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Math Help - y = sqrt{1 + (9x/4)} dx...Arc Length

  1. #1
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    y = sqrt{1 + (9x/4)} dx...Arc Length

    Find arc length given
    y = x^(3/2) - 1 over [0,1].

    Let INT be the integral symbol.

    I was able to construct the integral but after plugging the
    limits x = 0 to x = 1, I get a different answer every time.

    Here is the integral:

    INT sqrt{1 + (9x/4)} dx from
    x = 0 to x = 1.

    The correct answer is 1.43971 according to the wolfram website.

    Can someone do this one for me step by step?
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    Re: y = sqrt{1 + (9x/4)} dx...Arc Length

    well the method in general is just

    $\ell = \displaystyle{\int_0^1}\sqrt{1+(f^\prime(x))^2}~dx $

    here $f(x)=x^{3/2}-1$

    $f^\prime(x)=\dfrac 3 2 x^{1/2}$

    $(f^\prime(x))^2= \dfrac 9 4 x$

    so we must find the integral

    $\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx$

    $u=\dfrac 9 4 x$

    $du=\dfrac 9 4 ~dx$

    $\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx\Rightarrow \dfrac 4 9 \displaystyle{\int_0^{9/4}}\sqrt{1+u}~du $

    $\ell = \dfrac 4 9 \dfrac 2 3 (1+u)^{3/2}|^{9/4}_0=\dfrac 8 {27} \left(\left(1+\dfrac 9 4\right)^{3/2} - 1\right)\approx 1.43971$
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    Re: y = sqrt{1 + (9x/4)} dx...Arc Length

    I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
    I have trouble knowing what to substitute for u in some of these problems.

    I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.
    Last edited by nycmath; April 29th 2014 at 09:07 PM.
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    Re: y = sqrt{1 + (9x/4)} dx...Arc Length

    Quote Originally Posted by nycmath View Post
    I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
    I have trouble knowing what to substitute for u in some of these problems.

    I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.
    You can use that substitution, in fact, it would be easier
    Thanks from nycmath
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    Re: y = sqrt{1 + (9x/4)} dx...Arc Length

    Quote Originally Posted by Prove It View Post
    You can use that substitution, in fact, it would be easier
    Can you show me by letting u be [1+(9x/4)]?
    I find indefinite integrals a lot easier to play with. It is so easy to make mistakes with definite integrals.
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    Re: y = sqrt{1 + (9x/4)} dx...Arc Length

    Quote Originally Posted by nycmath View Post
    I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
    I have trouble knowing what to substitute for u in some of these problems.

    I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.
    that should have worked just as well. I suspect you didn't modify the integration limits properly.
    Thanks from nycmath
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