anti-derivative of {2y-1}{1+(1/(8y-4))} {}=square root?

Question

finding surface area where curve x={2y-1}

I've done a CD for 1/(8y-4) and then a u substitution for {2y-1} to cancel a term, but that still leaves me with an y term in my new integral.

I know writing the whole answer out on a computer is tedious, so any suggestions on how to go about solving this would really be appreciated.

Frank, can it simplify further than:

{2y-1}{(8y-3)/(8y-4)}

oh duh,
i'm being really dumb on this
didn't realize that {(8y-3)/(8y-4)} is the same as {8y-3}/{8y-4}

Answer 1

sqrt of
[(2y - 1)(1 + 1/(8y - 4))] =

sqrt of
(2y - 1)(8y - 4 + 1)/4(2y-1)] =
sqrt of
[(8y - 3)/4 =

sqrt of
(2y - 3/4) =

(2y - 3/4)^1/2

put 2y - 3/4 = u
2dy = du
dy = 1/2 du

integral of sqrt(2y - 3/4) = integral of 1/2[ u^1/2 du
1/2 u^(3/2)/3/2 + c
1/3 u^(3/2) + c
1/3(2y - 3/4)^(3/2) + c

Answer 2

Offhand, this looks like a Udv=UV-Vdu -type problem. There is no guarantee that your Vdu term that you wind up with may itself have to be integrated.

Answer 3

It relies upon despite in case you define "sq. root of four" as "the effective type whose sq. is 4" or "any type whose sq. is 4". i think of that interior the question "discover the sq. root of four", the 1st definition is implicit, or you need to have been asked "discover all sq. roots of four". If the instructions have been then the respond is two. The surd image, utilized to a non-detrimental real type x, continuously returns the non-detrimental real type y that satisfies y 2 = x.

Answer 4

The expression really simplifies.