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Definite Integral Confusion

  1. Apr 20, 2006 #1
    Hi,

    So this might be overwhelmingly stupid... But the fundamental theorem of calculus states:
    [tex]\int_{a}^{b}f(x)dx=F(b)-F(a)[/tex]
    Where F is any antiderivative of f.

    So I have this very simple integral that I'm trying to solve...:

    [tex]2\pi\int_{0}^{2}x^3\sqrt{1+9x^4}dx\rightarrow \ u=1+9x^4[/tex]
    [tex]=\frac{\pi}{18}*\frac{2}{3}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{1}^{145}[/tex]

    For the answer, I'm given the following:
    [tex]\frac{\pi}{27}(145\sqrt{145}-1)[/tex]

    However, when I use the fundamental theorem of calculus, I get something nasty:
    [tex]\frac{\pi}{27}[(1+9(145)^4)^{3/2}-(1+9(1)^4)^{3/2}][/tex]
    [tex]\mbox{calculator gives}=\frac{2}{27}[(1989227813\sqrt{1989227813}-5\sqrt{5})*\pi*\sqrt{2}]}[/tex]

    I don't know why I keep getting the wrong thing. I'm obviously making some sort of stupid error. Any suggestions?
     
    Last edited: Apr 20, 2006
  2. jcsd
  3. Apr 20, 2006 #2

    0rthodontist

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    Plug the two expressions (yours and the calculator's) into your calculator to evaluate them as numbers.
     
  4. Apr 20, 2006 #3
    The long one is 2.91x10^13
    The shorter one with the sqrt(145) comes to 203.044...
     
  5. Apr 20, 2006 #4

    dav2008

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    Gold Member

    Why are you evaluating from 1 to 145? Those are the limits for u, not x.

    Either keep it in terms of u and evaluate from 1 to 145 or change it back to x (extra work) and evaluate from 0 to 2:

    [tex]=\frac{\pi}{27}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{0}^{2}[/tex]

    In that line it's pointless to change back to x since it's a definite integral. Just evaluate u from 1 to 145 and you're done.
     
    Last edited: Apr 20, 2006
  6. Apr 20, 2006 #5
    Ah.

    There's the confusion. ...For some reason I guess I thought I had to replace the u by x as with indefinite integrals.

    Thanks much for helping me realize my mistake.


    Thanks,
    Heather
     
  7. Apr 20, 2006 #6

    dav2008

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    Gold Member

    There's nothing stopping you from rewriting it in terms of x, but like I said you have to be careful to write the limits in terms of x. Plus, it's just easier to keep things in u.
     
  8. Apr 20, 2006 #7
    Here's another problem where I'm screwing up the substitution... I've got the limits straight, (thanks to you), but I don't know why I'm getting this wrong.

    [tex]2\pi\int_{4}^{9}\sqrt{x+\frac{1}{4}dx}\rightarrow \ u=x+\frac{1}{4} \\ = 2\pi\int_{17/4}^{37/4}\sqrt{u}du \\ =2\pi*\frac{2}{3}u^{3/2}\mid_{17/4}^{37/4}[/tex]

    Somewhere along the line, my integral is supposed to come out to this:
    [tex]2\pi*\frac{1}{6}(4x+1)^{3/2}[/tex]

    ...Incorrect substitution...??
    u=x+1/4
    du=dx

    Either way, my answer is off by 1/2 when I get down to it with what I've done.

    My answer:
    [tex]\frac{\pi}{3}(37\sqrt{37}-17\sqrt{17})[/tex]
    The book's answer:
    [tex]\frac{\pi}{6}(37\sqrt{37}-17\sqrt{17})[/tex]
     
  9. Apr 20, 2006 #8
    Nevermind. I figured this out. Quick multiplication happened as a result of me not moving the denominators out of the expression with the limits plugged in.

    With the correction, I get the correct pi/6 multiplier.

    I would however like to know how the book and how maple is getting the answer with the 1/6 multiplier... Can someone explain that?
    ...It's different from what I got.
     
  10. Apr 20, 2006 #9

    t!m

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    Is
    [tex]2\pi*\frac{1}{6}(4x+1)^{3/2}[/tex]
    what you got, or what your book/maple got?
     
  11. Apr 20, 2006 #10
    what the book/maple got for the integral w/out limits
     
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