SpiralFlex
Well-Known Member
- Joined
- Dec 18, 2010
- Messages
- 6,960
- Gender
- Female
- HSC
- N/A
Re: HSC 2013 4U Marathon
Sorry didn't see that substitution.
Sorry didn't see that substitution.
How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.Evaluating the definite integral is well within the scope of MX2, the substitution
kills it.
To be fair, who would?Sorry didn't see that substitution.
Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.
The substitution u=arctanx doesn't work if it was an indefinite integral but it works very well here.How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.
ah okay, I see. nice.Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.
The first way I solved this question though was via a longer, different, less difficult to spot method...differentiation under the integral sign. But this is not MX2 material and it was longer than the substitution anyway.
Let L=sqrt(6+sqrt(6+sqrt(6+...
Difficulty: Easy
i) y=f(x) intersect y=x at least onceMuch easier than most of the questions I post here, but still quite a cool result.
Denote as the k'th triangular number (surprise, surprise).
Nice work. I'll try make a question with triangular number properties or something considering how much you like them heh.Denote as the k'th triangular number (surprise, surprise).
So we obtain:
as the sum of the multiples of 3.
So we obtain:
as the sum of multiples of 5.
Now we will double count all the multiples of 15, so using the same logic:
as the sum of multiples of 15.
Now we add and subtract some shit:
Noting that
Awesome question. I got the answer but I'll let others have a go first, lets try and have different people answering besides the usual few. If no one answers this by tonight I'll put up my solution though.