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anonymous

  • 5 years ago

Evaluating Indefinite Integrals

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  1. anonymous
    • 5 years ago
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    \[\int\limits 2(2x+4)^5dx\] u=2x+4

  2. anonymous
    • 5 years ago
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    du=2dx \[\int\limits_{}^{}u^{5}du\]

  3. anonymous
    • 5 years ago
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    He's doing well, let him continue. :P

  4. anonymous
    • 5 years ago
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    i got that far lol im stuck with the rest... I know the answer is 1/6(2x+4)^6+c

  5. anonymous
    • 5 years ago
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    Okay, so you've got that u = 2x+4. du = 2 * dx, right? Now, put it in terms of dx : dx = du/2.

  6. anonymous
    • 5 years ago
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    In your original integral, replace all (2x+4) with u, and all dx with du/2. The original 2 in front will cancel out, and you'll be left integrating u^5 du. :) Integrate, and plug the value for u back into it.

  7. anonymous
    • 5 years ago
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    im not sure if i am supposed to distribute the first "2" or put it in front of the integral

  8. anonymous
    • 5 years ago
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    The two can stay inside or outside of the integral, but it would make life a lot easier to not distribute it through your polynomial. :P Do you understand the substitution?

  9. anonymous
    • 5 years ago
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    yep... i have \[\int\limits u^5 du \]

  10. anonymous
    • 5 years ago
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    There you go. :) Integrate now, and when you're finished, all you have to do is sub 2x+4 back inside for u.

  11. anonymous
    • 5 years ago
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    thats where i am having trouble lol i HATE integration... is it u^6/6 +c?

  12. anonymous
    • 5 years ago
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    Yep, that's it.

  13. anonymous
    • 5 years ago
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    wonderful! i got it now... plug 2x+4 in for x and solve! thank you so much!

  14. anonymous
    • 5 years ago
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    Power rule for integration: \[\int\limits \ \ x^n \ dx = \frac{x^{n+1}}{n+1}+c\]

  15. anonymous
    • 5 years ago
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    i wish this site had a friends feature lol youd be a lifesaver... literally (not like me)

  16. anonymous
    • 5 years ago
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    Lol, no problem.

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