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# C4 Substitution watch

1. This question's beginning to grate on my nerves, and the markscheme is absolutely useless.

The markscheme's actually just completely messed up all my workings out and I can't think straight

I get 50/x once you sub in root h, but I get confused when you take out dx and put in dh.

Or am I just completely wrong?

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2. Perhaps it is easier when you have found what dh/dx is equal to, multiply both sides of that equation by dx. Then you have dh=-2(20x-x) dx. You can then substitute that straight into your integrand to replace dh.

3. ... and you can take the rest of it from there
4. Normally for substitution you do integral of the original function (with subbed in u's) * dx/du (worked out from 1/du/dx) du
So the du's cancel out.

I think this one's a bit weird because they are substituting h (which is in their original function, and not something else like u). So in this case they have done the integral of the original function (with x's subbed in) * dh/dx dx, you dont have to do 1/dh/dx.

Sorry if this wasn't very helpful.
5. (Original post by HitTheSwitch)
Normally for substitution you do integral of the original function (with subbed in u's) * dx/du (worked out from 1/du/dx) du
So the du's cancel out.

I think this one's a bit weird because they are substituting h (which is in their original function, and not something else like u). So in this case they have done the integral of the original function (with x's subbed in) * dh/dx dx, you dont have to do 1/dh/dx.

Sorry if this wasn't very helpful.
I think maybe this is what's confusing me, I do dh/dx and i get -2(20-x), so isn't replacing dx for dh also bring in 1/-2(20-x)?

So there's a -2(20-x) on the bottom of the integral as well? Or not?
6. (Original post by ArchedEdge)
I think maybe this is what's confusing me, I do dh/dx and i get -2(20-x), so isn't replacing dx for dh also bring in 1/-2(20-x)?

So there's a -2(20-x) on the bottom of the integral as well? Or not?
The original function is with respect to h (dh), so to make it a function with respect to x (dx), you multiply the function by dh/dx and add dx at the end, so the dx's sort of "cancel out" but your function now has x's in it instead of h's.

I don't really understand what you are asking...

They are multiplying by the -2(20-x).
7. (Original post by HitTheSwitch)
The original function is with respect to h (dh), so to make it a function with respect to x (dx), you multiply the function by dh/dx and add dx at the end, so the dx's sort of "cancel out" but your function now has x's in it instead of h's.

I don't really understand what you are asking...

They are multiplying by the -2(20-x).
Ah wait, totally stupid mistake, I was reading the original equation with dx at the end and replacing it with dh etc

Thanks for the help.
8. Not problem

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