Verification neededproblem seems odd

[gochi]

Hi,

I have to find the integral of 1/square root(x-3)dx

this is what i did

let u=sqr root(x-3)
du=1/2*square root(x-3)dx
dx=2*square root(x-3)du

the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

..ok now what i did is plug back u= square root of(x-3) in the denom
that way I am only left with 2du.

Then i integrated that to 2x2/2 which is equal to x^2.

Ultimatley I got the integral to equal X ! which can't be right!

But doesn't this make sense, I know i did not do anything wrong...

 

[neutrino]

\frac{d}{dx}\left(\sqrt{x-3}\right) \neq \frac{1}{2}\left(\sqrt{x-3}\right)

 

[SiddharthM]

..ok now what i did is plug back u= square root of(x-3) in the denom
that way I am only left with 2du.

that move is NOT legit.

the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

neither is that one, wen u change variables, ur new equation must have only that variable - or constants of course...if you write root(x-3) IN TERMS of u that is fine, but the way it is written doesn't make sense.

Hint: let u=x-3.

 

[gochi]

\frac{d}{dx}\left(\sqrt{x-3}\right) \neq \frac{1}{2}\left(\sqrt{x-3}\right)

I meant that the derivitive of sqroot(x-3)=1/2sqroot(x-3)

 

[neutrino]

I meant that the derivitive of sqroot(x-3)=1/2sqroot(x-3)

And I'm saying that it is not.

\frac{d}{dx}\left({x-3}\right)^\frac{1}{2} = \frac{1}{2}\left(\frac{1}{\sqrt{x-3}}\right)

 

[SiddharthM]

"1/2sqroot(x-3)", i think this is 1 OVER 2 times the square root of x minus 3.

 

[atqamar]

Your first part is flawless. Integrate by parts:
\int xdu = xu-\int udx
set x=2 and u=\sqrt{x-3}. Therefore du=\frac{1}{2\sqrt{x-3}}. Obviously dx=0.
Plug in your values:
\int 2(\frac{1}{2\sqrt{x-3}}) = 2(\sqrt{x-3})-\int 0

Simplify and you get:
\int \frac{1}{\sqrt{x-3}} = 2\sqrt{x-3}

As you do more integrations, this process will become easier.

 

[SiddharthM]

u-substituting x-3 is a lot easier!

 

[HallsofIvy]

"1/2sqroot(x-3)", i think this is 1 OVER 2 times the square root of x minus 3.

The problem is that most people would interpret it as (1/2) sqroot(x-3), not 1/(2 sqroot(x-3)). USE PARENTHESES!

 

[SiddharthM]

that he didn't make that mistake is clear. Look at his next step:

dx=2*square root(x-3)du

 

[neutrino]

Ah... I apologise, gochi. But as Halls emphasised, please make use of parentheses. :)