HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

leehuan · Jun 4, 2015

Re: MX2 2015 Integration Marathon

(-1)^k is the same as (-1)^(2+k)

The expression looks decent. I can't be bothered figuring out my combinatorics right now, but I found this:

$\frac { 1 }{ n! } \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) { \left( -1 \right) }^{ k }\ln { \left| x+k \right| } }$

Drsoccerball · Jun 4, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
(-1)^k is the same as (-1)^(2+k)

The expression looks decent. I can't be bothered figuring out my combinatorics right now, but I found this:

$\frac { 1 }{ n! } \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) { \left( -1 \right) }^{ k }\ln { \left| x+k \right| } }$

There should be 1/(n-k)! due to the partial fractions coefficients change

EDIT: and not so sure about the bionomial coefficients that you had ?

leehuan · Jun 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
There should be 1/(n-k)! due to the partial fractions coefficients change

EDIT: and not so sure about the bionomial coefficients that you had ?

The binomial coefficients spit out a 1/(n-k)! quantity anyway.
I suppose, if the binomial coefficients were expanded I'd have ended up with a 1/(k!) * 1/(n-k)!

I checked on Wolfram Alpha and I realised that the terms followed Pascal's triangle, which was what lead me to the assumption of the binomial coefficient being related.

$\left( \begin{matrix} n \\ k \end{matrix} \right) =\frac { n! }{ k!\left( n-k \right) ! }$

Drsoccerball · Jun 4, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
The binomial coefficients spit out a 1/(n-k)! quantity anyway.
I suppose, if the binomial coefficients were expanded I'd have ended up with a 1/(k!) * 1/(n-k)!

I checked on Wolfram Alpha and I realised that the terms followed Pascal's triangle, which was what lead me to the assumption of the binomial coefficient being related.

$\left( \begin{matrix} n \\ k \end{matrix} \right) =\frac { n! }{ k!\left( n-k \right) ! }$

hmmmm... what method did you use

leehuan · Jun 4, 2015

Re: MX2 2015 Integration Marathon

Partial fractions on n=0, 1, 2, 3 then I checked them on WolframAlpha. Didn't realise a pattern with the binomial coefficients until I saw the Mathematica solutions though. Then I just plugged n=4 and n=5 in and it turned out to be the same. I couldn't be bothered doing over n=3

Drsoccerball · Jun 4, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Partial fractions on n=0, 1, 2, 3 then I checked them on WolframAlpha. Didn't realise a pattern with the binomial coefficients until they put it up though. Then I just plugged n=4 and n=5 in and it turned out to be the same. I couldn't be bothered doing over n=3

yeah i did the same
$A_1=\frac{1}{(1)(2)...(n)}$

$A_2 =\frac{-1}{(n-1)!}$

$A_3= \frac{1}{(n-2!)}$

Kaido · Jun 4, 2015

Re: MX2 2015 Integration Marathon

Both your answers are the same

Missing one bit doe :L

InteGrand · Jun 4, 2015

Re: MX2 2015 Integration Marathon

I'm getting the same as leehuan.

$\frac{1}{x(x+1)(x+2)\cdots (x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\frac{A_2}{x+2}+\cdots \frac{A_n}{x+n}=\sum _{k=0}^n \frac{A_k}{x+k}$ , for some suitable A_k (partial fractions).

Using the Heaviside cover-up method,

$A_k = \frac{1}{\underbrace{-k(-k+1)(-k+2)\cdots (-k+(k-1))}_{k\text{ terms from which we will take }-1 \text{ out as a factor}}\times (-k+k+1)(-k+k+2)\cdots (-k+n)}$

$= \frac{1}{(-1)^k k(k-1)(k-2)\cdots (k-(k-1))\times 1\cdot 2 \cdot \cdots \cdot (n-k)}$ (taking out the k factors of -1)

$=\frac{(-1)^k }{k!(n-k)!}$ (as $\frac{1}{(-1)^k}=(-1)^k$ ).

$\therefore \int\frac{1}{x(x+1)(x+2)\cdots (x+n)} \text{ d}x = \int \sum _{k=0}^n \frac{A_k}{x+k} \text{ d}x$ , where $A_k = \frac{(-1)^k}{k!(n-k)!}$

$=\sum _{k=0}^n \int \frac{A_k}{x+k} \text{ d}x$

$=\sum _{k=0}^n A_k \ln |x+k|+c$

$=\sum _{k=0}^n \frac{(-1)^k}{k!(n-k)!} \ln |x+k|+c.$

leehuan · Jun 5, 2015

Re: MX2 2015 Integration Marathon

Oh whoops I forgot a +C...
---------------
Have a break.
$\int { x\sin { x } \cos { x } \tan { x } \csc { x } \sec { x } \cot { x } dx }$

InteGrand · Jun 5, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Oh whoops I forgot a +C...
---------------
Have a break.
$\int { x\sin { x } \cos { x } \tan { x } \csc { x } \sec { x } \cot { x } dx }$

$\frac{x^2}{2} + c$

Drsoccerball · Jun 5, 2015

Re: MX2 2015 Integration Marathon

$\int{sin(xcosx)}dx$

InteGrand · Jun 5, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{sin(xcosx)}dx$

Can't do it with elementary functions

leehuan · Jun 6, 2015

Re: MX2 2015 Integration Marathon

Just asking out of curiosity, do you make up integrals on the spot Drsoccerball? When I do that I usually plug them into mathematica or something to make sure they work. Friendly advice

---------------------------
$\int { \sec ^{ 7 }{ x } } dx\\ $You may use a reduction formula$$

Drsoccerball · Jun 7, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Just asking out of curiosity, do you make up integrals on the spot Drsoccerball? When I do that I usually plug them into mathematica or something to make sure they work. Friendly advice
---------------------------
$\int { \sec ^{ 7 }{ x } } dx\\ $You may use a reduction formula$$

The reduction formula is:
$I_n = \frac{tanx\cdot sec^{n-2}x +(n-2)I_{n-2}}{n-1}$

$$ Find $ I_7$

leehuan · Jun 7, 2015

Re: MX2 2015 Integration Marathon

$$Show that $\int _{ 0 }^{ 1 }{ { \left( \ln { x } \right) }^{ n }dx } ={ \left( -1 \right) }^{ n }n!$

Ekman · Jun 7, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$$Show that $\int _{ 0 }^{ 1 }{ { \left( \ln { x } \right) }^{ n }dx } ={ \left( -1 \right) }^{ n }n!$

$$Let $ I_{n} = \int_{0}^{1} (\ln{x})^n dx$

$$Via IBP $ I_{n}=-nI_{n-1} $ for $ n > 1$

$$Note: $ I_{0}=1$

$\therefore I_{n} = (-1)^n n! $ for $ n\geq0$

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

$\int { \frac { dx }{ { x }^{ 7 }-x } }$

Ekman · Jun 8, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { dx }{ { x }^{ 7 }-x } }$

$\int \frac{1}{x^7-x} dx = \int \frac{1}{x(x^6-1)} dx$

$$ Multiply top and bottom by $x^5 : \int \frac{x^5}{x^6(x^6-1)}dx$

$$ Let $u=x^6 \therefore \frac{1}{6} \int \frac{1}{u(u-1)}dx = \frac{1}{6}\ln{\left|\frac{x^6-1}{x^6}\right|} +c$

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

$\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx$

$$Hence show that $\frac { 22 }{ 7 } >\pi$

InteGrand · Jun 8, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx$

$$Hence show that $\frac { 22 }{ 7 } >\pi$

Integral is positive and comes out to be something interesting, as found by a quite tedious computation (involving polynomial division etc.).

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