Question

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

Answer 1

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

Answer 2

oh, also: \[0\le b \le1/2\]

Answer 3

Thx, I'll take a look.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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