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    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?

    Help please!
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    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.
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    h=\left(20-x\right)^2

    dh=-2\left(20-x\right)dx

    \displaystyle \int_{0}^{100}\dfrac{50}{20-\sqrt{h}}dh = -2\displaystyle \int_{20}^{10}\dfrac{50}{20-\sqrt{\left(20-x\right)^2}}\left(20-x\right)dx = 100\displaystyle \int_{20}^{10}\dfrac{x-20}{x}dx

    = 100\displaystyle \int_{20}^{10}\left(1-\dfrac{20}{x}\right)dx

    ... and you can take the rest of it from there
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    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.
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    (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?
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    (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).
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    (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.
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    Not problem
 
 
 
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