_viotl
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what will be the integration of 1/(50-m)^2
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Pangol
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(Original post by _viotl)
what will be the integration of 1/(50-m)^2
Write it as (50 - m)^(-2) and use the reverse chain rule.

You'll get better help with a better subject line. There seem to be so many posts lately just called "maths" or suchlike. It's not difficult to be a bit more specific!
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_viotl
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(Original post by Pangol)
Write it as (50 - m)^(-2) and use the reverse chain rule.

You'll get better help with a better subject line. There seem to be so many posts lately just called "maths" or suchlike. It's not difficult to be a bit more specific!
can you solve me this on step by step
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Pangol
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(Original post by _viotl)
can you solve me this on step by step
Not without giving everything away, and the aim of this forum is to help people solve their own problems. Do you know what the reverse chain rule is?
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_viotl
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(Original post by Pangol)
Not without giving everything away, and the aim of this forum is to help people solve their own problems. Do you know what the reverse chain rule is?
i dont know it completely i tried checking some videos and gives no sense at all
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Pangol
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Before you can use the reverse chain rule, you need to be really on top of differentiating using the chain rule. The idea is that you will notice integrals that might have come about from differentiating using the chain rule.

For example:

If we differentiate (x + 4)^7 we get 7(x + 4)^6. So if we integrate 7(x + 4)^6 we get (x+4)^7. And by extension, if we integrate (x+4)^6, we get (1/7)(x+4)^7.

Your problem is very similar. What do you think you would have to differentiate to end up with (50 - m)^(-2) ? Your first guess only needs to be approximately right, you usually have to do a bit of fine-tuning to get the constant at the start right.

But for people having problems with the reverse chain rule, the best thing to do is chain rule differentiation until you start to spot the form of the differentials. It is then much easier to go backwards.
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RDKGames
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(Original post by _viotl)
i dont know it completely i tried checking some videos and gives no sense at all
It's a mental shortcut for a linear substitution u=50-m. So just use the sub if you cant do it mentally.
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idontkn
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The integration rule is add 1 to the power and divide by the new power.
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Pangol
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(Original post by idontkn)
The integration rule is add 1 to the power and divide by the new power.
Yes, but this is not an integral of a simple power of the variable to be integrated, so a little more care is needed.
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idontkn
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(Original post by Pangol)
Yes, but this is not an integral of a simple power of the variable to be integrated, so a little more care is needed.
What does that mean, do you mean it’s complicated because there’s a letter (m) involved?
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Pangol
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(Original post by idontkn)
What does that mean, do you mean it’s complicated because there’s a letter (m) involved?
No, I mean that the rule whereby you integrate (x)^n by adding 1 to the n and then dividing by the new power only works in the situation where the thing in the bracket is only the variable to be integrated. The variable in this case is m rather than x, but that rule on its own will not work in this case becasue the bit in the bracket is 50 - m, not just m.
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RDKGames
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(Original post by idontkn)
What does that mean, do you mean it’s complicated because there’s a letter (m) involved?
The answer is not \dfrac{(50-m)^{-1}}{-1} because differentiating this yields -(50-m)^{-2} rather than (50-m)^{-2}.

So your rule of 'increase exponent by 1 and divide by it' only applies to a small range of expressions we integrate, namely only integrals of the form \displaystyle \int (x+a)^n .dx = \dfrac{(x+a)^{n+1}}{n+1} + c where a is some contstant.
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idontkn
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(Original post by RDKGames)
The answer is not \dfrac{(50-m)^{-1}}{-1} because differentiating this yields -(50-m)^{-2} rather than (50-m)^{-2}.

So your rule of 'increase exponent by 1 and divide by it' only applies to a small range of expressions we integrate, namely only integrals of the form \displaystyle \int (x+a)^n .dx = \dfrac{(x+a)^{n+1}}{n+1} + c where a is some contstant.
Oh I get it, sorry I didn’t do A2 maths, only AS
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