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Intergration by substitution

  • Thread starter thomas49th
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  • #1
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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:

Answers and Replies

  • #2
70
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Right lets rearrange u in terms of x (i think that's how you say it):

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

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

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.
 
Last edited:
  • #3
655
0
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 :)
 
  • #4
70
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Haha no worries, least you found out where you went wrong
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,794
924

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]
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

[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]
 

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