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Re: MX2 2015 Integration Marathon NEW Q: $ (i)\, Suppose that $x$ is a positive real number and that $n$ is a positive integer. Show that $\frac{1}{1+x^n}\,<\,1$. \\ (ii)\, Let $I_n = \int_{0}^{1} \frac{1}{1+x^n}\text{ d}x$ where $n$ is a positive integer and $n\,\geq\,2$. Show that...

What you get in past final exams will pretty much be similar in difficulty level to your final (there is nothing too hard), so practice those papers and time yourself.

Literally 0, you will not find this difficulty level of q's in your final exam, the questions will be much easier to solve, only hard thing is time really, if the exam was 3 hours long (and not 2) a lot of people will be able to get much higher to nearly full marks for the final.

Don't bother with it, they will never ask this sort of q in any exam and they probably put it in there just for LOLs. I doubt most of the cohort wouldn't even know what a spherical triangle is so they probably didn't even touch this question at all.

Re: MX2 2015 Integration Marathon I can assure you it is not a typo this time :p

Haha, is this like the last question of the Chapter? I'm pretty sure I didn't do this q at all, when I looked at it, I just closed my book and when I asked my tutor how to do this q, he just ran away :)

Re: MX2 2015 Integration Marathon NEW Q: $ If $I_n = \int_{0}^{1} x^{n}\sqrt{1-x}\text{ d}x$ show that $I_n = \frac{2n}{2n+3}I_{n-1}$ for $n\,\geq\,1$. \\ Hence, show that $I_n = \frac{n!(n+1)!}{(2n+3)!}4^{n+1}$ $

Re: HSC 2015 4U Marathon $ Since $P(x)=x^3+x+1$ is an increasing function over the reals, this means that only one root of $P(x)$ exists. Note that $P(-1)\,<\,0\,<\,P(0)$, so our root must lie in the open interval $(-1, 0)$. But by the rational root theorem, there exists no such ordered pair...

Re: MX2 2015 Integration Marathon I don't know why, I always make some stupid typo when typing up these questions :(

You posted and answered your own question LOL.

Re: MX2 2015 Integration Marathon .

Re: MX2 2015 Integration Marathon Change the x^2 into -(1-x^2-1) and see how you can go from there...

Re: MX2 2015 Integration Marathon Lol my bad, I misinterpreted your question.

Re: MX2 2015 Integration Marathon EASIER RECURRENCE Q: $Let $I_n = \int_{0}^{1} (1-x^2)^{\frac{n}{2}}\text{ d}x$, where $n\,\geq0$ is an integer.$ \\ $(i)$ \, $Show that $I_n=\frac{n}{n+1}I_{n-2}$ for every integer $n$\,\geq2.$ \\ $(ii)$ \, $Evaluate $I_5$ $

Re: MX2 2015 Integration Marathon SOLUTION $ $I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ \\ = $[\frac{x^2}{2}(1-x^3)^n]_{0}^{1}+\frac{3n}{2}\int_{0}^{1} x^4(1-x^3)^{n-1}\text{ d}x$ \\ = $\frac{3n}{2}\int_{0}^{1} x(1-(1-x^3))(1-x^3)^{n-1}\text{ d}x$ \\ = $\frac{3n}{2}(I_{n-1}-I_n)$ \\ So $I_n...

Re: MX2 2015 Integration Marathon Yes, it is a standard recurrence question that may require a trick or two to solve, but is probably one of the more easier recurrence questions I have seen. There are definitely much harder ones out there.

MATH1151 folder is now updated! :) I have put up some sample solutions for the 2013 Algebra Ver 2A so I hope that helps with your revision for the upcoming Algebra Quiz 1 in Week 6. Let me know if there are any errors or something you're not happy with in the solutions so I can fix it up. The...

Re: MX2 2015 Integration Marathon What exactly is troubling you?

Re: MX2 2015 Integration Marathon NEW Q: $ If $I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ for $n\,\geq 0$, show that $I_n = \frac{3n}{3n+2}I_{n-1}$ for $n$\,\geq 1. $

Re: MX2 2015 Integration Marathon .