HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

kawaiipotato · Jul 21, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$When differentiating $u$, you forgot to lower the power.$

$Should be $\text{d}u=\frac{n}{x^2}\left(1-\frac{1}{x} \right)^{n-1}\text{ d}x$.$

oops fixed

InteGrand · Jul 21, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
oops fixed

That error carried through to your working in subsequent lines. If you start again from the IBP part, you should get to the correct answer.

Drsoccerball · Jul 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^1{\frac{1}{1-log_2(1-x)}}$

BUMP

InteGrand · Jul 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
BUMP

Requires use of the exponential integral according to WolframAlpha, unlikely to come up in the HSC...

http://www.wolframalpha.com/input/?i=\int_0^1{\frac{1}{1-\log_2(1-x)}}

leehuan · Jul 21, 2015

Re: MX2 2015 Integration Marathon

${ I }_{ 2n-1 }=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2n-1 } }{ \sqrt { 1-{ x }^{ 2 } } } } \\ $Show that ${ I }_{ 2n+1 }=\frac { { 2 }^{ n } n! }{ (1)(3)(5) \dots \left( 2n+1 \right) }$

porcupinetree · Jul 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
${ I }_{ 2n-1 }=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2n-1 } }{ \sqrt { 1-{ x }^{ 2 } } } } \\ $Show that ${ I }_{ 2n+1 }=\frac { { 2 }^{ n } n! }{ (1)(3)(5) \dots \left( 2n+1 \right) }$

Sy123 · Jul 24, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
This question is based on your ability to approximate sums by integrals, hence its location in this marathon thread.

Prove that there exist positive constants $C_1,C_2$ such that

$C_1\leq\frac{n!}{n^{n+1/2}e^{-n}}\leq C_2$

for all positive integers $n.$

(In fact, the ratio in this question tends to an exact constant, but proving this convergence without guidance is perhaps a bit much to ask. It is a reasonable enough followup exercise though to calculate this exact constant, given that the ratio does in fact converge.)

$\\ $Let$ \ u_n = \frac{n!}{n^{n+1/2}e^{-n}}$

$\\ $Using the trapezium rule we, and that all components of$ \ u_n \ $are positive, then$ \ 0 < u_n < e, \ (1)$

$\\ \frac{u_{n+1}}{u_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^{n+1/2}}$

$\\ $Let$ \ f(x) = \ln \left(1 + \frac{1}{x} \right) - \frac{1}{x+1/2} , (x \geq 1) \\ \Rightarrow \ f'(x) = \frac{-1}{x(x+1)(2x+1)^2} < 0$

$\\ \Rightarrow f(x) \ $is monotone decreasing$ \ \Rightarrow \ f(x) > \lim_{x \to \infty} f(x) = \ln(1) - 0 = 0 \\ \\ \therefore \ f(x) > 0$

$\\ \Rightarrow \left(x+\frac{1}{2} \right)\ln \left(1 + \frac{1}{x} \right) > 1 \\ \Rightarrow \left(1 + \frac{1}{n} \right)^{n + 1/2} > e \ $for$ \ n \geq 1$

$\\ \Rightarrow 1 > \frac{e}{\left(1 + \frac{1}{n} \right)^{n+1/2}} = \frac{u_{n+1}}{u_n} \\ \Rightarrow \ u_{n+1} < u_{n} \ $for$ \ n \geq 1, \ (2)$

$\\ $Hence, using propositions$ \ (1) \ $and$ \ (2) \ $we see that$ \ u_n \ $converges to a finite limit$$

$\\ $Let$\ \lim_{n \ to \infty} u_n = \lim_{n \to \infty} u_{2n} = l$

$\\ $Consider$ \ x_n = \frac{u_{2n}}{u_n^2} = \frac{1}{\sqrt{2} 4^n} \binom{2n}{n} \cdot \sqrt{n}$

$\\ \therefore \ \lim_{n \to \infty} x_n = \frac{l}{l^2} = \frac{1}{l}, \ (3)$

$\\ $Now, consider$ \ i_n = \int_0^{\pi /2} \sin^{2n}x \ dx \ $using IBP, we see that$ \ i_n = \frac{1}{4^n} \binom{2n}{n} \frac{\pi}{2}$

$\\ $Hence,$ \ \frac{\sqrt{2}}{\pi} \sqrt{n} i_n = \frac{1}{4^n\sqrt{2}} \binom{2n}{n} \sqrt{n} = x_n$

$x_n^2 = \frac{2}{\pi^2} n i_n^2$

$\\ \therefore \ \frac{\pi^2}{2} \frac{1}{l^2} = \lim_{n \to \infty} n i_n^2, \ (4)$

$\\ $Let$ \ j_n = \int_{0}^{\pi /2} \sin^n x \ dx$

$\\ j_n = \frac{n-1}{n} j_{n-1} \\ \Rightarrow \ n j_{n} j_{n-1} = (n-1)j_{n-1} j_{n-2}$

$\\ $Letting$ \ k_n = n j_n j_{n-1} \ $we see that$ \ k_n = k_{n-1} \ $for$ \ n \geq 2$

$\\ \therefore \ k_n = k_1 = j_1 j_0 = \frac{\pi}{2} \\ \therefore \ nj_{n} j_{n-1} = \frac{\pi}{2}, \ (5)$

$\\ $Since$ \ j_{n} - j_{n-1} = \int_0^{\pi/2} \sin^{n-1}x (\sin x -1) \ dx < 0 \Rightarrow \ j_n < j_{n-1}$

$\\ $Hence using$ \ (5) \ $we get$ \ nj_n^2 < \frac{\pi}{2} < nj_{n-1}^2$

$\\ \Rightarrow \ \frac{n-1}{n} \frac{\pi}{2} < (n-1) j_{n-1}^2 \Rightarrow \ \frac{n}{n+1} \frac{\pi}{2} < nj_n^2 \\ \\ $Hence, finally$ \ \frac{n}{n+1} \frac{\pi}{2} < nj_n^2 < \frac{\pi}{2}$

$\\ \therefore \ \lim_{n \to \infty} n j_n^2 = \frac{\pi}{2}$

$\\ $Since$ \ i_n = j_{2n} \ $so,$ \ \lim_{n \to \infty} (2n) i_n^2 = \frac{\pi}{2} \\ \therefore \ \lim_{n \to \infty} ni_n^2 = \frac{\pi}{4}$

$\\ $Using$ \ (4) \ $we see that$ \ \frac{\pi^2}{2} l^2 = \frac{\pi}{4}$

$\\ $Therefore$ \ \frac{1}{l} = \frac{1}{\sqrt{2\pi}}$

$\\ $Therefore$ \ \lim_{n \to \infty} \frac{n!}{n^{n+1/2} e^{-n}} = \sqrt{2\pi}$

leehuan · Aug 4, 2015

Re: MX2 2015 Integration Marathon

$\int { { \left( \cos { \left[ \tan ^{ -1 }{ \left\{ \sin { \left( \cot ^{ -1 }{ \left( x \right) } \right) } \right\} } \right] } \right) }^{ 2 }dx }$

leehuan · Aug 5, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Two more questions
$\int_0^1 \log^{log_ex}(2) dx$

$\\ $The integral is improper.$ \\ $Note $ \int { { a }^{ x }dx } =\frac { { a }^{ x } }{ \log { a } } \quad \left( c={ e }^{ \log { c } } \right) \\ $Let $ x={ e }^{ u }\Rightarrow dx={ e }^{ u }du\\ \int _{ 0 }^{ 1 }{ { \left( \log { 2 } \right) }^{ \log { x } }dx } \\ =\int _{ -\infty }^{ 0 }{ { \left( e\log { 2 } \right) }^{ u }du } \\ =\lim _{ k\rightarrow -\infty }{ { \left[ \frac { { \left( e\log { 2 } \right) }^{ u } }{ \log { { \left( e\log { 2 } \right) } } } \right] }_{ k }^{ 0 } }$

$\\ =\frac { 1 }{ \log { { \left( e\log { 2 } \right) } } } -0\\ =\frac { 1 }{ \log { { \left( e\log { 2 } \right) } } } \\$

kawaiipotato · Aug 5, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { { \left( \cos { \left[ \tan ^{ -1 }{ \left\{ \sin { \left( \cot ^{ -1 }{ \left( x \right) } \right) } \right\} } \right] } \right) }^{ 2 }dx }$

$$ let $ cot^{-1}(x) = \alpha$

$cot \alpha = x$

$\therefore $ I $ = \int (cos[tan^{-1}(sin \alpha )])^2 dx$

$$ I $ = \int (cos[tan^{-1}(\frac{x}{\sqrt{1+x^2 }})])^2 dx$

$$ let $ tan^{-1}(\frac{x}{\sqrt{1+x^2 }}) = \theta$

$tan \theta = (\frac{x}{\sqrt{1+x^2 }})$

$\therefore $ I $ = \int (cos \theta )^2 dx$

$$ I $ = \int \frac{1+x^{2}}{1+2x^{2}} dx$

$= \frac{1}{2} \int \frac{1 + 2x^{2}}{1+2x^{2}} + \frac{1}{1 + 2x^{2}} dx$

$= \frac{1}{2} [ x + \frac{1}{\sqrt{2}}tan^{-1}(\sqrt{2}x)] + $ C $$

kawaiipotato · Aug 5, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ $The integral is improper.$ \\ $Note $ \int { { a }^{ x }dx } =\frac { { a }^{ x } }{ \log { a } } \quad \left( c={ e }^{ \log { c } } \right) \\ $Let $ x={ e }^{ u }\Rightarrow dx={ e }^{ u }du\\ \int _{ 0 }^{ 1 }{ { \left( \log { 2 } \right) }^{ \log { x } }dx } \\ =\int _{ -\infty }^{ 0 }{ { \left( e\log { 2 } \right) }^{ u }du } \\ =\lim _{ k\rightarrow -\infty }{ { \left[ \frac { { \left( e\log { 2 } \right) }^{ u } }{ \log { { \left( e\log { 2 } \right) } } } \right] }_{ k }^{ 0 } }$

$\\ =\frac { 1 }{ \log { { \left( e\log { 2 } \right) } } } -0\\ =\frac { 1 }{ \log { { \left( e\log { 2 } \right) } } } \\$

nice solution, I had this in mind:

$Using x^{logy} = y^{logx}$

$(log2)^{log(x)} = x^{log(log2)}$

$$ Hence, $ I = \int_0^1 x^{log(log2)} dx$

$= [\frac{x^{log(log2) + 1}}{log(log2)+1}]_0^1$

$= \frac{1}{log(log2) + 1)}$

leehuan · Aug 5, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
nice solution, I had this in mind:

$Using x^{logy} = y^{logx}$

$(log2)^{log(x)} = x^{log(log2)}$

$$ Hence, $ I = \int_0^1 x^{log(log2)} dx$

$= [\frac{x^{log(log2) + 1}}{log(log2)+1}]_0^1$

$= \frac{1}{log(log2) + 1)}$

Nifty. I never knew such a property existed. I suppose it's easy to verify it because just taking log of both sides yields log(x)log(y)=log(y)log(x)

leehuan · Aug 5, 2015

Re: MX2 2015 Integration Marathon

$\\ $Show that $\int _{ 0 }^{ \infty }{ \frac { { x }^{ 29 } }{ { \left( { 5x }^{ 2 }+49 \right) }^{ 17 } } dx } =\frac { 14! }{ 2\cdot { 49 }^{ 2 }\cdot { 5 }^{ 15 }\cdot 16! }$

Ekman · Aug 5, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ $Show that $\int _{ 0 }^{ \infty }{ \frac { { x }^{ 29 } }{ { \left( { 5x }^{ 2 }+49 \right) }^{ 17 } } dx } =\frac { 14! }{ 2\cdot { 49 }^{ 2 }\cdot { 5 }^{ 15 }\cdot 16! }$

$$ Sub in: $ x = \frac{7}{\sqrt{5}}\tan{\theta}$

$\therefore \int_{0}^{\infty} \frac{x^{29}}{(5x^2 + 49)^{17}} dx = \int_{0}^{\frac{\pi}{2}} \frac{\frac{49^{15}}{5^{15}} \tan^{29}{\theta} \sec^{2}{\theta}}{49^{17}(1+\tan^{2}{\theta})^{17}} d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{\tan^{29}{\theta}}{5^{15}\cdot 49^2 \cdot \sec^{32}{\theta}} d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{\sin^{29}{\theta} \cos^{3}{\theta}}{5^{15}\cdot 49^2 } d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{\sin^{29}{\theta} \cos{\theta}(1-\sin^{2}{\theta})}{5^{15}\cdot 49^2} d\theta$

$= \frac{1}{5^{15}\cdot 49^2} \left(\frac{\sin^{30}{\theta}}{30} -\frac{sin^{32}{\theta}}{32} \right)_{0}^{\frac{\pi}{2}}$

$= \frac{1}{5^{15}\cdot 49^2} \cdot \frac{2}{30 \cdot 32} = \frac{1}{5^{15}\cdot 49^2 \cdot 2 \cdot 15 \cdot 16} = \frac{14!}{5^{15}\cdot 49^2 \cdot 2\cdot 16! }$

Librah · Aug 8, 2015

Re: MX2 2015 Integration Marathon

$\\ \int _{ -a }^{ \ a }{ \frac { xdy }{ { \left( { x }^{ 2 }+ {y^{2} \right) }^{ 3/2 } } }$

I dunno, i can't seem to do it lol.

InteGrand · Aug 8, 2015

Re: MX2 2015 Integration Marathon

Librah said:
$\\ $Show that $\int _{ -a }^{ \ a }{ \frac { xdy }{ { \left( { x }^{ 2 }+ {y^{2} \right) }^{ 3/2 } } }$

I dunno, i can't seem to do it lol.

Show that it what? You forgot the RHS.

However, you should be able to do it with a trig. substitution $y = x\tan \theta$ .

braintic · Aug 8, 2015

Re: MX2 2015 Integration Marathon

Librah said:
$\\ $Show that $\int _{ -a }^{ \ a }{ \frac { xdy }{ { \left( { x }^{ 2 }+ {y^{2} \right) }^{ 3/2 } } }$

I dunno, i can't seem to do it lol.

Are x and y dependent variables?

Librah · Aug 8, 2015

Re: MX2 2015 Integration Marathon

Forget the show that part, meant to just find the integral.

braintic said:
Are x and y dependent variables?

Nope.

InteGrand · Aug 8, 2015

Re: MX2 2015 Integration Marathon

Librah said:
Nope.

Just use trig. sub. then.

Librah · Aug 8, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Just use trig. sub. then.

Ah yeah, that works thx.

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