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Math Help - Proofs: definite integrals

  1. #1
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    Proofs: definite integrals

    May I have a little help here please?

    a.) Prove that \int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).

    b.) Show that \int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).
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  2. #2
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    Quote Originally Posted by Hellbent View Post

    a.) Prove that \int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).
    You are going to need a substitution here me thinks. Maybe x = \tan \theta


    Quote Originally Posted by Hellbent View Post

    b.) Show that \int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).
    With this guy make some partial fractions i.e. x^2-5x-6 = (x-6)(x+1)
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    Quote Originally Posted by Hellbent View Post
    May I have a little help here please?

    a.) Prove that \int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).
    \int_0^2 x^2(x^2 + 4)^\frac{1}{2}

    \text {let u} = x^2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = (x^2 + 4)^\frac{1}{2}

    v = \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = 2x

    =\frac{2x^2}{3}(x^2 + 4)^\frac{3}{2} = \int \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,2x

    \text{let u} = 2x\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = \frac{2}{3}(x^2 + 4)^\frac{3}{2}

    v = \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 2

    = \frac{2x^2}{15}(x^2 + 4)^\frac{3}{2} - \frac{8x}{15}(x^2 + 4)^\frac{5}{2} \int \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\dot\,\,2

    \text{let u} = 2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = ....

    v = \frac{8}{15.7}(x^2 + 4)^\frac{7}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 0

    Gave up here. I believe I'm going up a blind alley.
    Last edited by Hellbent; October 28th 2010 at 06:33 PM.
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    Quote Originally Posted by pickslides View Post
    With this guy make some partial fractions i.e. x^2-5x-6 = (x-6)(x+1)
    \text{Show that} \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \ln\bigg(\frac{32}{11}\bigg)
    \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \frac{A}{(x - 6)} + \frac{B}{(x + 1)}

    Ax + A + Bx - 6B
    A + B = 0, A - 6b = 7
    A = -B\,\,\,\,\,(-B) - 6B = 7
    \therefore A = 1\,\,\,\,\,-7B = 7\,\,\,B = -1

    \frac{1}{(x - 6)} + \frac{1}{x + 1}
    <br />
\bigg[\ln|x - 6| - \ln|x + 1|\bigg]_7^{10} + C

    \ln |\frac{x - 6}{x + 1}| - \ln |\frac{x - 6}{x + 1}|
    \ln |\frac{4}{11}| - \ln |\frac{1}{8}|

    \ln \bigg(\frac{4}{11}\div{\frac{1}{8}}\bigg)

    \ln\bigg(\frac{4}{11} \times 8\bigg) = \ln \bigg(\frac{32}{11}\bigg)
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  5. #5
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    Quote Originally Posted by Hellbent View Post
    a.) Prove that \int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).
    Let x = 2\sinh{\varphi}, then dx = 2\cosh{\varphi}\;{d\varphi}. When x = 0, ~ \varphi = 0, and when [LaTeX ERROR: Convert failed] .
    Thus I = \int_{0}^{2} x^2\sqrt{x^2+4}\;{dx} = \int_{\varphi = 0}^{\text{arcsinh(1)}}(2\sinh{\varphi})^2\sqrt{(2\  sinh{\varphi})^2+4}(2\cosh{\varphi})\;{d\varphi}
    [LaTeX ERROR: Convert failed] (by by-parts).
     = 6\sqrt{2} - \ln(3 + 2\sqrt{2}) (by using the identity \text{arcsinh(t)} = \ln\left\{t + \sqrt{t^2 + 1}\right\} etc). Done!

    b.) Show that \int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).
    Since x^2-5x-6 = (x-6)(x+1), let t = \frac{x-6}{x+1}, then \frac{dt}{dx} = \frac{7}{(x+1)^2} ~ \Rightarrow dx = \frac{(x+1)^2}{7}dt.
    When x = 7, ~ t = \frac{1}{8}. When x = 10, t = \frac{4}{11}. Thus \int_{7}^{10}\frac{7}{(x-6)(x+1)}\;{dx} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{7(x+1)^2}{7  (x-6)(x+1)}\;{dt}
    = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{x+1}{x-6}\;{dt} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{1}{t}\;{dt} = [\ln{t}]_{\frac{1}{8}}^{\frac{4}{11}}  = \ln\left(\frac{4}{11}\right)-\ln\left(\frac{1}{8}\right) =  \ln\left(\frac{4}{11}\times\frac{8}{1}\right) = \ln\left(\frac{32}{11}\right).
    Last edited by TheCoffeeMachine; October 29th 2010 at 08:05 PM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Hellbent View Post
    \int_0^2 x^2(x^2 + 4)^\frac{1}{2}

    \text {let u} = x^2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = (x^2 + 4)^\frac{1}{2}

    v = \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = 2x

    =\frac{2x^2}{3}(x^2 + 4)^\frac{3}{2} = \int \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,2x

    \text{let u} = 2x\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = \frac{2}{3}(x^2 + 4)^\frac{3}{2}

    v = \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 2

    = \frac{2x^2}{15}(x^2 + 4)^\frac{3}{2} - \frac{8x}{15}(x^2 + 4)^\frac{5}{2} \int \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\dot\,\,2

    \text{let u} = 2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = ....

    v = \frac{8}{15.7}(x^2 + 4)^\frac{7}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 0

    Gave up here. I believe I'm going up a blind alley.
    pickslides suggested a trig substitution, the appropriate one is x = 2 \tan \theta, so do it that way. Or if you've learned about hyperbolic trig substitution, TheCoffeeMachine gave you the solution.
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    Quote Originally Posted by Jhevon View Post
    pickslides suggested a trig substitution, the appropriate one is x = 2 \tan \theta.
    Yes, I saw it. I can't do it his way. What's the reasoning behind giving me x = 2 \tan \theta instead of using the already given x = \tan \theta?
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    Quote Originally Posted by TheCoffeeMachine View Post
    Since x^2-5x-6 = (x-6)(x+1), let t = \frac{x-6}{x+1}, then \frac{dt}{dx} = \frac{7}{(x+1)^2} ~ \Rightarrow dx = \frac{(x+1)^2}{7}dt.
    When x = 7, ~ t = \frac{1}{8}. When x = 10, t = \frac{4}{11}. Thus \int_{7}^{10}\frac{7}{(x-6)(x+1)}\;{dx} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{7(x+1)^2}{7  (x-6)(x+1)}\;{dt}
    = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{x+1}{x-6}\;{dt} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{1}{t}\;{dt} = [\ln{t}]_{\frac{1}{8}}^{\frac{4}{11}}  = \ln\left(\frac{4}{11}\right)-\ln\left(\frac{1}{8}\right) =  \ln\left(\frac{4}{11}\times\frac{8}{1}\right) = \ln\left(\frac{32}{11}\right).
    Why did you substitute \frac{x-6}{x+1} for t?
    Neither do I get this part: \frac{dt}{dx} = \frac{7}{(x+1)^2}?!
    It seems like you pulled it out of thin air. Where did the 7 and (x + 1)^2 come from?

    Thanks.
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    Quote Originally Posted by Hellbent View Post
    Yes, I saw it. I can't do it his way. What's the reasoning behind giving me x = 2 \tan \theta instead of using the already given x = \tan \theta?
    Because the objective here is to get rid of the radical, you need to find a function that makes the
    expression under the radical a square. If you put x = \tan{t}, then the expression under the radical
    becomes (\tan{t})^2+4 = \tan^2{t}+4 , but x = 2\tan{t} gives 4\tan^2{t}+4 = 4(\tan^2{t}+1)
    which is equal to 2^2\sec^2{t} and \sqrt{2^2\sec^2{t}} = 2\sec{t}... it gets rid of the \sqrt (the other one doesn't).

    Quote Originally Posted by Hellbent View Post
    Why did you substitute \frac{x-6}{x+1} for t?
    It's just a substitution that works and makes an alternative to partial fractions. Whenever you have
    integral of the form \int\frac{k}{(ax+b)(cx+d)}\;{dx}, you can let t = \frac{ax+b}{cx+d} or t = \frac{cx+d}{ax+b} (doesn't matter which).

    Neither do I get this part: \frac{dt}{dx} = \frac{7}{(x+1)^2}?!
    It seems like you pulled it out of thin air. Where did the 7 and (x + 1)^2 come from?
    You have t = \frac{x-6}{x+1}. Differentiating it with respect to x using the quotient rule, we have:

    \displaystyle \frac{dt}{dx} = \frac{(x-6)'(x+1)-(x-6)(x+1)'}{(x+1)^2} = \frac{x+1-(x-6)}{(x+1)^2} = \frac{7}{(x+1)^2}.
    Last edited by TheCoffeeMachine; October 30th 2010 at 11:38 PM.
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  10. #10
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    Thank you, devourer of coffee
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