Question

evaluate the improper integral if it is convergent:
∫1/((2x-1)^(3/2)) dx from 0 to ∞

Answer 1

from 1 to ∞, sorry

Answer 2

\[\int_1^{\infty } \frac{1}{(2 x-1)^{3/2}} \, dx\]

Answer 3

quick question... how did u make the integral sign with the limits?

Answer 4

\int_1^{\infty } \frac{1}{(2 x-1)^{3/2}} \, dx

Answer 5

put between \[

Answer 6

\[\int_1^r \frac{1}{(2 x-1)^{3/2}} \, dx\]

Answer 7

i got u=2x-1 and du=2dx

Answer 8

using u sub , it evaluate to
\[1-\frac{1}{\sqrt{-1+2 r}}\]

Answer 9

as r goes to infinity, second term goes to 0. Leaving only 1

Answer 10

can you use b instead of r please?

Answer 11

of course

Answer 12

how did you get 1-1/(√-1+2b)?

Answer 13

u=2x-1

du= 2 dx

Answer 14

ok...

Answer 15

i got the lim b -> ∞ is -1/(√2b-1) + 1 which I think is what you said

Answer 16

hold on

Answer 17

\[\frac{1}{2} \int \frac{1}{u^{3/2}} \, du\]

\[-\frac{1}{\sqrt{u}}\]

Answer 18

yep got that

Answer 19

\[-\frac{1}{\sqrt{2x-1}}\]

Answer 20

does x = b?

Answer 21

i got what you got but -1 at the end

Answer 22

because i had to subtract when u=1 from when u=2b-1

Answer 23

\[-\frac{1}{\sqrt{2b-1}}-\left(-\frac{1}{\sqrt{2(1)-1}}\right)\]

Answer 24

\[1-\frac{1}{\sqrt{-1+2 b}}\]

Answer 25

\[\lim_{b\to \infty } \, \left(1-\frac{1}{\sqrt{2 b-1}}\right)\]

Answer 26

okay im following

Answer 27

as b approaches infinity , the second term goes to 0

Answer 28

the second term goes to 0 right? because the number you are dividing x by is getting greater and greater

Answer 29

yes

Answer 30

thank you!!!