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anonymous

  • 5 years ago

Evaluate: I will post problem

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{-1}^{3} (3x-5)^4\]

  2. anonymous
    • 5 years ago
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    I get to 1/24(3x-5)^8

  3. anonymous
    • 5 years ago
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    this is before I substitute

  4. anonymous
    • 5 years ago
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    1st I get u=3x-5 3dc= 1/3 du

  5. anonymous
    • 5 years ago
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    the dc is dx

  6. anonymous
    • 5 years ago
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    Where are you getting 1/24 and ^8?

  7. anonymous
    • 5 years ago
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    never mind the 1/24 and 1/8 I looked at something else as I was writing. I will redo this one first

  8. anonymous
    • 5 years ago
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    how about 1/15(3x-5)^5 does that look better? I really unsure about what I am doing.

  9. anonymous
    • 5 years ago
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    this answer is before I substitute

  10. anonymous
    • 5 years ago
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    Yeah, that looks right

  11. anonymous
    • 5 years ago
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    \[\int\limits (3x-5)^4 \, dx = -\frac{1}{15} (5-3 x)^5 \] \[\int\limits_{-1}^3 (3x-5)^4 \, dx= \frac{11264}{5} \] \[\left(-\frac{1}{15} (5-3 x)^5\text{/.}x\to 3\right)-\left(-\frac{1}{15} (5-3 x)^5\text{/.}x\to -1\right)=\frac{11264}{5} \]

  12. anonymous
    • 5 years ago
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    Good example to follow. (You miswrote the problem.

  13. anonymous
    • 5 years ago
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    why did you switch to 5-3x and not keep it as 3x-5

  14. anonymous
    • 5 years ago
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    typo

  15. anonymous
    • 5 years ago
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    where did I miswrite the problem? The only thing I left off was the dx at the end

  16. anonymous
    • 5 years ago
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    Not you Robto

  17. anonymous
    • 5 years ago
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    ok

  18. anonymous
    • 5 years ago
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    now I think I should have 1/15(3(3)-5)^4 - 1/15(3(-1)-5)^4 1/15(16) - 1/15(4096) is that correct so far??

  19. anonymous
    • 5 years ago
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    Processing chaguanas's statement.

  20. anonymous
    • 5 years ago
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    ?

  21. anonymous
    • 5 years ago
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    are you still there

  22. anonymous
    • 5 years ago
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    what's wrong?

  23. anonymous
    • 5 years ago
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    did you see the last post I made What I think I should have after substitution

  24. anonymous
    • 5 years ago
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    Your process is right. I don't have to check the minutia, that part is all arithmetic. Put you are inputting -1. The writing is very small on the original problem is the low end point 1 or -1?

  25. anonymous
    • 5 years ago
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    -1 I was just checking because the problem asks to express as a decimal, approximate to one decimal place. After I do the problem I get -272. I thought I might be doing it wrong.

  26. anonymous
    • 5 years ago
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    Approximate to one decimal place, tells you nothing of the answer. It is instructions on how to write your answer. Assuming that is the right answer, to one decimal place is -272.0

  27. anonymous
    • 5 years ago
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    ok I was just thinking the answer would be different

  28. anonymous
    • 5 years ago
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    mom: Way to go mom. No ambiguity in your problem statement. chaguanas: \[-\frac{1}{15} (5-3 x)^5 = \frac{1}{15} (3 x-5)^5 \] I am using the Mathematica program, version 8, to solve this and that on this web site. With Mathematica at my disposal I am not about to solve anything related to mathematics with pencil and paper. Mathematica has been in development for some years now. In the early years the developers made some decisions regarding the input language construction and what they would deliver for output forms (answers). One thing that "does not look right" is that their polynomial answers are printed with the exponents, in the exponential terms, ordered low to high, not high to low as presented on school chalk/white boards. At first that was annoying for me but soon one adjusts and excepts their formulations. Probably the tendency for some inexperienced math students is to conclude that because a polynomial as written "doesn't look right", it must be inherently wrong and conveys the wrong intent.

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