MX2 Integration Marathon 2021 (1 Viewer)

stupid_girl

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I made a typo in Wolfram earlier. It does give a nice answer.
 
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s97127

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Let :



Notice that is the derivative of and hence the integral is in the form: . Thus,


(using the well-known identity )
how come a medical student like yourself still has so much interest in math after finishing HSC? :)
 

YonOra

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Med is science, so how could u not get bored o_O
 

vernburn

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I have a feeling @vernburn likes maths more than the stuff he is going to do in med.
It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).

The ultimate tedious integral:

I may be bored but not enough to attempt this monster!
 

Qeru

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It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).
Doesn't look like you were bored of maths considering you got a 99 for 4U. There had to be some interest there lol
 

CM_Tutor

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The ultimate tedious integral:

This can easily be converted to a much more recognisable integral with the substitution:



From which it follows that:



and the problem is reduced to a very (very) messy partial fractions problem. According to the integral calculator, and after cleaning up its awful formatting / presentation, the indefinite integral is:







 

stupid_girl

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I wrote this in the past but I have now forgotten how to solve it.


Edit: I remember how to solve it now. 😈
 
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CM_Tutor

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I wrote this in the past but I have now forgotten how to solve it.
Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?
 

stupid_girl

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Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?
The integral is 1 irrespective of the value of k.
 

Qeru

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can you post the solution here? This gives me a headache for the last few days :(
Yeah lol, some factoring trick I imagine, or King Rule but I dont see how that is used.
 

stupid_girl

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can you post the solution here? This gives me a headache for the last few days :(
What tricks have you tried?

The following techniques are NOT required. 😈
integration by parts
differentiation under the integral sign
hyperbolic function
 

s97127

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What tricks have you tried?

The following techniques are NOT required. 😈
integration by parts
differentiation under the integral sign
hyperbolic function
I've tried all the techniques above. I think the problem will be solved by substituting x with another variable t and we got I + I = 2 so I = 1.
Am i on the right path?
 

stupid_girl

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By adjusting the coefficient of k to 1, you should be able to take out log_2 (2), which is just 1. Integrating dx from 0 to 1 gives you 1.

The remaining factors have the same structure and you can offset them by suitable substitution.
 

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