}{3}x\sqrt{|x|}$$​

Last edited: Dec 12, 2006
4. Dec 12, 2006

### somebody-nobody

thank you

thanks to both of you