2011 MIT Integration Bee Qualifying Test

Printable View

Show 40 post(s) from this thread on one page

January 27th 2011, 05:48 AM
Krizalid

haha, i saw that video once, it's a fairly easy integral, so i consider they are trying to prove your intelligence and speed altogether.
January 27th 2011, 08:28 AM
FernandoRevilla

Quote:

Originally Posted by TheCoffeeMachine

It took the winner about 30 seconds to do $\int \frac{x^4}{1-x^2}\;{dx}$ . What do you think?

Some books (for example Demidovich) include the following formula in the Table for immediate integrals:

$\displaystyle\int \dfrac{dx}{a^2-x^2}=\dfrac{1}{2a}\ln \left |{\dfrac{a+x}{a-x}}\right |+C\quad (a\neq 0)$

In that case, and taking into account that we needn't an elegant development, and that the euclidean division is almost immediate, the integral can be computed in less than 20 seconds. :)

Fernando Revilla
January 27th 2011, 11:42 AM
tom@ballooncalculus

Nice case of the 'addition/subtraction' method in no. 14.

$\displaystyle{\ \ \frac{1}{x^2(x^4 + 1)^{3/4}}}$

$\displaystyle{=\ \frac{(x^4 + 1) - x^4}{x^2(x^4 + 1)^{3/4}}$

$\displaystyle{=\ \frac{(x^4 + 1)^{3/4}(x^4 + 1)^{1/4} - x^4}{x^2(x^4 + 1)^{3/4}}$

$\displaystyle{=\ \frac{(x^4 + 1)^{1/4}}{x^2}\ -\ \frac{x^2}{(x^4 + 1)^{3/4}}$

As with no.s 2 and 16, start parts with one half of it and it's all over 'too' soon...

http://www.ballooncalculus.org/draw/parts/eight.png

... where (key in spoiler) ...

Spoiler:

http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
January 27th 2011, 12:28 PM
tom@ballooncalculus

Is this party over? Oh well, anyway... no.11

http://www.ballooncalculus.org/draw/parts/nine.png

... where (key in spoiler) ...

Spoiler:

http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
January 28th 2011, 04:14 AM
Danny

2) $\displaystyle \int 2 \ln x + \ln^2 x\, dx = \int \left( x \ln^2x \right)'\,dx$
January 28th 2011, 02:16 PM
carrot

$\int \sin[(n+1)x] \sin^{n-1}x {dx} = \int [\sin(nx)\cos x+\cos(nx)\sin x ] \sin^{n-1}x {dx}$
$= \frac{1}{n} \int \sin(nx) d\sin^n x+\int \cos(nx) \sin^n x {dx} =\frac{1}{n} \sin(nx)\sin^n x + c$
January 31st 2011, 01:48 AM
TheCoffeeMachine

I thought the following three integrals were very similar to those in the test:

1. $\int\frac{2x\cos{x}+\sin{x}}{4\sqrt{x}\sqrt{\sin{x }\sqrt{x}}}\;{dx}.$

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

3. $\int\frac{1}{x+x\log^2{x}}\;{dx}$ .

All three can be solved without explicit substitution (the last being the easiest).
January 31st 2011, 12:21 PM
Drexel28

Quote:

Originally Posted by TheCoffeeMachine

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

Is this an accident? It equals $\displaystyle \pi\int\frac{dx}{\sqrt{1-x^2}}=\pi\arcsin(x)+C$
February 1st 2011, 01:59 AM
TheCoffeeMachine

Quote:

Originally Posted by Drexel28

Is this an accident? It equals $\displaystyle \pi\int\frac{dx}{\sqrt{1-x^2}}=\pi\arcsin(x)+C$

Doh, I didn't notice that! (Doh) I was thinking more along the lines of:

$\begin{aligned} \displaystyle \int \frac{2\sin^{-1}{x}+2\cos^{-1}{x}}{\sqrt{1-x^2}}\;{dx} & = \int \frac{2\sin^{-1}{x}}{\sqrt{1-x^2}}+\frac{2\cos^{-1}{x}}{\sqrt{1-x^2}}\;{dx} \\& = \int 2\sin^{-1}{x}(\sin^{-1}{x})' -2\cos^{-1}{x}(\cos^{-1}{x})'\;{dx} \\& = \int \left[\left(\sin^{-1}{x}\right)^2\right]'-\left[\left(\cos^{-1}{x}\right)^2\right]' \;{dx} \\& = \left(\sin^{-1}{x}\right)^2-\left(\cos^{-1}{x}\right)^2+k. \end{aligned}$

Show 40 post(s) from this thread on one page