## anonymous 3 years ago Please explain this to me: $\frac{1}{2}\int_{b-1/2}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$ $\frac{1}{2}\int_{b-1/2}^{0}\frac{1}{\sqrt{|x|}}dx+\frac{1}{2}\int_{0}^{b+1/2}\frac{1}{\sqrt{|x|}}dx=$ $\sqrt{1/2-b}+\sqrt{1/2+b}$

1. anonymous

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

2. anonymous

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

3. anonymous

Thx, I'll take a look.

4. anonymous

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

5. anonymous

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

6. hartnn

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

7. anonymous

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