- #1

- 655

- 0

## Homework Statement

Using the substitution u² = 2x - 1, or otherwise, find the exact value of

[tex]\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx[/tex]

## The Attempt at a Solution

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

[tex] x = \frac{u^{2} - 1}{2}[/tex]

And now get an expression for dx

[tex]u = (2x-1)^{\frac{1}{2}}[/tex]

use the chain rule on it to give

[tex]\frac{dx}{du} (2x-1)^{-\frac{1}{2}}[/tex]

= [tex]dx = \frac{du}{(2x-1)^{-\frac{1}{2}}}[/tex]

Right now substitute that in

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

now according to the mark scheme

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

can be simplified

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

shouldn't the bit here:

[tex]\frac{3u^{2} - 3}{2u}[/tex]

be

[tex]\frac{3u^{2} - 3}{6u}[/tex]

Last edited: