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anonymous

  • one year ago

Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.

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  1. anonymous
    • one year ago
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    i've found the area bounded by the x-axis and f(x) its 3.771123

  2. anonymous
    • one year ago
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    would i have to integrate with respect to the y -axis on this one?

  3. IrishBoy123
    • one year ago
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    have you plotted this can you see the area enclosed by these curves?

  4. anonymous
    • one year ago
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    yep iv graphed it, i know what it looks like

  5. IrishBoy123
    • one year ago
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    you are looking for the area bounded by the curves so you need to know where they meet to establish the limits of integration

  6. IrishBoy123
    • one year ago
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    you could draw it using the Draw function here

  7. anonymous
    • one year ago
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    okay dont know how to draw function but the graphs seem to come together at x =1 and x=-1

  8. IrishBoy123
    • one year ago
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    so firstly integrate the functions within those limits and see what comes out

  9. blackstreet23
    • one year ago
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    It seems you need to use the method of slicing

  10. anonymous
    • one year ago
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    which function f(x)or g(x)

  11. blackstreet23
    • one year ago
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    this looks like a disk

  12. anonymous
    • one year ago
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    is their a formula to this method ?

  13. IrishBoy123
    • one year ago
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    @blackstreet23 are you deliberately trying to take this of course? this is a simple integration.

  14. blackstreet23
    • one year ago
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    no i just want to help

  15. welshfella
    • one year ago
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    |dw:1438814403525:dw|

  16. blackstreet23
    • one year ago
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    ohh yeah i am sorry it is not revolving haha

  17. welshfella
    • one year ago
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    intersections at (-1'1) and (1,1)

  18. anonymous
    • one year ago
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    yes okay i got that now what ....?

  19. blackstreet23
    • one year ago
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    \[\int\limits_{?}^{?}( 2-x^2)-(x^4)\]

  20. blackstreet23
    • one year ago
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    if i am right this is because it is suppose the areas under the curve

  21. IrishBoy123
    • one year ago
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    |dw:1438814688554:dw| "just" integrate

  22. anonymous
    • one year ago
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    okay so for the boundris it would just be -1 to 1 right ?

  23. blackstreet23
    • one year ago
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    if you see on the drawing that one of the users drew the upper function is 2-x^2an

  24. blackstreet23
    • one year ago
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    and the lower function is X^4

  25. anonymous
    • one year ago
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    yay okay that gave me 44/15

  26. IrishBoy123
    • one year ago
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    yes, that is right limits are \(-1 \le x \le +1\)

  27. welshfella
    • one year ago
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    integrate 2 - x^2 between the limits 0 and 1 integrate x^4 between limits 0 and 1 subtract the second result from the first then as the area is symmetrical about the y-axis multiply this result by 2

  28. anonymous
    • one year ago
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    why do i have to multiply by 2 ?

  29. welshfella
    • one year ago
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    because i integrate between 0 and 1 but its more direct to use limits -1 and 1

  30. blackstreet23
    • one year ago
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    but why should it be integrated separately if the bounded region is above the x-axis?

  31. IrishBoy123
    • one year ago
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    he changed the limits from -1, 1 to 0,1 as they are symmetrical, then x2 : but you can just do -1, 1 and fly with it

  32. anonymous
    • one year ago
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    okay the answer is 44/15 either way

  33. welshfella
    • one year ago
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    yes - i dont know why i did it the longer way Irish Boy is right both methods will gave same answer

  34. anonymous
    • one year ago
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    okay thank you very much :D

  35. welshfella
    • one year ago
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    yw

  36. IrishBoy123
    • one year ago
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    \(\huge A = \int_{-1}^{1} 2 - x^2 - x^4 \ dx\) \( \huge = [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{-1}^{1}\) \(\huge = 2 [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{0}^{1}\) \(\huge = 2 [2 - \frac{1}{3} - \frac{1}{5} ] = 2(\frac{30 - 5 - 3}{15}) = 2( \frac{22}{15}) = \frac{44}{15}\) \(\large \checkmark\)

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