Simplifying Integration by Substitution

[thomas49th]

Homework Statement

Using the substitution u² = 2x - 1, or otherwise, find the exact value of
\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx

The Attempt at a Solution

Right let's rearrange u in terms of x (i think that's how you say it):

x = \frac{u^{2} - 1}{2}And now get an expression for dx

u = (2x-1)^{\frac{1}{2}}

use the chain rule on it to give

\frac{dx}{du} (2x-1)^{-\frac{1}{2}}

= dx = \frac{du}{(2x-1)^{-\frac{1}{2}}}
Right now substitute that in

\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

now according to the mark scheme

\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

can be simplified

\int^{5}_{1} \frac{3u^{2} - 3}{2u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

shouldn't the bit here:

\frac{3u^{2} - 3}{2u}

be

\frac{3u^{2} - 3}{6u}

 

[lukas86]

Right let's rearrange u in terms of x (i think that's how you say it):

x = \frac{u^{2} - 1}{2}

Should be...

x = \frac{u^{2} + 1}{2}

Shouldn't it?

I managed to get the integral with 2u on the bottom and looking the same, but let me know how you're making out.

 

[thomas49th]

woops my bad i did 3 x EVERYTHING On the fraction (including the denominator). didn;t really have my brain in gear

sorry

cheers tho :)

 

[lukas86]

Haha no worries, least you found out where you went wrong

 

[HallsofIvy]

Homework Statement

Using the substitution u² = 2x - 1, or otherwise, find the exact value of
\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx

The Attempt at a Solution

Right let's rearrange u in terms of x (i think that's how you say it):

x = \frac{u^{2} - 1}{2}


And now get an expression for dx

u = (2x-1)^{\frac{1}{2}}

It's much simpler not to rearrange u in terms of x (or "solve for x"). Instead, from u²= 2x- 1, 2udu= 2dx so udu= dx.

use the chain rule on it to give

\frac{dx}{du} (2x-1)^{-\frac{1}{2}}

= dx = \frac{du}{(2x-1)^{-\frac{1}{2}}}



Right now substitute that in

\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

now according to the mark scheme

\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

can be simplified

\int^{5}_{1} \frac{3u^{2} - 3}{2u}\frac{du}{(2x-1)^{-\frac{1}{2}}}

shouldn't the bit here:

\frac{3u^{2} - 3}{2u}

be

\frac{3u^{2} - 3}{6u}