HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

leehuan · May 27, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$I = \int_0^1 \frac{x^4(1-x^2)^5}{(1+x^2)^{10}}$d$x$

$$Find $\frac{1}{I}$

$\frac{1}{I}=1260$ where $16I=\int_0^\frac{\pi}{4}{\sin^4{2\theta}\cos^5{2 \theta }\,d\theta}$ which can be evaluated in a variety of ways$$

Paradoxica · May 27, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\frac{1}{I}=1260$ where $16I=\int_0^\frac{\pi}{4}{\sin^4{2\theta}\cos^5{2 \theta }\,d\theta}$ which can be evaluated in a variety of ways$$

I suggest double angle and then immediately use complex exponential without hesitation. Time will be of the essence for 2016ers.

juantheron · Jun 3, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int_{0}^{\infty}\frac{\ln (x)}{x^2+9}dx$

$Evaluation of $\int_{-1}^{1}\cot^{-1}(\frac{1}{\sqrt{1-x^2}})\cot^{-1}(\frac{x}{\sqrt{1-(x^2)^{|x|}}})dx$

$Evaluation of $\int_{\frac{\pi}{4}}^{\frac{65\pi}{4}}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx$

Paradoxica · Jun 3, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int_{0}^{\infty}\frac{\ln (x)}{x^2+9}dx$

$Evaluation of $\int_{-1}^{1}\cot^{-1}(\frac{1}{\sqrt{1-x^2}})\cot^{-1}(\frac{x}{\sqrt{1-(x^2)^{|x|}}})dx$

$Evaluation of $\int_{\frac{\pi}{4}}^{\frac{65\pi}{4}}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx$

The first one is easily handled by a remapping (substitution) of x -> 9/x

The second integrand is odd.

The third integrand has period 2π. This means the integral is identically equal to 8 times the integral of the integrand over an arbitrary interval of length 2π.
Perform the following remappings: x -> x - π, x -> x - π/2, x -> x + π/2
Shifting the borders so that they are all identical, say, 0 to 2π, add the four variations of the integral together.
The four integrands simplify to 1. So we have four times the integral is eight times the integral from 0 to 2π.
Thus, the integral is 4π.

seanieg89 · Jun 4, 2016

Re: MX2 2016 Integration Marathon

Compute:

$I_n:=\int_{-\pi}^\pi \frac{\sin(nx)}{(2^x+1)\sin(x)}\, dx.$

Paradoxica · Jun 4, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
Compute:

$I_n:=\int_{-\pi}^\pi \frac{\sin(nx)}{(2^x+1)\sin(x)}\, dx.$

If n is odd: π

If n is even: 0

juantheron · Jun 6, 2016

Re: MX2 2016 Integration Marathon

But Paradoxica answer of my second Integral is $\neq 0$

Paradoxica · Jun 9, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
But Paradoxica answer of my second Integral is $\neq 0$

That's not possible. Unless you interpret the exponentiation the other way around (which I cannot, as you have clearly and explicitly wrapped it in brackets), the integral is zero.

Mathematica also returns zero.

Paradoxica · Jun 14, 2016

MX2 2016 Integration Marathon

$\lim_{n \to \infty}\int_0^1 \sqrt{1 + n^2 x^{2n-2}} $d$ x$

Hint: Geometry

nanoputian · Jun 30, 2016

Re: MX2 2016 Integration Marathon

Answer is 1

As n approaches infinity, 2n-2 approaches 0 much faster than n^2 goes to infinity. Therefore as n approaches infinity, n^2.x^(2n-2) approaches 0. Hence, the integral simply becomes 1.

Btw, this is my first time posting, how do you format the equations?

InteGrand · Jun 30, 2016

Re: MX2 2016 Integration Marathon

nanoputian said:
Btw, this is my first time posting, how do you format the equations?

This is done using LaTeX. You can find a Short Guide to LaTeX here, which outlines how to use it on this site: http://community.boredofstudies.org/14/mathematics-extension-2/234259/short-guide-latex.html .

InteGrand · Jul 1, 2016

Re: MX2 2016 Integration Marathon

NEW QUESTION

$\noindent Find $\int _0^1 x^2 \sqrt{1 -x^2}\text{ d}x$.$

jathu123 · Jul 1, 2016

Re: MX2 2016 Integration Marathon

latex code crashed half way through

I might have did the long way

leehuan · Jul 1, 2016

Re: MX2 2016 Integration Marathon

jathu123 said:
View attachment 33308 latex code crashed half way through I might have did the long way

Lol rip

But most certainly correct

jathu123 · Jul 1, 2016

Re: MX2 2016 Integration Marathon

$\int_{0}^{\frac{\pi}{2}}\frac{\mathrm{d} x }{1+\tan ^{\sqrt{2}}x}$

not sure if this question was already posted, but a fun integral to try. It's from the Putnam 1980 competition

InteGrand · Jul 1, 2016

Re: MX2 2016 Integration Marathon

$\noindent Also, for $a\in \mathbb{R}$, find $\int _0 ^\frac{\pi}{2} \frac{1}{1+\tan ^a x}\text{ d}x$ (determining for which real $a$ we have convergence).$

(Essentially same method as previous Q., except also need to determine for which a the integral converges.)

(I have a feeling Paradoxica asked this some time in the past, maybe on the 2015 Integration thread.)

Actually it's fairly easy to show it converges for all real a (there's no question at all in fact as the denominator is always greater than 1, so no chance of a singularity). Also discussing convergence (or improper integrals in the first place) isn't in the HSC syllabus, so you usually wouldn't need to worry too much about proving convergence in HSC 4U.

leehuan · Jul 1, 2016

Re: MX2 2016 Integration Marathon

Is it really that easy
$\begin{align*}I&=\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^a{x}}\\ &=\lim_{R \to \frac{\pi}{2}} \int_0^{R}\frac{\cos^ax}{\cos^ax+\sin^ax}dx\\ &= \lim_{R \to \frac{\pi}{2} } \int_0^{R} \frac{\sin^ax}{\sin^ax+\cos^ax}dx\\ \implies 2I &= \lim_{R \to \frac{\pi}{2}} \int_0^{R}dx\\ \iff I&=\frac{\pi}{4} \end{align*}$

Also what would you compare it with?
$$As $x \to \frac{\pi}{2}, \tan x \le ?$

InteGrand · Jul 1, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Is it really that easy
$\begin{align*}I&=\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^a{x}}\\ &=\lim_{R \to \frac{\pi}{2}} \int_0^{R}\frac{\cos^ax}{\cos^ax+\sin^ax}dx\\ &= \lim_{R \to \frac{\pi}{2} } \int_0^{R} \frac{\sin^ax}{\sin^ax+\cos^ax}dx\\ \implies 2I &= \lim_{R \to \frac{\pi}{2}} \int_0^{R}dx\\ \iff I&=\frac{\pi}{4} \end{align*}$

Also what would you compare it with?
$$As $x \to \frac{\pi}{2}, \tan x = o(?)$

Naturally inclined to use the "order-bound" comparison test my lecturer taught

No need to even take limits, because the integrand is a continuous function on the domain of integration (if we want we can define appropriate left- or right-hand values that make the function continuous there; there isn't any problem because the denominator will always be greater than 1, so there's no singularity).

And yes, answer is pi/4.

seanieg89 · Jul 1, 2016

Re: MX2 2016 Integration Marathon

nanoputian said:
Answer is 1

As n approaches infinity, 2n-2 approaches 0 much faster than n^2 goes to infinity. Therefore as n approaches infinity, n^2.x^(2n-2) approaches 0. Hence, the integral simply becomes 1.

Btw, this is my first time posting, how do you format the equations?

This is not correct. (The flaw in your logic is that when x=1 at the endpoint of the interval, n^2.x^(2n-2) does NOT tend to zero, in fact it gets very large!) This endpoint dragging your function up is enough to make the limiting value of the integral more than 1.

In fact, the limit is 2. A nonrigorous but convincing way of seeing this is observing that the arc length of a curve y=f(x) between 0 and 1 is given by

$\int_0^1 \sqrt{1+f'(x)^2}\, dx.$

(You can prove this by summing the lengths of the infinitesimal line segments $\sqrt{\delta x^2+\delta y^2}$ and using the definition of the Riemann integral.)

Then the integral in this question is just the arc length of y=x^n between x=0 and x=1.

For large n, this curve gets smaller away from x=1, and increases more sharply to reach the endpoint (1,1). Visually, for large n, it tends to the mirrored L-shape consisting of the line segment between (0,0) and (1,0) and the line segment between (1,0) and (1,1), each of which has length 1.

I.e it's arclength tends to 2.

This argument would be convincing to any mathematician because of how well behaved the curve y=x^n is, but in general you have to be quite careful when making statements about the limiting arc lengths of a sequence of curves that approximates a limit curve, so it is a worthwhile exercise to try to prove that the limit is 2 analytically. (I.e. using inequalities etc.)

InteGrand · Jul 1, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
This is not correct. (The flaw in your logic is that when x=1 at the endpoint of the interval, n^2.x^(2n-2) does NOT tend to zero, in fact it gets very large!) This endpoint dragging your function up is enough to make the limiting value of the integral more than 1.

In fact, the limit is 2. A nonrigorous but convincing way of seeing this is observing that the arc length of a curve y=f(x) between 0 and 1 is given by

$\int_0^1 \sqrt{1+f'(x)^2}\, dx.$

(You can prove this by summing the lengths of the infinitesimal line segments $\sqrt{\delta x^2+\delta y^2}$ and using the definition of the Riemann integral.)

Then the integral in this question is just the arc length of y=x^n between x=0 and x=1.

For large n, this curve gets smaller away from x=1, and increases more sharply to reach the endpoint (1,1). Visually, for large n, it tends to the mirrored L-shape consisting of the line segment between (0,0) and (1,0) and the line segment between (1,0) and (1,1), each of which has length 1.

I.e it's arclength tends to 2.

This argument would be convincing to any mathematician because of how well behaved the curve y=x^n is, but in general you have to be quite careful when making statements about the limiting arc lengths of a sequence of curves that approximates a limit curve, so it is a worthwhile exercise to try to prove that the limit is 2 analytically. (I.e. using inequalities etc.)

Are there any useful sufficient conditions that'd make the arc length of the limit curve equal the limit of the arc lengths?

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