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Math Help - finding the exact value of an integral which contains log and e

  1. #1
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    Exclamation finding the exact value of an integral which contains log and e

    Can someone please help me?
    find the exact value of:
    ln7

    ⌡ int e^(6x)(e^(6x) + 10)^7/2 dx
    ln5


    I have a brief idea of how to solve this, but I'm not too sure about the specific parts.
    You can make e^6x=u, differentiate this to find dx, but I'm not too sure about the new limits and how to solve the rest. Is there any easier way to solve this, other than the substitution method?

    any help would be great, thanks
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  2. #2
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    Perhaps with just a little more experience, you will see that this one is screaming for a substitution. Try u = e^(6x) + 10.
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  3. #3
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    Hi!

    I remember at school there used to be this formula that we blindly used as a tool whenever something like this popped up, then we learnt that it's in fact a form substitution method:

    .

    Can you know why? Just remember that

    And remember, by this so called change of variable method we're now integrating w.r.t. u and not x. So if and the limits are x = 0, 1, the limits of u are 1, 3.

    That plus TKHunny's hint, can you complete this?
    Last edited by mr fantastic; April 24th 2011 at 03:01 PM. Reason: Applied temporary fix for latex.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rebghb View Post
    Hi!

    I remember at school there used to be this formula that we blindly used as a tool whenever something like this popped up, then we learnt that it's in fact a form substitution method:
     \int u'(x) . u^n (x) dx = \int u^n du .

    ⌡ u'(x) . u^n(x) dx =

    ⌡ u^n du
    Can you know why? Just remeber that u' =  \frac{du}{dx} u' = du/dx.
    And remeber, by this so called change of variable method we're now integrating w.r.t. u and not x. So if u = 2x+1 and the limits are x=0, 1 x = 0, 1, the limits of u are 1, 3 1, 3.
    That plus TKHunny's hint, can you complete this?

    Editing Note: Sorry, there's a problem with the LaTeX compiler


    Let u = e^{6x} + 10. Your lower limit for x is ln(5), so plug x = ln(5) into the u formula to get the lower limit for u...

    For the rest of it, you can easily solve for e^{6x} from the u equation. So you can find the integrand in terms of u. Finally you can find du = ( ) dx. Solve for dx and plug that into the integral. You will get



    Can you take it from there?

    -Dan
    Last edited by topsquark; April 25th 2011 at 05:40 AM.
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    Quote Originally Posted by topsquark View Post

    You will get


    I get everything else, but would you please explain how you found that equation in terms of u? thanks
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    Forum Admin topsquark's Avatar
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    Let u = e^{6x} + 10. Then, in the integrand



    Now we need du in terms of dx.


    We need to put the e^{6x} in terms of u. But we know that e^{6x} = u - 10. So


    So


    Now sub dx into the original integral and the result follows.

    -Dan
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  7. #7
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    Hello all!
    Oh dear, topsquark, I don't know why you want to go there! Since e^{6x} + 10 isn't just any regular function, why not going as follows:
    u = e^{6x}+10 and noticing that 6e^{6x}=\frac{du}{dx}. Then:
    \displaysyle \int e^{6x} (e^{6x}+10)^{7/2}dx= \int \frac{du}{dx}u^{7/2}dx
    = \displaystyle \int u^{7/2} du = (2/7) u^{9/2} between your 2 modified limits.

    Other version:

    ⌡(1/6). u'(x) . u^n(x) dx =

    ⌡(1/6) u^(7/2) du = (1/6) / (9/2) u ^ (9/2) = (1/27) . (e^(6x) + 10) ^ (9/2)

    Somebody please do something with the LaTeX compiler!
    Last edited by rebghb; April 25th 2011 at 07:24 AM.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rebghb View Post
    Hello all!
    Oh dear, topsquark, I don't know why you want to go there! Since e^{6x} + 10 isn't just any regular function, why not going as follows:
    u = e^{6x}+10 and noticing that 6e^{6x}=\frac{du}{dx}. Then:
    \displaysyle \int e^{6x} (e^{6x}+10)^{7/2}dx= \int \frac{du}{dx}u^{7/2}dx
    = \displaystyle \int u^{7/2} du = (2/7) u^{9/2} between your 2 modified limits.
    (shrugs) It's not like it's that hard. And it's a simple and direct application of the substitution method. But you like your way and I like mine. No problems.

    Quote Originally Posted by rebghb View Post
    Somebody please do something with the LaTeX compiler!
    See post 8 in this thread.

    -Dan
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