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jackkim4

  • 3 years ago

Find the interval [a,b] for which the value of the integral b s(2+x-x^2)dx a is maximum

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  1. jackkim4
    • 3 years ago
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    help me :'(

  2. UnkleRhaukus
    • 3 years ago
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    is this your question ?\[\int\limits_a^b(2+x-x^2)\text dx\]

  3. jackkim4
    • 3 years ago
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    yes!!! sry first time here....

  4. jackkim4
    • 3 years ago
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    is the answer [-1, 2] or [-1,1/2] ??

  5. helder_edwin
    • 3 years ago
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    as @UnkleRhaukus pointed out, you have maximize \[ \large F(a,b)=\int_a^b(2+x-x^2)\,dx \]

  6. helder_edwin
    • 3 years ago
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    which one gives u the greatest area?

  7. helder_edwin
    • 3 years ago
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    *value, sorry.

  8. jackkim4
    • 3 years ago
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    doesnt [-1, 1/2] wait no... it's [-1, 2]

  9. UnkleRhaukus
    • 3 years ago
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    what did you get when you integrated

  10. jackkim4
    • 3 years ago
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    27/6 ???

  11. helder_edwin
    • 3 years ago
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    which interval?

  12. jackkim4
    • 3 years ago
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    [-1,2]

  13. UnkleRhaukus
    • 3 years ago
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    \[\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b\]\[=2(b-a)+\frac{(b-a)^2}2-\frac{(b-a)^3}{3}\] right?

  14. jackkim4
    • 3 years ago
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    yes!

  15. UnkleRhaukus
    • 3 years ago
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    actually i think i made a mistake

  16. jackkim4
    • 3 years ago
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    it's separated right

  17. jackkim4
    • 3 years ago
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    [2b+b^2/2-b^3/3]-[2a+2a^2/2-a^3/3]

  18. UnkleRhaukus
    • 3 years ago
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    \[\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b\]\[\qquad\qquad=\left({2b}+\frac{b^2}2-\frac{b^3}3\right)-\left({2a}+\frac{a^2}2-\frac{a^3}3\right)\]yeahs

  19. UnkleRhaukus
    • 3 years ago
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    so when is this maximum

  20. jackkim4
    • 3 years ago
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    at a= -1 b= 2 ! right?

  21. sirm3d
    • 3 years ago
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    Maximize F(b) by derivative, minimize F(a) by derivative.

  22. UnkleRhaukus
    • 3 years ago
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    \[=b\left({2}+\frac{b}2-\frac{b^2}3\right)+a\left(-{2}-\frac{a}2+\frac{a^2}3\right)\]

  23. UnkleRhaukus
    • 3 years ago
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    \[=6b\left(12+3b-2b^2\right)+6a\left(2a^2-3a-12\right)\]

  24. sirm3d
    • 3 years ago
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    \[\large F \prime (b)=2+b-b^2=0\]\[\large \text{max at } b=2\]\[\large F \prime (a)=2+a-a^2=0\]\[\large \text{max at } a=-1\] interval: [-1, 2]

  25. sirm3d
    • 3 years ago
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    wow, my answer is among the choices given.

  26. jackkim4
    • 3 years ago
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    omg lol

  27. sirm3d
    • 3 years ago
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    typo mistake. it should be min at a = -1

  28. jackkim4
    • 3 years ago
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    yayyyyyy tytytyty! :')

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