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Math Help - indefinite integral substitution

  1. #1
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    indefinite integral substitution

    Hi, I think this should be fairly basic for you guys(/girls?).

    By using the substitution (a) u=2x+7 and (b) u^2=2x+7 find the indefinite integral of x(2x+7)^1/2 dx. I'm fine with the process of finding (a) but there is no examples in the text book for finding (b), so I don't understand what it wants me to do? There is only a single answer in the back of the book, not an answer for (a) and (b). What the? And thank you in advance for your time.
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  2. #2
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    Well, a and b both give you the same answer. You're just using two techniques to solve for it.

    Honestly, though, i've never done a problem like this before. What does your answer say?

    \int nu^{n-1}du = \frac{u^{n+1}}{n+1}

    \int x\sqrt{2x+7}dx = 1/4 \int u^{3/2}du - 7/4 \int udu

    1/10u^{5/2} - 7/2u^2

    \frac{(2x+7)^{5/2}}{10} - \frac{7(2x+7)^2}{2}

    Wow, i'm kinda off with what I got. xp
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  3. #3
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    In the answer, the denominator in the second fraction is 6 and also in the second fraction (2x+7) is raised to the power of 3
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  4. #4
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    Wow, i'm really off. =p You mean, 3/2 for that last power, though, hopefully? Otherwise i'm completely lost.

    Ok, so let u = 2x + 7, now replace all your x values with u.

    \sqrt{2x+7} = \sqrt u ... x = \frac{u-7}{2} ... dx = 2du

    So that your new equation looks like this:

    1/2 \int \frac{u-7}{2} \sqrt u du

    Now go ahead and multiply everything out:

    1/2 \int \frac{u^{3/2}du - 7\sqrt udu}{2} = 1/4 \int (u^{3/2}du - 7\sqrt udu)

    So that's

    1/4 \int u^{3/2}du - 1/4 \int 7u^{1/2}du

    Now we simply raise the power and divide by that power.

    \frac{1}{4} \frac{2}{5}u^{5/2} - \frac{1}{4} \frac{2}{3}7u^{3/2} = \frac{u^{5/2}}{10} - \frac{7u^{3/2}}{6}
    Last edited by derfleurer; April 20th 2009 at 05:23 PM.
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  5. #5
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    Thanks, what you've got is correct. I'm just going to take the next few minutes to try and digest what you have done. This is new to me. Cheers
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  6. #6
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    Hey, it's new to me, too. So sorry if my work is vague. I wasn't sure if it was right so I wasn't about to explain it like it was. =p

    Let me know and i'll gladly expand on it.
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  7. #7
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    thanks derfleurer. I get it now. I had know idea I could rearrange u=2x+7 to get x and then sub that in to the original equation. I was that x that was stuffing me up. One thing, when you said u and u^2 would give me the same answer, I don't understand why? Will there some times be questions when I will need to recognise I need to sub u^2 instead of just u?
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  8. #8
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    Sometimes one substitution is easier than another, yes. But again, they still both work. So it's not like you have to recognize one over the other.

    Look at the original equation:

    x\sqrt {2x+7}

    With u = 2x + 7, our substituted formula was

    \frac {u - 7}{2} \sqrt udu

    However, with u^2 = 2x+7, we have to rearrange the formula according to new rules. So, if u^2 = 2x + 7, what is u? u is \sqrt {2x+7}.

    So in this case, we replace \sqrt {2x+7} with just u. We replace x with \frac {u^2 - 7}{2}. And our du = \frac {1}{\sqrt {2x+7}} (i'll let you figure out what dx is =p)
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  9. #9
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    is dx= udu?
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  10. #10
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    Yes. Just remember, your integral won't look quite the same until you replace your u and du.
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