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

clbaker · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Alternatively, try the self-inverse substitution $\,u=\frac{1-x}{1+x}$

SpiralFlex said:
Hint for others, exact value is a dead give away especially with zero at the lower limits.

$\int_{0}^{a}f(x)\;dx = \int_{0}^{a}f(a-x)\;dx$

Combine this with a clever substitution (trig).

Librah · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{e^{x^2}}$

Totally legit solution,
(e^x^2)/2x +C

Drsoccerball · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Librah said:
Totally legit solution,
(e^x^2)/2x +C

bestest solution

nightweaver066 · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Err okay, were students expected to produce a non-elementary result? Kinda defeats the purpose of an integration bee.

The question was in a unique team event Integration Bee paper, specially designed for one team haha.

Of those questions, some probably required uni maths.

seanieg89 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Here is a primitive for $e^{x^2}$ , defined on the whole real line:

$\sum_{k=0}^\infty \frac{x^{2k+1}}{k!(2k+1)}.$

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Evaluate $\int_{\frac{1}{t}}^{t} \frac{\tan^{-1}x}{x}dx$$

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Was the answer pi/4 ln(t^2)

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Good job

. Answer could also be expressed as $\frac{\pi}{2}\ln{|t|}$ .

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Good job . Answer could also be expressed as $\frac{\pi}{2}\ln{|t|}$ .

Was there a long process? mine took a lot of steps

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int_{\frac{1}{t}}^{t} \frac{\tan^{-1}x}{x}dx$, \forall t > 0$

We note that if $f(x)=\frac{\tan^{-1}x}{x},x\neq 0$ , then $f(-x)=\frac{\tan^{-1}(-x)}{-x}=\frac{-\tan^{-1}(x)}{-x}=\frac{\tan^{-1}(x)}{x}=f(x)$ . Hence the integrand is even, and our answer for positive t will be the same for negative t. Also note the inverse trig. identity $\tan ^{-1} x = \frac{\pi}{2} - \tan ^{-1} \frac{1}{x}\text{ }\forall x >0$

Now, let $I=\int_{\frac{1}{t}}^{t} \frac{\tan^{-1}x}{x}\text{ d}x$ . Let $u=\frac{1}{x}\Leftrightarrow x=\frac{1}{u}$ . Then $\mathrm{d}x = -\frac{1}{u^2}\text{ d}u$ . Then changing the limits appropriately, we get

$I = \int _t ^{\frac{1}{t}} \frac{\tan ^{-1} \frac{1}{u}}{\frac{1}{u}}\cdot \frac{-1}{u^2}\text{ d}u$

$= \int _{\frac{1}{t}} ^t \frac{\frac{\pi}{2} - \tan ^{-1}u}{u}\text{ d}u$

$=\int _{\frac{1}{t}} ^t \frac{\frac{\pi}{2}}{u}\text{ d}u - I$

$\Rightarrow 2I = \frac{\pi}{2}\left(\ln |t| - \ln \left| \frac{1}{t}\right | \right) = \pi \ln |t|$

$\Rightarrow I = \frac{\pi}{2}\ln |t| \text{ }\forall t\in \mathbb{R}:t \neq 0$ .

Edit: Wait, for negative t, the answer should be negative the answer for positive t, since we'd be integrating the area "the opposite way", right? In that case I think $I = \mathrm{sgn}(t)\cdot \frac{\pi}{2}\ln |t|,t\in \mathbb{R}:t\neq 0$ ?

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

Find $\int \sqrt{\frac{1+x}{1-x}} \text{ d}x$ .

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int_{\frac{1}{t}}^{t} \frac{\tan^{-1}x}{x}dx$$

$\frac{\pi}{4} $ ln(t^{2}) ? $ \\ $ I did it but it's very long. I think there's probably a different way. $ \\ $ Let I = \int_{\frac{1}{t}}^{t} {\frac{tan^{-1}(x)}{x}} dx \\ $ Let x = tan(\theta) $ \\ $ dx = sec^{2}(\theta)d \theta \\ $ Borders are now from tan^{-1}{\frac{1}{t}} $ $ to tan^{-1}{t} $ \\ $ I $ $ = \int_{tan^{-1}{\frac{1}{t}}}^{tan^{-1}{t}} \frac{\theta sec^{2}(\theta)d \theta}{tan(\theta)} \\ $ Using the fact that tan^{-1}(a) + tan^{-1}(\frac{1}{a}) = \frac{\pi}{2} \\ $ The integral can now be represented as: $ \\ \int_{{\frac{\pi}{2} - tan^{-1}(t)}}^{tan^{-1}{t}} \frac{\theta sec^{2}(\theta)d \theta}{tan(\theta)} \\ $ Let $ \theta = \frac{\pi}{2} - u \\ d \theta = -du \\ $ Then the integral is now $ \\ = \int_{{\frac{\pi}{2} - tan^{-1}(t)}}^{tan^{-1}{t}} \frac{(\frac{\pi}{2} - \theta) cosec^{2}(\theta)d \theta}{cot(\theta)} \\ Add it up with the other 'I' for '2I', you get \frac{\pi}{2} int_{\frac{\pi}{2} - tan^{-1}(t)}^{tan^{-1}(t)} \frac{1}{0.5sin(2\theta)} d\theta$
Screw it im just gonna post up a pic. www.imgur.com/7uAvxcA Was this method ok?

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
NEW QUESTION:

Find $\int \sqrt{\frac{1+x}{1-x}} \text{ d}x$ .

Multiply the integrand by $\sqrt{\frac{1+x}{1+x}}$ , then split the fraction into two components. This will get it into standard form, answer is $\frac{\pi+2}{2}$ .

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$\frac{\pi}{4} $ ln(t^{2}) ? $ \\ $ I did it but it's very long. I think there's probably a different way. $ \\ $ Let I = \int_{\frac{1}{t}}^{t} {\frac{tan^{-1}(x)}{x}} dx \\ $ Let x = tan(\theta) $ \\ $ dx = sec^{2}(\theta)d \theta \\ $ Borders are now from tan^{-1}{\frac{1}{t}} $ $ to tan^{-1}{t} $ \\ $ I $ $ = \int_{tan^{-1}{\frac{1}{t}}}^{tan^{-1}{t}} \frac{\theta sec^{2}(\theta)d \theta}{tan(\theta)} \\ $ Using the fact that tan^{-1}(a) + tan^{-1}(\frac{1}{a}) = \frac{\pi}{2} \\ $ The integral can now be represented as: $ \\ \int_{{\frac{\pi}{2} - tan^{-1}(t)}}^{tan^{-1}{t}} \frac{\theta sec^{2}(\theta)d \theta}{tan(\theta)} \\ $ Let $ \theta = \frac{\pi}{2} - u \\ d \theta = -du \\ $ Then the integral is now $ \\ = \int_{{\frac{\pi}{2} - tan^{-1}(t)}}^{tan^{-1}{t}} \frac{(\frac{\pi}{2} - \theta) cosec^{2}(\theta)d \theta}{cot(\theta)} \\ Add it up with the other 'I' for '2I', you get \frac{\pi}{2} int_{\frac{\pi}{2} - tan^{-1}(t)}^{tan^{-1}(t)} \frac{1}{0.5sin(2\theta)} d\theta$
Screw it im just gonna post up a pic. www.imgur.com/7uAvxcA Was this method ok?

I think the final answer needs to be multiplied by the sign of t.

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Multiply the integrand by $\sqrt{\frac{1+x}{1+x}}$ , then split the fraction into two components. This will get it into standard form, answer is $\frac{\pi+2}{2}$ .

Lol right method, except that the integral was an indefinite one.

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
NEW QUESTION:

Find $\int \sqrt{\frac{1+x}{1-x}} \text{ d}x$ .

$\int \frac{1+x}{\sqrt{1-x^{2}}} dx \\ \frac{-1}{2} \int {\frac{-2}{\sqrt{1-x^{2}} + \frac{-2x}{\sqrt{1-x^{2}}} \\ {\frac{-1}{2}} (-2sin^{-1}(x) + 2 \sqrt{1-x^{2}})$

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Lol right method, except that the integral was an indefinite one.

My bad, answer should then be $\sin^{-1}x-\sqrt{1-x^2}+C$

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

pls someone fix my eqn. im gonna cry with latex why is it invalid

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
I think the final answer needs to be multiplied by the sign of t.

What did u mean by sign of t?

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Evaluate $\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x$$

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