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Re: MX2 2015 Integration Marathon The substitution (x = sec(u)) morphs the integrands to ∫tanu/(|tanu|) du and the sign is dependent on the values of x or bounds of the original integral Either I've overlooked something or you overlooked something lol

Re: MX2 2015 Integration Marathon Was it +-sec^-1 x + C?

Re: MX2 2015 Integration Marathon Yeah, the mathematical intuition required to see that division is quite spectacular!

Re: MX2 2015 Integration Marathon I see the consequence of it (Ekman's soln) after some manipulation of the integrand but I've seen people use it straight away. I guess they've completed the manipulation mentally..

This is so interesting... going to think about this tonight

Re: MX2 2015 Integration Marathon How do people know to substitute something like: u =x + \frac{1}{x}

How about would'nt've ;)

I'll have agent 47 eliminate him.

Re: MX2 2015 Integration Marathon sqrt(x) = int of x' / (2*sqrt(x)) dx Thats formula

Re: HSC 2015 2U Marathon You mean polynomial. Consider f(x) = sin (x)...

Re: HSC 2015 2U Marathon solve?

Re: HSC 2015 2U Marathon By inspection ln(x) = -1 ln(x) = 3

the best way is to immerse yourself through writing to force yourself to produce coherent and cogent sentences. this way you apply the word(s) and improve vocab real quick. if u dont apply, then the memory will vanish

r u srs? why?

acting begins when you enter the premises you need to show that you cannot focus at anything when he is talking, force yourself to think about something else so you lose focus on his words. this shows you cannot listen and easily distracted try to interrupt him as he speaks do bits of...

Do you think a doctor will prescribe someone that if they don't have ADHD (or supreme acting skills)?

Going too far m8. Learn to concentrate by yourself without artificial stimulants.

If f' is increasing, then f is increasing is a logical error because that is the incorrect relation. f' can be negative, hence f is not always increasing. Example, y = x^2 and y' = 2x. y' is increasing on all x but y isn't increasing on all x. As for f(x) = x^3, f' is not increasing. It is...

Yes, always. But with respect to f(x), not f'(x).

By inspecting the curve, the gradient function (f') itself is decreasing (starts off with high gradient then lowers down near 0), so we can say f is increasing at a decreasing rate. Example values, x = 0.01, y' = 100 x = 0.2, y' = 5 x = 5, y' = 0.2 x = 100, y' = 0.01 Since y' is clearly...