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# Taking constants out of integration... Watch

1. Here's something that really confused about taking constants out of integration...

There's the question... (I have used int for the integral sign as I can't work out how to do it)

Here's what I did:

because u = 3x-1

du/dx = 3

so you can replace the dx with du/3

And to put the x in terms of u: x = (u/3) - 1

so it looks like:

int (u/3 - 1) u^0.5 du/3

int (u^1.5)/3 - u ^0.5 du/3 by doing the multiplication

Obviously you can take the 1/3 that 'goes with' the du as its effectively multiplied by everything, but in the mark scheme they seem to have taken out the 1/3 that 'goes with' the u^1.5 as well. I don't understand how you can do that, as it only applies to one of the terms, not the term its added to as well, and when the mark scheme completes the question it multiples the u^0.5 term by it as well.

Here's the mark scheme:

Any help would be much appreciated!
2. (Original post by eddhann)
Here's something that really confused about taking constants out of integration...

There's the question... (I have used int for the integral sign as I can't work out how to do it)

Here's what I did:

because u = 3x-1

du/dx = 3

so you can replace the dx with du/3

And to put the x in terms of u: x = (u +1)/3
so it looks like:

int (u/3 - 1) u^0.5 du/3

int (u^1.5)/3 - u ^0.5 du/3 by doing the multiplication

Obviously you can take the 1/3 that 'goes with' the du as its effectively multiplied by everything, but in the mark scheme they seem to have taken out the 1/3 that 'goes with' the u^1.5 as well. I don't understand how you can do that, as it only applies to one of the terms, not the term its added to as well, and when the mark scheme completes the question it multiples the u^0.5 term by it as well.

Here's the mark scheme:

Any help would be much appreciated!
see bold
3. x=(u+1)/3 you fool!
4. thanks everyone

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