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anonymous

  • one year ago

I can't believe sometimes that English is my first language. What does this question mean? "Here is a plot of f[x] = x^3 on [-2,2] How does the plot give away the value of Integrate[x^3, {x, -2, 2}] ?​" pictures coming ...

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  1. anonymous
    • one year ago
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    image attached

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  2. UsukiDoll
    • one year ago
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    ok... so we are given a function \[\LARGE f(x) = x^3\] graphed between the intervals of -2 to 2

  3. anonymous
    • one year ago
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  4. UsukiDoll
    • one year ago
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    and then we are asked how the plot of that graph gave away the value when we are integrating from -2 to 2

  5. UsukiDoll
    • one year ago
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    you wouldn't mind if we integrate first right? do you know anti-derivatives?

  6. ganeshie8
    • one year ago
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    Hint : if \(f(x)\) is an odd function, \[\int\limits_{-a}^a f(x)\,dx = 0\]

  7. anonymous
    • one year ago
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    I dont mind, but negative on the anti derivatives (I think)

  8. UsukiDoll
    • one year ago
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    when we are taking the antiderivative, we add one to the exponent and divide by the new exponent.

  9. UsukiDoll
    • one year ago
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    so since your exponent is 3. What's 3 +1 ?

  10. anonymous
    • one year ago
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    so do they mean, I am to look at the plot, and then work out the value of the integration equation?

  11. anonymous
    • one year ago
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    hmmm.. 5-1?

  12. anonymous
    • one year ago
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    j/k 4

  13. UsukiDoll
    • one year ago
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    |dw:1436955882543:dw| now evaluate f(b)-f(a) or f(2)-f(-2)

  14. anonymous
    • one year ago
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    when the question says.. the value of ... do they mean that the equation itself is a value, or that the equation comes out to some value?

  15. ganeshie8
    • one year ago
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    Notice that the graph is symmetric about origin, so f(x) is an odd function. You don't need to do any further work, the definite integral between a symmetric interval would be simply 0 because the positive area and negative area kill each other out.

  16. anonymous
    • one year ago
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    oh, so by looking at the image.. I should see that it is symmetrical about zero, and therefore the sum of it's parts will be a negative area + an equal positive area = zero.

  17. UsukiDoll
    • one year ago
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    yes it is an odd function, so we do have symmetry around the origin . When we graph x^3 only we should have just |dw:1436956009479:dw|

  18. UsukiDoll
    • one year ago
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    but keep in mind we are restricted from -2 to 2 as well

  19. ganeshie8
    • one year ago
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    Exactly!

  20. anonymous
    • one year ago
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    thank you both.. gotcha, I just wasn't sure why they would ask me about.. 'giving away' .. I was like.. "giving away what?"

  21. anonymous
    • one year ago
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    The author of this program uses so many strange metaphors to me

  22. UsukiDoll
    • one year ago
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    |dw:1436956101513:dw| well the result does come to 0 if we do evaluate after taking the anti-derivative.

  23. UsukiDoll
    • one year ago
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    so there is cancellation...

  24. ganeshie8
    • one year ago
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    Indeed, the graph does give away the fact the the given function is "odd" |dw:1436956075476:dw|

  25. UsukiDoll
    • one year ago
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    so by having an odd function, there's symmetry around the origin, the negative and positive cancel out and that leads to a 0 which is the same answer if we were able to evaluate that integral. By taking the anti-derivative and evaluating... we will have 0 (though I like to work backwards....easier)

  26. UsukiDoll
    • one year ago
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    maybe the cancellations gave out the answer to that integral ?

  27. ganeshie8
    • one year ago
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    You could also work it analytically : \(f(-x) = (-x)^3 = -x^3 = -f(x)\) \(f(-x) = -f(x)\) means \(f(x)\) is an odd function.

  28. anonymous
    • one year ago
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    gotcha, I haven't got to anti derivatives yet.. I think they're coming up.. and they just want me to look at the graph.. there's another identical one next..

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  29. UsukiDoll
    • one year ago
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    another odd function... thanks to symmetry on the origin

  30. ganeshie8
    • one year ago
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    same thing, just check if it is an odd one

  31. anonymous
    • one year ago
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    hmmm, they might want me to do it analytically too..

  32. UsukiDoll
    • one year ago
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    I don't think we need integration by parts on this one... just u-substitution.

  33. ganeshie8
    • one year ago
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    |dw:1436956504186:dw|

  34. anonymous
    • one year ago
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    Well there is that f[-x] == -f[x] symmetry going on.

  35. anonymous
    • one year ago
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    yeah, you must be telepathic

  36. UsukiDoll
    • one year ago
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    xD!

  37. anonymous
    • one year ago
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    thank you for your instant help guys.. I dont think it was 30 seconds and you had me covered.

  38. anonymous
    • one year ago
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    please someone approve usukiDoll for qualified helper status.

  39. UsukiDoll
    • one year ago
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    I was going to do my nails when I saw the QH Questions change color, so I was like someone just asked a question.

  40. UsukiDoll
    • one year ago
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    Good thing I didn't have wet polish otherwise I would have to do it over XD

  41. anonymous
    • one year ago
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    Do all odd functions center on origin? or is this only because x^3 is fairly basic. And by odd did you mean the expression had degree 3, or odd means any function of an odd degree?

  42. UsukiDoll
    • one year ago
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    if the function is odd then there is symmetry at the origin and yes we need an odd exponent number like \[\large f(x) = x^3,f(x)=x^5.f(x)=x^7\]

  43. UsukiDoll
    • one year ago
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    there's also a test too.. that determines if the function is odd or not

  44. UsukiDoll
    • one year ago
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    now if we have an even function, that means that there is symmetry on the y-axis the parabola \[\large f(x) =x^2\] is the most common example for an even function

  45. UsukiDoll
    • one year ago
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    for trigonometry , sinx cosx, tanx sinx and tanx are odd functions cosx is the only even function

  46. anonymous
    • one year ago
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    So then it is only functions then that exhibit the property of -f[x] + f[x] == 0 that are considered odd?

  47. ganeshie8
    • one year ago
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    For simple polynomials we do need odd degree, but that is not so important. It would be more simple to just stick to the definition : \(f(x)\) is an odd function if \(f(-x) = -f(x)\)

  48. UsukiDoll
    • one year ago
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    erm.. you have to plug in -x in your x and yes the definitions of odd and even functions help a lot.

  49. UsukiDoll
    • one year ago
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    So, take the function and plug in -x for x and simplify should our result become \[\large f(-x)= f(x) \] the function is even and are signs are the same If our signs are the opposite (for example minus signs become plus signs or plus signs become minus signs) then we have an odd function \[\large f(-x)=-f(x) \] There is a chance that our function is neither odd nor even

  50. ganeshie8
    • one year ago
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    for example below graph is odd too; there is no algebraic expression to express it though |dw:1436958097638:dw|