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Yeah.

Because the limits and integrand both depend on t, we can't just use FTC simply like that (I'm assuming you subbed t for s in the integrand to get your dx/dt).

$\noindent Here's a generalised form Leibniz rule with variable limits too that you'll want to use:$ https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement . $\noindent To obtain something like that (for your one the lower limit is a constant), essentially set the upper limit to...

$\noindent a) Note that $\frac{\mathrm{d}}{\mathrm{d} x}\left(\int _0 ^x U(t) \text{ d}t\right)= U(x)$ from the FTC. Now,$ $$\begin{align*}V^\prime (x) &= -U(x) + \left(a-x\right)u(x) + U(x) \\ &= \left(a-x\right)u(x). \end{align*}$$ (So your first answer is right.) $\noindent b) Note...

Re: HSC Chemistry Marathon 2016 Uni-level Q's should probably be posted in the university Science threads in order to maximise chances of getting the question answered.

$\noindent (We'll ignore the influence of the $\sqrt{2}$ factor here since this doesn't affect the query.) This is because the function is of the form $f\left(2x + a\right)$, rather than $f(x+a)$ (where here, the function $f$ is the cosine function, i.e. $f(u) = \cos u$). If it was $f(x+a)$, you...

Not a priori, it's possible for something to be bounded between two values without actually attaining all values in between.

The question isn't asking for a maximum, it's asking for what is the total no. of possible values for m. So I guess you should technically justify why m can indeed attain each of the integer values from 6 to 54 without having any three points in a line, but for the purposes of an MCQ test like...

$\noindent 1) Here's some hints.$ $\noindent a) Use $\ddot{x} = \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{2} v^2 \right)$.$ $\noindent b) Note from a) $v^2 = 2x^2-6x+4$. At $x=1$, we have acceleration $\ddot{x} = 2-3=-1$, and also $v^2 = 2-6+4 = 0 \Rightarrow$ the velocity is $0$. For...

Yeah, can be any real number a > 0. This is a pretty classic fact about improper integrals in general (not just p-test), and is analogous with a result from infinite series (essentially you can start the series anywhere and it doesn't affect the convergence status; only the "long-run behaviour"...

Yeah.

The lower limit (a finite one) doesn't matter for these improper integrals if the function doesn't have unusual behaviour at the lower limit (or somewhere in between that and a different lower lower limit) like a singularity there (i.e. as long as the function's integrable over the interval...

Oops sorry, I was using ≥ for something else at the time and had that in mind and wrote ≥ by mistake. It should just be >.

It converges iff p > 1. To see this, note that if you transform the integral by using a substitution of u = ln x, it becomes a regular "p-integral", which converges iff p > 1.

The maths of the Third Law is pretty simple to apply. The laws' derivations are pretty interesting mathematically speaking, and how they can be derived from Newton's Law of Gravitation / conservation of angular momentum.

Re: Multivariable Calculus Basically, partials of r and theta wrt x mean we are thinking of r and theta as functions of x and y, and we are differentiating wrt x holding y constant. If you don't know about Jacobians, note r as a function of x and y is r = sqrt(x^2 + y^2), so...

Have you learnt Kepler's Third Law?

The Riemann Sum doesn't get rearranged to the answer, it gets to the final answer by taking a limit as n -> oo. Also, the final answer would have (j+1) on the denominator, not n (remember, n is nothing in the end, it goes away when taking the limit to infinity).

What have you gotten up to using my previous post?

Have you used any trigonometry? You should use trig. to work out the angle the strings make with the horizontal.