Please explain this to me:
$$\frac{1}{2}\int_{b-1/2}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$$
$$\frac{1}{2}\int_{b-1/2}^{0}\frac{1}{\sqrt{|x|}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$$
$$\sqrt{1/2-b}+\sqrt{1/2+b}$$

Question

Please explain this to me:
$$\frac{1}{2}\int_{b-1/2}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$$
$$\frac{1}{2}\int_{b-1/2}^{0}\frac{1}{\sqrt{|x|}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$$
$$\sqrt{1/2-b}+\sqrt{1/2+b}$$

anonymous · Answer

This is the fundamental theorem of calculus. Check out this video, there's an example similar to yours at 6:00 http://www.youtube.com/watch?v=PGmVvIglZx8

anonymous · Answer

oh, also: $$0\le b \le1/2$$

anonymous · Answer

Thx, I'll take a look.

anonymous · Answer

also 
$$\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx$$ is always true

anonymous · Answer

Ok, I get the first part. What happens when the integrals are calculated?

hartnn · Answer

$\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx$
true if 
$a \le c \le b$

anonymous · Answer

actually it is always true, so long as the integral exists