HSC 2015 MX2 Integration Marathon (archive) (7 Viewers)

Ekman · Sep 30, 2015

Re: MX2 2015 Integration Marathon

psyc1011 said:
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

But why does the tanu have absolute signs. Because whenever I had to trig sub something into an integral to get rid of a square root sign, I never had to consider the absolute signs. This is actually the first I have seen something like this

InteGrand · Sep 30, 2015

Re: MX2 2015 Integration Marathon

Technically you SHOULD consider absolute value signs to get a full solution generally speaking (because √(t²) = |t|), but somehow they don't seem to bother with it much in the HSC.

Paradoxica · Sep 30, 2015

Re: MX2 2015 Integration Marathon

A less impossible integral, seeing as the past few integrals have been symmetric substitution type integrals.

$\int{\frac{-\sin{2x}}{\sqrt{1+\cos^4{x}}}}dx$

Physicklad · Oct 1, 2015

Re: MX2 2015 Integration Marathon

sub u=cos^2(x) and its done

kawaiipotato · Oct 1, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$Evaluation of $\displaystyle \int\frac{2x^3-1}{x^6+2x^3+9x^2+1}dx\;\; $x$$

I don't think anyone tried to answer this

$= \int{ \frac{2x^3 - 1}{x^6 + 2x^3 + 9x^2 + 1}} $ dx $$

$= \int \frac{2x^{3}-1}{(x^3 + 1)^2 + 9x^{2}} $dx$$

$= \int \frac{2x - \frac{1}{x^2}}{(x^2 + \frac{1}{x})^2 + 9} $dx$$

$= \frac{1}{3} \tan ^{-1}\left( \frac{x^2 + \frac{1}{x}}{3} \right ) + $ C $$

kawaiipotato · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Saw this question online

$$Show $ \int_0^1 \frac{ (-x\ln x)^n}{n!} dx = (n+1)^{-(n+1)}$

idk why it doesn't work . Show \int_0^1 (-xlnx)^n / (n!) dx = (n+1)^{-(n+1)}

InteGrand · Oct 1, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Saw this question online

$Show $ \int_0^1 \frac{(-x \ln x)^{n}}{n!} \text{ d}x$ = $ (n+1)^{-(n+1)}$

Fixed.

InteGrand · Oct 1, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Saw this question online

$Show $ \int_0^1 \frac{ (-xlnx)^n}{n!} $ dx $ = $ (n+1)^{-(n+1)}$

idk why it doesn't work . Show \int_0^1 (-xlnx)^n / (n!) dx = (n+1)^{-(n+1)}

I think you messed up the $'s.

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Saw this question online

$$Show $ \int_0^1 \frac{ (-x\ln x)^n}{n!} dx = (n+1)^{-(n+1)}$

idk why it doesn't work . Show \int_0^1 (-xlnx)^n / (n!) dx = (n+1)^{-(n+1)}

$$ Let $ I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^n}{n!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^n}{n!(n+1)}\right)_{0}^{1} = I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!(n+1)} $ (Differentiate both sides if you want to confirm it) $$

$\therefore (n+1)I=\int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^{n-1}}{(n-1)!(n+1)} \right)_{0}^{1} = (n+1)I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!(n+1)}$

$\therefore (n+1)^2 I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!}$

$$ Hence by following the pattern: $ (n+1)^n I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{0}}{(0)!} = \int_{0}^{1} (-x)^n = \frac{(-1)^n}{n+1}$

$\therefore I = \frac{(-1)^n}{(n+1)^{n+1}}$

Can someone please tell me where I went wrong with the (-1)^n

kawaiipotato · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$$ Let $ I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^n}{n!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^n}{n!(n+1)}\right)_{0}^{1} = I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!(n+1)} $ (Differentiate both sides if you want to confirm it) $$

$\therefore (n+1)I=\int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^{n-1}}{(n-1)!(n+1)} \right)_{0}^{1} = (n+1)I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!(n+1)}$

$\therefore (n+1)^2 I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!}$

$$ Hence by following the pattern: $ (n+1)^n I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{0}}{(0)!} = \int_{0}^{1} (-x)^n = \frac{(-1)^n}{n+1}$

$\therefore I = \frac{(-1)^n}{(n+1)^{n+1}}$

Can someone please tell me where I went wrong with the (-1)^n

Were you using by parts?

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Were you using by parts?

Yes. Its a different form of IBP, where instead of establishing u, u' and v, v', you just find a function and shape it in a way such that when it is differentiated, one term will provide the integral desired.

Drsoccerball · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Did you just assume $xlnx \rightarrow =0$ ? as x approaches 0

Paradoxica · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Did you just assume $xlnx \rightarrow =0$ ? as x approaches 0

Care to prove it? I'm not complaining, since that's the only way you can have a definite area

kawaiipotato · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Care to prove it? I'm not complaining, since that's the only way you can have a definite area

L'hopital's would show it approaches 0^- but that's not really a proof

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Care to prove it? I'm not complaining, since that's the only way you can have a definite area

You can prove it graphically or you can use L'hopital's rule. The reason why I just made it equal to 0 straight away was because I already knew that the limit went to 0, but under exam conditions I would of proved it via L'hopital's to make the working out more mathematically correct.

Drsoccerball · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
You can prove it graphically or you can use L'hopital's rule. The reason why I just made it equal to 0 straight away was because I already knew that the limit went to 0, but under exam conditions I would of proved it via L'hopital's to make the working out more mathematically correct.

How is it possible to even prove that using L'hoptial's rule? It involves differntiating top and bottom and subbing in the limit

$\frac{\frac{d}{dx}(xlnx)}{0}$

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
How is it possible to even prove that using L'hoptial's rule? It involves differntiating top and bottom and subbing in the limit

$\frac{\frac{d}{dx}(xlnx)}{0}$

$\lim_{x \rightarrow 0} x\ln{x} = \lim_{x \rightarrow 0} \frac{\ln{x}}{\frac{1}{x}}$

$$ Using L'Hopital's: $ \lim_{x \rightarrow 0} \frac{\ln{x}}{\frac{1}{x}} = \lim_{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}} = \lim_{x \rightarrow 0} -x = 0$

Paradoxica · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$$ Let $ I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^n}{n!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^n}{n!(n+1)}\right)_{0}^{1} = I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!(n+1)} $ (Differentiate both sides if you want to confirm it) $$

$\therefore (n+1)I=\int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-1}}{(n-1)!}$

$\left(\frac{-(-x)^{n+1} (\ln{x})^{n-1}}{(n-1)!(n+1)} \right)_{0}^{1} = (n+1)I - \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!(n+1)}$

$\therefore (n+1)^2 I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{n-2}}{(n-2)!}$

$$ Hence by following the pattern: $ (n+1)^n I = \int_{0}^{1} \frac{(-x)^n (\ln{x})^{0}}{(0)!} = \int_{0}^{1} (-x)^n = \frac{(-1)^n}{n+1}$

$\therefore I = \frac{(-1)^n}{(n+1)^{n+1}}$

Can someone please tell me where I went wrong with the (-1)^n

When you differentiate the LHS of that IBP, the negative from the outside is cancelled out by the negative from the x term. Therefore, the LHS should be a sum of terms. This one mistake correction will telescope throughout your entire working and your answer should be correct.

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
When you differentiate the LHS of that IBP, the negative from the outside is cancelled out by the negative from the x term. Therefore, the LHS should be a sum of terms. This one mistake correction will telescope throughout your entire working and your answer should be correct.

Ah yes, makes sense now. Thanks!

Paradoxica · Oct 1, 2015

Re: MX2 2015 Integration Marathon

$\int_{0}^{1}{\frac{dx}{(1+x^n)^2 \sqrt[n]{1+x^n}}}$

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