Please help with part b of this integration question (1 Viewer)

Life'sHard

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For i)
Let x = a-u on the LHS then differentiate to solve for du
 

Life'sHard

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For b)
Whenever there's an x change it to pi-x I think.
 

CM_Tutor

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Applying the result in part (a) to the question in part (b), you get:



We have re-generated our original integral and can combine them to get:

 
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kimtuluu

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Applying the result in part (a) to the question in part (b), you get:



We have re-generated our original integral and can combine them to get:

That's perfect. Many thanks for your help. Note: there are couple of lines in the solution that have double du in it.
 

CM_Tutor

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FYI, when I see the proof in part (a), I expect to apply it to an integral that either regenerates the same integral or an equivalent that allows a major simplification of the problem.
 

kimtuluu

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FYI, when I see the proof in part (a), I expect to apply it to an integral that either regenerates the same integral or an equivalent that allows a major simplification of the problem.
I thought so too. Just didn't think that it would involve substitution as well as integration by parts where full working out is required for a 4 mark question.
 

kimtuluu

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This is my answer of part (a). Please let me know if it is acceptable
1625726612842.png
 

quickoats

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This is my answer of part (a). Please let me know if it is acceptable
View attachment 31027
No this isn’t the proper way to prove it using a substitution.
You start by saying let u = a - x. Then you find du/dx, find the new bounds, then sub everything back into the original equation to change it from being in terms of x to in terms of u.
This method should be familiar if you are doing MX1/2
 

stupid_girl

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I thought so too. Just didn't think that it would involve substitution as well as integration by parts where full working out is required for a 4 mark question.
This question does not require integration by parts.
 

CM_Tutor

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This is my answer of part (a). Please let me know if it is acceptable
View attachment 31027
As @quickoats has noted, this is not the standard way to do this proof. However, that doesn't mean that it is either wrong or unacceptable.

It would be more usual to start your approach with "Let be the primitive function of ", allowing you to go straight to



Having said that, I see no flaw in your method and it is a valid proof. I would give it full marks.

The more usual approach would be:

 

mr.habibbi

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is there an alternative way of proving this without needing the "in a definite integral, the variable is interchangeable" statement (because i lost a mark for not writing this statement in my proof)
 

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