HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

leehuan · Dec 5, 2015

Re: MX2 2016 Integration Marathon

Inquiry: Shouldn't the a and b be in braces instead of parentheses? Or does it actually not matter.

InteGrand · Dec 5, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Inquiry: Shouldn't the a and b be in braces instead of parentheses? Or does it actually not matter.

There shouldn't be brackets of any sorts at all actually.

Also, the integral diverges.

Paradoxica · Dec 5, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
There shouldn't be brackets of any sorts at all actually.

Also, the integral diverges.

$a,b \in \mathbb R^+$

$$Evaluate $ \int_{-\infty}^\infty d \left ( \frac{1}{a^x + b^{\frac{1}{x}}} \right )$

$\noindent a) Determine all possible cases where the integral is convergent$

$\noindent b) Determine a formula that covers all the convergent cases. You may use any elementary functions you deem necessary. E.g. Floor function, Absolute values, Sign function$

Drsoccerball · Jan 1, 2016

Re: MX2 2016 Integration Marathon

After all the integration we did in the 2015 Integration Marathon an ln integration question was asked in question 11. I think this means that this year there will be a few integration questions so continue this forum

leehuan · Jan 1, 2016

Re: MX2 2016 Integration Marathon

Nice and easy one for the 2016ers

$\\ \int { \ln { \left( 1+{ x }^{ 2 } \right) } } dx$

KingOfActing · Jan 2, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Nice and easy one for the 2016ers

$\\ \int { \ln { \left( 1+{ x }^{ 2 } \right) } } dx$

$\begin{align*} \int \ln(1+x^2)\, dx &= x\ln(1+x^2) - \int\frac{2x^2}{1+x^2}\,dx\;\; \fbox{I.B.P.} \\&=x\ln(1+x^2) - \int\left(2 - \frac{2}{1+x^2}\right)\,dx \\&=x\ln(1+x^2) - 2x + 2\arctan(x) + C\\\\\text{NEXT QUESTION: } \int \frac{x^2+x+1}{x^2-x+1}\,dx\end{align*}$

Paradoxica · Jan 2, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\text{NEXT QUESTION: } \int \frac{x^2+x+1}{x^2-x+1}\,\text{d}x$

$$\noindent$ \int \frac{x^2+x+1}{x^2-x+1}\text{d}x = \int \frac{x^2-x+1+2x-1+1}{x^2-x+1}\text{d}x = \int 1 + \frac{2x-1}{x^2-x+1} + \frac{1}{x^2-x+1} \text{d}x\\=\int \text{d}x + \frac{\text{d}(x^2-x+1)}{x^2-x+1}+\frac{\text{d}\left(x-\frac{1}{2}\right )}{\left(x-\frac{1}{2}\right )^2+\left(\frac{\sqrt{3}}{2}\right )^2}\\=x+\ln(x^2-x+1)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right )+C$

$\noindent a)$ \int \frac{\frac{x^n}{n!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx \\$b) What is the limit as $n \rightarrow \infty$?$

KX · Jan 2, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$ \int \frac{x^2+x+1}{x^2-x+1}\text{d}x = \int \frac{x^2-x+1+2x-1+1}{x^2-x+1}\text{d}x = \int 1 + \frac{2x-1}{x^2-x+1} + \frac{1}{x^2-x+1} \text{d}x\\=\int \text{d}x + \frac{\text{d}(x^2-x+1)}{x^2-x+1}+\frac{\text{d}\left(x-\frac{1}{2}\right )}{\left(x-\frac{1}{2}\right )^2+\left(\frac{\sqrt{3}}{2}\right )^2}\\=x+\ln(x^2-x+1)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right )+C$

$\noindent a)$ \int \frac{\frac{x^n}{n!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx \\$b) What is the limit as $n \rightarrow \infty$?$

$a)I= \int \frac{\frac{x^n}{n!}\pm \sum_{r=0}^{n-1}\frac{x^r}{r!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx=\int 1-\frac{\sum_{r=0}^{n-1}\frac{x^r}{r!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx=x-ln(\sum_{r=0}^{n}\frac{x^r}{r!}) \ + \ C \\ b)$as $n \to \infty , I\rightarrow x-ln(e^x)+C=C$

Paradoxica · Jan 2, 2016

Re: MX2 2016 Integration Marathon

KX said:
$a)I= \int \frac{\frac{x^n}{n!}\pm \sum_{r=0}^{n-1}\frac{x^r}{r!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx=\int 1-\frac{\sum_{r=0}^{n-1}\frac{x^r}{r!}}{\sum_{r=0}^{n}\frac{x^r}{r!}}dx=x-ln(\sum_{r=0}^{n}\frac{x^r}{r!}) \\ b)$as $n \to \infty , I\rightarrow x-ln(e^x)=0$

$Incorrect. The final answer could be anything. You forgot the $C$.$

$$\noindent a) Simplify: $\cot^{-1}\left(\frac{ab+1}{a-b}\right )\\$b) Hence, evaulate: $\int \cot^{-1} (x^2 + x + 1)$

glittergal96 · Jan 2, 2016

Re: MX2 2016 Integration Marathon

tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))

So if A-B is in (-pi/2,pi/2), we have:

A-B=arctan((tan(A)-tan(B))/(1+tan(A)tan(B)))

Let A=arctan(a) and B=arctan(b) and this simplifies to

arctan(a)-arctan(b)=arctan((a-b)/(1+ab))=arccot((1+ab)/(a-b))

Note again that this ia only valid when |LHS| < pi/2.

You can find a general expression casewise for arccot((1+ab)/(a-b)) when this criteria is not satisfied if you like, but I am too lazy to do this right now and we don't need this for b).

b) Let a=x+1, b=x and use our identity, noting that 0 < arctan(x+1)-arctan(x) < 2*arctan(1/2) < pi/2 for all x.

This reduces our integrand to arctan(x+1)-arctan(x).

So the end answer is 1/(1+(x+1)^2)-1/(1+x^2)+const.

You can combine the fractions if you want, but the resulting expresion is not much nicer.

KX · Jan 2, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$Incorrect. The final answer could be anything. You forgot the $C$.$

$$\noindent a) Simplify: $\cot^{-1}\left(\frac{ab+1}{a-b}\right )\\$b) Hence, evaulate: $\int \cot^{-1} (x^2 + x + 1)$

Whoops, I made a silly mistake! Cheers for pointing that out

InteGrand · Jan 2, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))

So if A-B is in (-pi/2,pi/2), we have:

A-B=arctan((tan(A)-tan(B))/(1+tan(A)tan(B)))

Let A=arctan(a) and B=arctan(b) and this simplifies to

arctan(a)-arctan(b)=arctan((a-b)/(1+ab))=arccot((1+ab)/(a-b))

Note again that this ia only valid when |LHS| < pi/2.

You can find a general expression casewise for arccot((1+ab)/(a-b)) when this criteria is not satisfied if you like, but I am too lazy to do this right now and we don't need this for b).

b) Let a=x+1, b=x and use our identity, noting that 0 < arctan(x+1)-arctan(x) < 2*arctan(1/2) < pi/2 for all x.

This reduces our integrand to arctan(x+1)-arctan(x).

So the end answer is 1/(1+(x+1)^2)-1/(1+x^2)+const.

You can combine the fractions if you want, but the resulting expresion is not much nicer.

$\noindent I think you accidentally differentiated at the end instead of integrated? If we have inverse tan's in the integrand, we should use Integration By Parts to get the primitive.$

$\noindent \textbf{NEW QUESTION}$

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

glittergal96 · Jan 2, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent I think you accidentally differentiated at the end instead of integrated? If we have inverse tan's in the integrand, we should use Integration By Parts to get the primitive.$

$\noindent \textbf{NEW QUESTION}$

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

Oh lol, yeah. Got a bit complacent there

.

The primitive of arctan(x) is x*arctan(x)-log(1+x^2)/2 + C (a straightforward IBP) so use that instead.

Cheers!

omegadot · Jan 3, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION}$

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

$$\noindent Let $x = u^2$, so that $dx = 2u \, du.$ Thus\\\begin{align*}\int \sqrt{\frac{x}{1- x}} \, dx &= \int \frac{2u^2}{\sqrt{1 - u^2}} \, du\\&= -2 \int \frac{(1 - u^2) - 1}{\sqrt{1 - u^2}} \, du\\&= 2 \int \frac{du}{\sqrt{1 - u^2}} - 2 \int \sqrt{1 - u^2} \, du\\&= 2 \sin^{-1} u - 2 \int \sqrt{1 - u^2} \, du.\end{align*}$

$$\noindent The integral to the right can be found using IBP. On doing so we find:$

$\begin{align*}\int \frac{2 u^2}{\sqrt{1 - u^2}} \, du &= 2u \sqrt{1 - u^2} - \int \frac{2u^2}{\sqrt{1 - u^2}} \, du + 2 \sin^{-1} u\\\Rightarrow 2 \int \frac{2 u^2}{\sqrt{1 - u^2}} \, du &= 2u \sqrt{1 - u^2} + 2 \sin^{-1} u\\\Rightarrow \int \frac{2 u^2}{\sqrt{1 - u^2}} \, du &= u \sqrt{1 - u^2} + \sin^{-1} u, \end{align*}$

$$\noindent or$\\\begin{align*}\int \sqrt{\frac{x}{1 - x}} \, dx &= \sin^{-1} (\sqrt{x}) - \sqrt{x - x^2} + \cal{C}.\end{align*}$

$\noindent \textbf{New Question} -- Goodbye 2015$

$$\noindent Evaluate $\int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2014} \, dx.$

Drsoccerball · Jan 3, 2016

Re: MX2 2016 Integration Marathon

Alternatively

$$ Let $ u=x-\frac{1}{2}$

$I=\int{\sqrt{\frac{\frac{1}{2}+u}{\frac{1}{2}-u}}}du$

$I= \int{(\frac{\frac{1}{2}}{\sqrt{\frac{1}{4}-u^2}} + \frac{u}{\sqrt{\frac{1}{4}-u^2}})} du$

$This is easy to integrate.$

InteGrand · Jan 3, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Alternatively

$$ Let $ u=x-\frac{1}{2}$

$I=\int{\sqrt{\frac{\frac{1}{2}+u}{\frac{1}{2}-u}}}du$

$I= \int{(\frac{\frac{1}{2}}{\sqrt{\frac{1}{4}-u^2}} + \frac{u}{\sqrt{\frac{1}{4}-u^2}})} du$

$This is easy to integrate.$

Other method is to just rationalise the numerator straight away and then have a square root of quadratic in the denominator and a linear term in the numerator, which can easily be split up and integrated (bit tedious).

Drsoccerball · Jan 3, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Other method is to just rationalise the numerator straight away and then have a square root of quadratic in the denominator and a linear term in the numerator, which can easily be split up and integrated (bit tedious).

Yes, I think making the substitution makes it easier to see.

InteGrand · Jan 5, 2016

Re: MX2 2016 Integration Marathon

$\noindent Here's a pretty nice result that's easy to prove.$

$\underline{QUESTION}$

$\noindent Let $f$ be a function that is differentiable and invertible on $[a,b]$ and suppose $f(a)=a$ and $f(b)=b$ (\small{i.e. $a$ and $b$ are \textit{fixed points} of $f$}). Show that $\int _a ^b \left(f(x) + f^{-1} (x) \right)\text{ d}x = b^2 - a^2$.$

$\sim \sim \sim$

$\noindent (Knowing or deriving this result can make it possible to do certain definite integrals that we may not be able to find antiderivatives for.)$

Drsoccerball · Jan 5, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{New Question} -- Goodbye 2015$

$$\noindent Evaluate $\int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2014} \, dx.$

Im guessing

$$ Let $ u=x+1$

$I=\int_1^2{(\frac{2-u^2}{u})^{2014}}du$

$$ Using binomial theorem : $ I= \int_1^2(\sum_{k=0}^{2014}(2^k \cdot (-1)^k \cdot x^{2014-k} \cdot \binom{2014}{k}))dx$

$Then integrate each term and find a formula then sub in values?$

Paradoxica · Jan 5, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Im guessing

$$ Let $ u=x+1$

$I=\int_1^2{(\frac{2-u^2}{u})^{2014}}du$

$$ Using binomial theorem : $ I= \int_1^2(\sum_{k=0}^{2014}(2^k \cdot (-1)^k \cdot x^{2014-k} \cdot \binom{2014}{k}))dx$

$Then integrate each term and find a formula then sub in values?$

It can't be that messy... can it? My intuition tells me there is a very simple way to do it, and it is killing me because I don't see anything that can make it a three-liner.

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