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Argonx16

  • 3 years ago

Integration (See inside)

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  1. Argonx16
    • 3 years ago
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    Integrate: \[\int\limits_{1}^{3} \frac {1}{x+1} dx \] I'm brain farting hard right now. Probably only need like 2 steps worked out.

  2. Argonx16
    • 3 years ago
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    EDIT: I totally meant \[\int\limits_{1}^{3} \frac {x}{x+1}dx\]

  3. zepdrix
    • 3 years ago
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    It's a simple U-substitution with a little trick involved. It's very easy to get confused on a problem like this one. :)

  4. Dido525
    • 3 years ago
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    You have to do a little trick with u-substitution in this case.

  5. zepdrix
    • 3 years ago
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    Let \(\large u=x+1\), solve for x giving us \(\large x=u-1\). Then differentiating gives us, \(\large du=dx\).

  6. Dido525
    • 3 years ago
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    I guess I will let Zepdrix finish it off :P .

  7. zepdrix
    • 3 years ago
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    Understand what's going on argon? :O It's just that little piece \(\large x=u-1\) that is easy to forget about.

  8. Argonx16
    • 3 years ago
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    @Dido525 Thanks anyways c: @zepdrix Ok, let me go through it, one sec.

  9. Dido525
    • 3 years ago
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    If you want I can show you another way to do this question other than what Zepdrix is doing.

  10. Argonx16
    • 3 years ago
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    @zepdrix So I looked at my work, and I have tried u-substitution, and I did get the fact that du=dx (u=x+1, du=1 dx, du=dx), but how does that fit into the big picture? @Dido525 Thanks, but I'm fine for now :D

  11. zepdrix
    • 3 years ago
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    So what will happen is, you'll have the addition/subtraction on TOP instead of the bottom of the fraction when you perform this substitution. That's good news, cause we'll be able to break it up into a couple of fractions from there.

  12. zepdrix
    • 3 years ago
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    Before we had that messy term x+1 in the bottom, so we couldn't do much with it.. but with a u-1 in the top, we can split it up nicely.

  13. zepdrix
    • 3 years ago
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    \[\large \int\limits \frac{x}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{u-1}{u}du\]I think we get something like this yes? Which we can write as,\[\large \int\limits \frac{u}{u}-\frac{1}{u}du\]

  14. Dido525
    • 3 years ago
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    Long division would yield the same thing by the way. Just a note.

  15. zepdrix
    • 3 years ago
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    Ah good call :3

  16. Argonx16
    • 3 years ago
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    Wow... that's some clever manipulation! (I GET IT!) @Dido525 Now that you mention it, I think my teacher did say that long division was necessary on some of the questions.

  17. zepdrix
    • 3 years ago
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    Yah that makes more sense actually, instead of doing a substitution. Just add and subtract 1 from the top.\[\large \int\limits \frac{x+1-1}{x+1}dx \qquad \rightarrow \qquad \int\limits \frac{x+1}{x+1}-\frac{1}{x+1}dx\]

  18. Dido525
    • 3 years ago
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    Well it's not NECESSARY but it's DEFINITELY Easier :P .

  19. Argonx16
    • 3 years ago
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    Ok. Thanks so much guys! I was so puzzled as to why Wolfram Alpha had \[\int\limits_ {}^{} 1- \frac{1}{x+1}\]

  20. Dido525
    • 3 years ago
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    Long division again.

  21. Argonx16
    • 3 years ago
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    Yep!

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