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anonymous

  • 5 years ago

ok how to find the area ??

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  1. anonymous
    • 5 years ago
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  2. amistre64
    • 5 years ago
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    integrate :) the trick is, does your material say to keep (-) areas negative or not....

  3. anonymous
    • 5 years ago
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    ok i did that

  4. anonymous
    • 5 years ago
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    but i get confuse what points show i use

  5. amistre64
    • 5 years ago
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    \[F(x) = \int\limits_{0}^{3} (x^3 -4x) dx \rightarrow F(x)=\frac{x^4}{4} - \frac{4x^2}{2} |_{0}^{3}\]

  6. anonymous
    • 5 years ago
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    ok

  7. amistre64
    • 5 years ago
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    Area = F(3) - F(0) F(0) = 0 F(3) = 81/4 - 36

  8. amistre64
    • 5 years ago
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    22'3/4 - 36 = -13' 1/4..... so that should be the area

  9. amistre64
    • 5 years ago
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    it aint cause I forgot how to multiply lol

  10. amistre64
    • 5 years ago
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    (3)^4 ---- - 2(3)^2 4

  11. amistre64
    • 5 years ago
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    22' 3/4 - 18 = 4' 3/4...right?

  12. anonymous
    • 5 years ago
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    i have -135/4

  13. amistre64
    • 5 years ago
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    ack!!...someone get me a calculator lol

  14. anonymous
    • 5 years ago
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    lol

  15. amistre64
    • 5 years ago
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    2.25.... and thats my final offer :) How you coming up with -135/4?

  16. anonymous
    • 5 years ago
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    i mixed my #s lol but i have 9/4

  17. amistre64
    • 5 years ago
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    20' 1/4 ------ 4 | 81 -8 ----- 1 -0 --- 1 <- remainder

  18. amistre64
    • 5 years ago
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    9/4 is good :)

  19. anonymous
    • 5 years ago
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    ok:D

  20. amistre64
    • 5 years ago
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    20' 1/4 -18 -------- 2' 1/4 = 9/4

  21. anonymous
    • 5 years ago
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    sweet! :D

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