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Math Help - 2011 MIT Integration Bee Qualifying Test

  1. #31
    Math Engineering Student
    Krizalid's Avatar
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    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.
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  2. #32
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    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
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  3. #33
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    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...



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

    Spoiler:


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



    ... 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!
    Last edited by mr fantastic; January 28th 2011 at 01:18 PM. Reason: Fixed a typo (as requested by Tom).
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  4. #34
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    Is this party over? Oh well, anyway... no.11



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

    Spoiler:


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



    ... 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!
    Follow Math Help Forum on Facebook and Google+

  5. #35
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    2) \displaystyle  \int 2 \ln x + \ln^2 x\, dx = \int \left( x \ln^2x \right)'\,dx
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  6. #36
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     \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
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  7. #37
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    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).
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  8. #38
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    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
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  9. #39
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    Quote Originally Posted by Drexel28 View Post
    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! 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}
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