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Math Help - Is this correct?

  1. #1
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    Is this correct?

    \int(2x+7)\sqrt{4x-5}
    \frac{1}{4}\int(2x+7)\sqrt{4x-5}du
    \frac{1}{4}\int\frac{2u+12}{4}u^{1/2}du
    \frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du
    \frac{1}{16}[\frac{4}{5}(4x-5)u^{5/2}+8(4x-5)^{3/2}]+c
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by gammaman View Post
    \int(2x+7)\sqrt{4x-5}
    \frac{1}{4}\int(2x+7)\sqrt{4x-5}du
    \frac{1}{4}\int\frac{2u+12}{4}u^{1/2}du
    \frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du
    \frac{1}{16}[\frac{4}{5}(4x-5)u^{5/2}+8(4x-5)^{3/2}]+c
    u=4x-5 \Rightarrow x=\frac{u+5}{4}

    So 2x+7=\frac{u+5}{2}+7=\frac{u+5+14}{2}=\frac{u+19}{  2} \neq \frac{2u+12}{4}
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  3. #3
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    ok but how did the u+5/4 become u+5/2? did the 4 reduce because of the 2x?
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  4. #4
    Moo
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    Quote Originally Posted by gammaman View Post
    ok but how did the u+5/4 become u+5/2? did the 4 reduce because of the 2x?
    Yes

    Otherwise, your method is correct, though :
    \frac{1}{16}[2(\frac{2}{5})u^{5/2}+12(\frac{2}{3})u^{3/2}]du
    this line could be skipped (it makes no sense to write this du)
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  5. #5
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    what do I do from here

    \frac{1}{4}[(\frac{u+19}{2})u^{1/2}]du
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  6. #6
    Moo
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    Quote Originally Posted by gammaman View Post
    what do I do from here

    \frac{1}{4}[(\frac{u+19}{2})u^{1/2}]du
    It's \frac 14\int \frac{u+19}{2} u^{1/2} ~du
    =\frac 14 \int \frac 12 u^{3/2} ~du+\frac 14 \int \frac{19}{2} u^{1/2} ~du

    just do what you did in the first time !
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  7. #7
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    Is this the answer?

    \frac{1}{20}[(4x-5)^{5/2}+\frac{19}{3}(4x-5)^{3/2}]+c
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  8. #8
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    Why don't you differentiate the result? That should yield the integrand.
    Last edited by mr fantastic; February 21st 2009 at 02:04 PM. Reason: No edit - just flagging the reply as having been moved from a duplicate thread
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