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xartaan

  • 2 years ago

Check my work on changing the order of integration? \[\int\limits_{0}^{1} \int\limits_{0}^{x^3} e^{x}\sin(y)dydx\]

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  1. xartaan
    • 2 years ago
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    \[\int\limits_{0}^{1} \int\limits_{0}^{y^{1/3}} e^{x}\sin(y)dxdy\]

  2. xartaan
    • 2 years ago
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    I just drew a picture, and since I had y=x^3 I figured solving for x gives y^(1/3) .. Am I close, or way off?

  3. xartaan
    • 2 years ago
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    I guess it didn't copy well in the initial post, here is the original ntegral: \[\int\limits_{0}^{1} \int\limits_{0}^{x^3} e^{x}\sin(y)dydx\]

  4. zepdrix
    • 2 years ago
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    |dw:1362707277183:dw|

  5. xartaan
    • 2 years ago
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    Ok cool, that is what my sketch looked like as well. So I did it right?

  6. zepdrix
    • 2 years ago
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    \[\huge \large \int\limits_{y=0}^1\quad\int\limits_{x=y^{1/3}}^1\quad e^x \;\sin y \;dx\;dy\] Hmm like this I think?

  7. xartaan
    • 2 years ago
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    Isn't x going from 0 to y^1/3 not from 1? I guess I am missing something

  8. zepdrix
    • 2 years ago
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    |dw:1362707659927:dw|So this is what our original setup looks like. yes? Y is going from y=0 up to the function.

  9. xartaan
    • 2 years ago
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    yep

  10. zepdrix
    • 2 years ago
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    We're switching the limits, so now our X will relate to the function.|dw:1362707796144:dw|

  11. xartaan
    • 2 years ago
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    oooooooh, I think I see! so x goes from the curve, to the line x=1

  12. zepdrix
    • 2 years ago
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    ya c:

  13. xartaan
    • 2 years ago
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    Ok, last question then. Looking at this, how do I decide wich is at the top of the integral and which is on bottom? My guess is y^1/3 is on top because it is father left than the line x=1, so its smaller values?

  14. xartaan
    • 2 years ago
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    Wait, smaller value would put it on the bottom.

  15. zepdrix
    • 2 years ago
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    We always integrate \(\large \text{left to right}\) and \(\large \text{bottom to top}\). Which in turn will always be smallest to largest, like you said. :)

  16. xartaan
    • 2 years ago
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    <3 Thank you so much for clearing this up for me!

  17. zepdrix
    • 2 years ago
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    We can see from the picture, that x=y^(1/3) is SMALLER that x=1 in the given interval right? :) Yay glad I could help.

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