HSC 2017 MX2 Integration Marathon (archive) (5 Viewers)

wu345 · Mar 31, 2017

Re: HSC 2017 MX2 Integration Marathon

stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$

leehuan · Mar 31, 2017

Re: HSC 2017 MX2 Integration Marathon

wu345 said:
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$

The reverse quotient rule isn't obvious.

The denominator involves an odd power. Try making it an even power somehow.

Paradoxica · Mar 31, 2017

Re: HSC 2017 MX2 Integration Marathon

wu345 said:
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$

RPR

omegadot · Apr 1, 2017

Re: HSC 2017 MX2 Integration Marathon

wu345 said:
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$

$\noindent As Paradoxica says, the reverse product rule can be used. In particular the rule$

$\int e^{f(x)} [f'(x) g(x) + g'(x)] \, dx = g(x) e^{f(x)} + \mathcal{C},$

$\noindent is very useful to know. In the case for your integral $f(x) = -x$ and $g(x) = -\frac{1}{x^2}$ giving$

$\int \frac{(2 + x) e^{-x}}{x^3} \, dx = \int \left [-1 \cdot -\frac{1}{x^2} + \frac{2}{x^3} \right ] e^{-x} \, dx = -\frac{e^{-x}}{x^2} + \mathcal{C}.$

stupid_girl · Apr 15, 2017

Re: HSC 2017 MX2 Integration Marathon

$\text{Define }<f,g>=\int_0^1 f(s)g(s)ds\text{ and }<f,g>_w=\int_0^1 f(t)g(t)w(t)dt\text{. Find }\frac{<f,g>_w}{<f,g>}\text{ when}\\f(x)=x^{a-1},g(y)=(1-y)^{b-1}\text{ and }w(z)=(z+k)^{-a-b}\text{ for positive real constants a,b and k.}$

Paradoxica · Apr 15, 2017

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
$$\noindent Define $ f \circ g =\int_0^1 f(x)g(x) \, \text{d}x $ and $ f \stackrel{w}{\circ} g = \int_0^1 f(x)g(x)w(x) \, \text{d}x \\\\ $Find $\frac{f \stackrel{w}{\circ} g}{f \circ g}$ when:$ \\\\ f(x) = x^{a-1}, g(x) = (1-x)^{b-1} $ and $ w(x) = (x+k)^{-a-b} $ for positive real constants $a,b$ and $k.$

Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$$\noindent Evaluate $\displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.

Kingom · Apr 15, 2017

Re: HSC 2017 MX2 Integration Marathon

Paradoxica said:
Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$$\noindent Evaluate $\displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.

$Let\quad u=\frac { (k+1)x }{ x+k } \\ du=\frac { k(k+1)\quad dx }{ { (x+k) }^{ 2 } } \\ x=0,\quad u=0\\ x=1,\quad u=1\\ \frac { 1-x }{ x+k } =\frac { 1-u }{ k } \\ \frac { x }{ x+k } =\frac { u }{ k+1 } \\ \therefore \int _{ 0 }^{ 1 }{ \frac { { x }^{ a-1 }{ (1-x) }^{ b-1 } }{ { (x+k) }^{ a+b } } dx=\int _{ 0 }^{ 1 }{ \frac { { u }^{ a-1 }{ (1-u) }^{ b-1 } }{ { k }^{ b }{ (k+1) }^{ a } } du } } \\ \therefore \frac { { \int _{ 0 }^{ 1 }{ { x^{ a-1 }{ (1-x) }^{ b-1 } } } }dx }{ \int _{ 0 }^{ 1 }{ \frac { { x }^{ a-1 }{ (1-x) }^{ b-1 } }{ { (x+k) }^{ a+b } } dx } } ={ k }^{ b }{ (k+1) }^{ a }\\$

stupid_girl · Apr 17, 2017

Re: HSC 2017 MX2 Integration Marathon

Paradoxica said:
Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$$\noindent Evaluate $\displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.

It is purposefully obfuscated but I think average MX2 student should be able to interpret it. End of the day, <f,g>_w and <f,g> are just two real numbers. The answer would be the same if I call them star and moon, and ask for star/moon.

By the way, I think the condition a,b,k being positive cannot be dropped.

Paradoxica · Apr 17, 2017

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
It is purposefully obfuscated but I think average MX2 student should be able to interpret it. End of the day, <f,g>_w and <f,g> are just two real numbers. The answer would be the same if I call them star and moon, and ask for star/moon.

By the way, I think the condition a,b,k being positive cannot be dropped.

Wrong. k can be strictly less than -1, provided a and b are positive rationals with odd numerator.

Under the standard real definition of roots.

Paradoxica · Apr 17, 2017

Re: HSC 2017 MX2 Integration Marathon

$$\noindent a) $\int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 - 1}} \\\\ $b) $ \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 + 1}}$

Kingom · May 13, 2017

Re: HSC 2017 MX2 Integration Marathon

Paradoxica said:
$$\noindent a) $\int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 - 1}} \\\\ $b) $ \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 + 1}}$

$a) \quad Let\quad I=\int { \frac { dx }{ x\sqrt { { x }^{ 10 }\pm { x }^{ 5 }+1 } } } \\ Let\quad u={ x }^{ 5 }\\ du=5{ x }^{ 4 }dx\\ I=\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { u^{ 2 }\pm u-1 } } } \\ \quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }-\frac { 5 }{ 4 } } } }\quad$
$Let\quad I=\quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } } } \\ Let\quad w=u\pm \frac { 1 }{ 2 } \\ dw=du\\ I=\frac { 1 }{ 5 } \int { \frac { dw }{ (w\mp \frac { 1 }{ 2 } )\sqrt { { w }^{ 2 }+\frac { 3 }{ 4 } } } } \\ =\frac { 1 }{ 5 } (ln\left| w\mp \frac { 1 }{ 2 } \right| -ln\left| 2\sqrt { { w }^{ 2 }+\frac { 3 }{ 4 } } \pm w+\frac { 3 }{ 2 } \right| )+C\quad \\ =\frac { 1 }{ 5 } (ln\left| u \right| -ln\left| 2\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } \pm u\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ =\frac { 1 }{ 5 } (ln\left| { x }^{ 5 } \right| -ln\left| 2\sqrt { { ({ x }^{ 5 }\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } \pm { x }^{ 5 }\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ =\frac { 1 }{ 5 } (ln\left| { x }^{ 5 } \right| -ln\left| 2\sqrt { { x }^{ 10 }\pm { x }^{ 5 }+1 } \pm { x }^{ 5 }\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ \\$
$b)\quad Let\quad I=\int { \frac { dx }{ x\sqrt { { x }^{ 10 }\pm { x }^{ 5 }-1 } } } \\ Let\quad u={ x }^{ 5 }\\ du=5{ x }^{ 4 }dx\\ I=\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { u^{ 2 }\pm u-1 } } } \\ \quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }-\frac { 5 }{ 4 } } } } \\ Let\quad w=u\pm \frac { 1 }{ 2 } \\ dw=du\\ I=\frac { 1 }{ 5 } \int { \frac { dw }{ (w\mp \frac { 1 }{ 2 } )\sqrt { { w }^{ 2 }-\frac { 5 }{ 4 } } } } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { \frac { w }{ 2 } \mp \frac { 5 }{ 4 } }{ \sqrt { { w }^{ 2 }-\frac { 5 }{ 4 } } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { \frac { u }{ 2 } \pm \frac { 1 }{ 4 } \mp \frac { 5 }{ 4 } }{ \sqrt { { u }^{ 2 }\pm u-1 } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { u\pm 2 }{ 2\sqrt { { u }^{ 2 }\pm u-1 } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { { x }^{ 5 }\mp 2 }{ 2\sqrt { { x }^{ 10 }\pm { x }^{ 5 }-1 } } )+C }$
though in reality these questions are trivial and can easily be done by inspection

Paradoxica · Jun 21, 2017

Re: HSC 2017 MX2 Integration Marathon

If b>a>0 evaluate:

$\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x$

Kingom · Jun 21, 2017

Re: HSC 2017 MX2 Integration Marathon

Paradoxica said:
If b>a>0 evaluate:

$\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x$

Trivial by inspection

$$\noindent$ I=\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x \\\\ $Let $ x=\dfrac{1}{y} \\\\ \text{d}x = \dfrac{-\text{d}y}{y^2} \\\\ I=\int _b^a \dfrac{\tan^{-1}{\tfrac{a}{y}}-\cot^{-1}{\tfrac{b}{y}}}{\tfrac{1}{y}} \dfrac{-\text{d}y}{y^2} \\\\ I =\int_a^b \dfrac {\tan^{-1}{\tfrac{a}{y}}- \cot^{-1}{\tfrac{b}{y}}}{y} \text{d}y \\\\ I =\int_a^b \dfrac{\cot^{-1}{\tfrac{y}{a}} - \tan^{-1}{\tfrac{y}{b}}}{y} \text{d}y \\\\ $Let $ y=abx \\\\I =\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\cot^{-1}{bx} - \tan^{-1}{ax}}{abx} \text{d}x \\\\ \quad =-ab\int _{ \tfrac{1}{b}}^{ \tfrac{1}{a}} \frac{\tan^{-1}{ax}-\cot^{-1}{bx}}{x} \text{d}x \\\\ \quad I=-abI\\\\ \therefore \quad I=0 $ since $ ab > 0 > -1$
Note the potential tautology if ab=-1 cannot occur as b>a>0.

EngineeringHelp · Jun 24, 2017

Re: HSC 4U Integration Marathon 2017

Paradoxica said:
$$ \noindent$ \int \frac{\sqrt{\cos{2x}}}{\sin{x}} \text{d}x = \int \frac{\sqrt{2\cos^2{x}-1}}{1-\cos^2{x}} \sin{x} \text{d}x \\\\ \sqrt{2} \cos{x} = \sec{\theta} \Rightarrow -\sqrt{2} \sin{x} \text{d}x = \sec{\theta} \tan{\theta} \text{d}\theta \\\\ \sqrt{2}\int \frac{\sqrt{(\sqrt{2} \cos{x})^2 -1}}{(2\cos{x})^2 -2} (-\sqrt{2} \sin{x} \text{d}x) \\\\ = \sqrt{2} \int \frac{\sec{\theta} \tan^2{\theta}}{\sec^2{\theta} -2} \text{d}\theta = \sqrt{2} \int \frac{\sin^2{\theta} \cos{\theta} \, \, \text{d}(\theta)}{(1-\sin^2{\theta}) (2\sin^2{\theta}-1)} \\\\ \sin^2{\theta} \equiv (2\sin^2{\theta}-1) + (1-\sin^2{\theta}) \\\\ \Rightarrow \frac{\sin^2{\theta}}{(1-\sin^2{\theta})(2\sin^2{\theta}-1)} \equiv \sec^2{\theta} + \frac{1}{2\sin^2{\theta}-1} \\\\ \sqrt{2}\int \sec{\theta} \, \text{d}\theta - \sqrt{2} \int \frac{\text{d}(\sin{\theta})}{1-2\sin^2{\theta}}$

$$\noindent$ = \sqrt{2} \log{(\sec{\theta} + \tan{\theta})} + \frac{1}{2} \log{\left(\frac{1-\sqrt{2}\sin{\theta}}{1+\sqrt{2}\sin{\theta}} \right)} + \mathcal{C} \\\\ \sec{\theta} = \sqrt{2}\cos{x},\, \tan{\theta} = \sqrt{\cos{2x}}, \sin{\theta} = \frac{\sqrt{\cos{2x}}}{\sqrt{2} \cos{x}} \\\\ \sqrt{2} \log(\sqrt{2}\cos{x} + \sqrt{\cos{2x}}) + \frac{1}{2} \log{\left( \frac{\cos{x} - \sqrt{\cos{2x}}}{\cos{x} + \sqrt{\cos{2x}}} \right)} + \mathcal{C}$

Simplifying the derivative in Mathematica confirms the result.

Hey, sorry for being so bad at this subject, but how do you know to use that substitution? Thinking process? What are some key things to look out for when attempting a question?

Paradoxica · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

EngineeringHelp said:
Hey, sorry for being so bad at this subject, but how do you know to use that substitution? Thinking process? What are some key things to look out for when attempting a question?

Eliminate irrational functions using whatever you have at your disposal.

The only three ways of doing so are:

Classical Trigonometric Substitution (High School level)

Hyperbolic Trigonometric Substitution (University level)

Euler Substitution (Extracurricular)

EngineeringHelp · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

Paradoxica said:
Eliminate irrational functions using whatever you have at your disposal.

The only three ways of doing so are:

Classical Trigonometric Substitution (High School level)

Hyperbolic Trigonometric Substitution (University level)

Euler Substitution (Extracurricular)

Ooh, so for the HSC all i should worry about is a Trig Sub? As it wouldn't be right to be asked questions requiring the other two, right?

EngineeringHelp · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

Also, does anyone have any 'not so complicated' reduction-formula questions? If so, please post. Im assuming Paradoxica will be supplying solutions xD <3

pikachu975 · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

EngineeringHelp said:
Also, does anyone have any 'not so complicated' reduction-formula questions? If so, please post. Im assuming Paradoxica will be supplying solutions xD <3

http://prntscr.com/fnxmsv

Paradoxica · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

pikachu975 said:
$\int_0^1 \frac{x^n}{\sqrt{x+1}} \, \text{d}x$

...ew

si2136 · Jun 25, 2017

Re: HSC 4U Integration Marathon 2017

pikachu975 said:
http://prntscr.com/fnxmsv

Lmao use Mathematical Induction to prove? Stuff In

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