HSC 2012-14 MX2 Integration Marathon (archive) (7 Viewers)

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

braintic said:
The fraction is still a unit that remains intact - it doesn't need to be broken up. As long as you think of the fraction as a variable itself, its exactly the same.
Seriously, if anyone sees expanding [1+5/(x+2)]^5 as being more difficult that (u+5)^5, they are going to struggle with a lot in Extension 2.

lol wow calm down, i understand the fraction doesn't have to be broken down, i'm just saying your expansion will look much messier than my expansion. i understand that the algebraic skill required for both expansions is similar...

braintic · Jun 9, 2013

Re: MX2 Integration Marathon

Makematics said:
lol wow calm down, i understand the fraction doesn't have to be broken down, i'm just saying your expansion will look much messier than my expansion. i understand that the algebraic skill required for both expansions is similar...

Calm down?? You're completely misreading me.

Sy123 · Jun 10, 2013

Re: MX2 Integration Marathon

$Prove without differentiating the Right Hand Side$

$\int \frac{dx}{\sqrt{(x-a)(x-b)}} = \ln \left |\frac{2x - (a+b)}{\sqrt{2(a^2+b^2)}} + \sqrt{\frac{4x^2-4x(a+b) - (a-b)^2}{2(a^2+b^2)}} \right |$

No room for the +c :/

HeroicPandas · Jun 10, 2013

Re: MX2 Integration Marathon

Makematics said:
Using long division followed by a simple application of the heaviside method, the answer is

$-\frac{1}{3}x^3-x+\frac{1}{2}\ln\left |\frac{x+1}{x-1} \right |+c$

yes very nice

Sy123 · Jun 10, 2013

Re: MX2 Integration Marathon

$Prove that$

$\lim_{n \to \infty} n \int_0^{\frac{\pi}{2}} (1-\sqrt[n]{\sin x}) \ dx = \frac{\pi}{2} \ln 2$

Sy123 · Jun 10, 2013

Re: MX2 Integration Marathon

$Prove, without differentiating the Right Hand Side$

$\int \frac{dx}{(x+1)(x+2)(x+3)\dots (x+n)} = \ln \left(\prod_{k=1}^{n} (x+k)^{\frac{(-1)^{k-1}}{(k-1)!(n-k)!}} \right) + c$

$Note that$

$\prod_{k=1}^{n} f(k) \equiv f(1) \cdot f(2) \cdot f(3) \dots f(n)$

asianese · Jun 10, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Prove, without differentiating the Right Hand Side$

$\int \frac{dx}{(x+1)(x+2)(x+3)\dots (x+n)} = \ln \left(\prod_{k=1}^{n} (x+k)^{\frac{(-1)^{k-1}}{(k-1)!(n-k)!}} \right) + c$

$Note that$

$\prod_{k=1}^{n} f(k) \equiv f(1) \cdot f(2) \cdot f(3) \dots f(n)$

Gonna need some intense heaviside there

Kipling · Jun 10, 2013

Re: MX2 Integration Marathon

Integrate: (7x-3)/(6+4x-x^2)^0.5

asianese · Jun 10, 2013

Re: MX2 Integration Marathon

Kipling said:
$\int \dfrac{7x-3}{\sqrt{6+4x-x^2}} \mathop{\mathrm{d}x}$

FTFY.

#procrastinating

seanieg89 · Jun 10, 2013

Re: MX2 Integration Marathon

The following question is set in http://en.wikipedia.org/wiki/Three-dimensional_space with the standard coordinates.

$Let $r>0$. \\Define $C(r)$ to be the solid of revolution formed by rotating the planar region $\{(x,0,z)\in\mathbb{R}^3:|x|\leq r,0\leq z \leq e^{-x^2}\}$ about the $z$-axis.\\ Define $S(r)$ to be the solid region $\{(x,y,z)\in\mathbb{R}^3: |x|\leq r,|y|\leq r,0\leq z\leq e^{-(x^2+y^2)}\}.$\\ \\ By considering the volumes of $C(r)$ and $S(r)$, evaluate:\\ \\ \displaystyle\lim_{r\rightarrow\infty} \int_{-r}^r e^{-t^2}\, dt.$

Sy123 · Jun 11, 2013

Re: MX2 Integration Marathon

seanieg89 said:
The following question is set in http://en.wikipedia.org/wiki/Three-dimensional_space with the standard coordinates.

$Let $r>0$. \\Define $C(r)$ to be the solid of revolution formed by rotating the planar region $\{(x,0,z)\in\mathbb{R}^3:|x|\leq r,0\leq z \leq e^{-x^2}\}$ about the $z$-axis.\\ Define $S(r)$ to be the solid region $\{(x,y,z)\in\mathbb{R}^3: |x|\leq r,|y|\leq r,0\leq z\leq e^{-(x^2+y^2)}\}.$\\ \\ By considering the volumes of $C(r)$ and $S(r)$, evaluate:\\ \\ \displaystyle\lim_{r\rightarrow\infty} \int_{-r}^r e^{-t^2}\, dt.$

$$By considering$ \ \ t^2=x^2+y^2$

$Looking at the x-y plane, for varying t, the curve looking downwards is always a circle, as t varies from$ \ \ -r \rightarrow r \ \ $according to the equation$

$z=e^{-t^2} \ \ $where z is the height of the curve at that specific point$$

$Since for the solid region$ \ S(r) \ $there is extra bits at the corners, and since$ \ \ (x^2+y^2=t^2) \ \ $is a circle$

$C(r) + \varepsilon(r) = S(r)$

$$However$ \ \ r \rightarrow \infty \ \ \varepsilon \rightarrow 0$

$Hence as r increases without bound, the volumes are the same$

$By cutting$ \ S(r) \ $perpendicular to the x-y plane, they are common cross sections$

$z= e^{-x^2} \times k(y)$

At some y, I think the notation is (x,k,z), then there will be a 2d curve $z=e^{-x^2} \times e^{-k^2}$

So treating y like a constant, we find the area of a general cross section.

$A(x) = e^{-y^2} \int_{-\infty}^{\infty} e^{-x^2} \ dx$

Then collecting all the cross sections:

$S = \int_{-\infty}^{\infty} e^{-y^2} \int_{-\infty}^{\infty}e^{x^2} \ dx \ dy$

$\sqrt{S} = \int_{-\infty}^{\infty} e^{-x^2} \ dx$ Since they are definite integrals of the same thing but different variables.

$$S=C at infinite$ \ \ r$

$\therefore \int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$

Is that all? No rigour issues or anything?

$C= \pi \ \ $from general computation$$

Is this right so far?

seanieg89 · Jun 12, 2013

Re: MX2 Integration Marathon

Sy123 said:
$$By considering$ \ \ t^2=x^2+y^2$

$Looking at the x-y plane, for varying t, the curve looking downwards is always a circle, as t varies from$ \ \ -r \rightarrow r \ \ $according to the equation$

$z=e^{-t^2} \ \ $where z is the height of the curve at that specific point$$

$Since for the solid region$ \ S(r) \ $there is extra bits at the corners, and since$ \ \ (x^2+y^2=t^2) \ \ $is a circle$

$C(r) + \varepsilon(r) = S(r)$

$$However$ \ \ r \rightarrow \infty \ \ \varepsilon \rightarrow 0$

$Hence as r increases without bound, the volumes are the same$

$By cutting$ \ S(r) \ $perpendicular to the x-y plane, they are common cross sections$

$z= e^{-x^2} \times k(y)$

At some y, I think the notation is (x,k,z), then there will be a 2d curve $z=e^{-x^2} \times e^{-k^2}$

So treating y like a constant, we find the area of a general cross section.

$A(x) = e^{-y^2} \int_{-\infty}^{\infty} e^{-x^2} \ dx$

Then collecting all the cross sections:

$S = \int_{-\infty}^{\infty} e^{-y^2} \int_{-\infty}^{\infty}e^{x^2} \ dx \ dy$

$\sqrt{S} = \int_{-\infty}^{\infty} e^{-x^2} \ dx$ Since they are definite integrals of the same thing but different variables.

$$S=C at infinite$ \ \ r$

$\therefore \int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$

Is that all? No rigour issues or anything?

$C= \pi \ \ $from general computation$$

Is this right so far?

Pretty much, I would prefer if you did the work to show the epsilon term tends to zero, this is not too hard.

Also it is probably better to describe your slices as being perpendicular to the y-axis (rather than the x-y plane), ie every point in a slice has the same y-coordinate, this makes it unambiguous.

What is nice is that these volume computations are standard mx2 work, even if the notation for the solids is slightly beyond.

twinklegal19 · Jun 13, 2013

Sy123 said:
$Prove without differentiating the Right Hand Side$

$\int \frac{dx}{\sqrt{(x-a)(x-b)}} = \ln \left |\frac{2x - (a+b)}{\sqrt{2(a^2+b^2)}} + \sqrt{\frac{4x^2-4x(a+b) - (a-b)^2}{2(a^2+b^2)}} \right |$

No room for the +c :/

I got a simpler result?

$\ln\left | x-\frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right |$

asianese · Jun 13, 2013

Re: MX2 Integration Marathon

If you differentiate twinkie's answer you get the integrand.

Sy123 · Jun 13, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
I got a simpler result?

$\ln\left | x-\frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right |$

Well then there you go lol, I did make a mistake.

asianese · Jun 14, 2013

Re: MX2 Integration Marathon

$\int \dfrac{\mathop{\mathrm{d}x}}{2+\sin x+\cos x}$

twinklegal19 · Jun 14, 2013

Re: MX2 Integration Marathon

asianese said:
$\int \dfrac{\mathop{\mathrm{d}x}}{2+\sin x+\cos x}$

just use the t-formula? Should end up with an inverse tan

petermik · Jun 15, 2013

Re: MX2 Integration Marathon

$\int (1+x^2)^{\frac{3}{2} \, dx$ Using integration by parts

asianese · Jun 15, 2013

Re: MX2 Integration Marathon

Somehow the dx flew up top.

petermik · Jun 16, 2013

Re: MX2 Integration Marathon

$\int (1+x^2)^{\frac{3}{2}} \, dx$ Using integration by parts (fixed it)..

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