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virtus

  • 2 years ago

find the definite integral y = x(1-x) from -2 to 2

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  1. LagrangeSon678
    • 2 years ago
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    you can multiply the x into the parenthese and obtain:|dw:1324593176181:dw|, then its just routine

  2. LagrangeSon678
    • 2 years ago
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    integrate and evaluate at the limits

  3. virtus
    • 2 years ago
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    could you please further elaborate thankx

  4. myininaya
    • 2 years ago
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    \[\int\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C\] \[\int\limits_{a}^{b}f(x) dx=F(x)|_a^b=F(b)-F(a)\]

  5. LagrangeSon678
    • 2 years ago
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    man, i caught up with something else, or else i would have explained it to you. But the one above me, did a fairly good job

  6. virtus
    • 2 years ago
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    thank you @ lagrangeson678 ^.^

  7. virtus
    • 2 years ago
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    myininaya what did you get as your answer?

  8. myininaya
    • 2 years ago
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    i found the antiderivative now all you have to do is: (just plug in upper limit) then write minus (then plug in lower limit)

  9. virtus
    • 2 years ago
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    oh because i got a negative answer

  10. virtus
    • 2 years ago
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    and according to the solution booklet it is a positive answer i.e 17/3 square units

  11. LagrangeSon678
    • 2 years ago
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    |dw:1324594394344:dw|

  12. LagrangeSon678
    • 2 years ago
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    |dw:1324594437592:dw|

  13. virtus
    • 2 years ago
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    yes and i got -16/3 square units

  14. LagrangeSon678
    • 2 years ago
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    Chek out the graph of this fucntion:http://www.wolframalpha.com/input/?i=integral+y+%3D+x%281-x%29+%5B-2%2C2%5D

  15. LagrangeSon678
    • 2 years ago
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    Do you notice anything?

  16. virtus
    • 2 years ago
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    yes the area is below the x axis, but how come in the solution book to this question, their answer is 17/3?

  17. virtus
    • 2 years ago
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    OH I SEE!!!!!!!!!! THANKS HEAPS

  18. LagrangeSon678
    • 2 years ago
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    but why were you saying 17/3

  19. virtus
    • 2 years ago
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    oh thats what the solution book says

  20. LagrangeSon678
    • 2 years ago
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    The answer is negative

  21. myininaya
    • 2 years ago
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    \[\int\limits\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C \] \[\int\limits_{-2}^{2}(x-x^2) dx=[\frac{x^2}{2}-\frac{x^3}{3}]^2_{-2}=[\frac{2^2}{2}-\frac{2^3}{3}]-[\frac{(-2)^2}{2}-\frac{(-2)^3}{3}]\] \[=[2-\frac{8}{3}]-[2+\frac{8}{3}]=2-\frac{8}{3}-2-\frac{8}{3}=2-2-\frac{8}{3}-\frac{8}{3}=0-\frac{16}{3}\]

  22. virtus
    • 2 years ago
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    THANK YOU @ MYININAYA

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