anonymous
  • anonymous
Check my work on changing the order of integration? \[\int\limits_{0}^{1} \int\limits_{0}^{x^3} e^{x}\sin(y)dydx\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{0}^{1} \int\limits_{0}^{y^{1/3}} e^{x}\sin(y)dxdy\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
I guess it didn't copy well in the initial post, here is the original ntegral: \[\int\limits_{0}^{1} \int\limits_{0}^{x^3} e^{x}\sin(y)dydx\]

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zepdrix
  • zepdrix
|dw:1362707277183:dw|
anonymous
  • anonymous
Ok cool, that is what my sketch looked like as well. So I did it right?
zepdrix
  • zepdrix
\[\huge \large \int\limits_{y=0}^1\quad\int\limits_{x=y^{1/3}}^1\quad e^x \;\sin y \;dx\;dy\] Hmm like this I think?
anonymous
  • anonymous
Isn't x going from 0 to y^1/3 not from 1? I guess I am missing something
zepdrix
  • zepdrix
|dw:1362707659927:dw|So this is what our original setup looks like. yes? Y is going from y=0 up to the function.
anonymous
  • anonymous
yep
zepdrix
  • zepdrix
We're switching the limits, so now our X will relate to the function.|dw:1362707796144:dw|
anonymous
  • anonymous
oooooooh, I think I see! so x goes from the curve, to the line x=1
zepdrix
  • zepdrix
ya c:
anonymous
  • anonymous
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?
anonymous
  • anonymous
Wait, smaller value would put it on the bottom.
zepdrix
  • zepdrix
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. :)
anonymous
  • anonymous
<3 Thank you so much for clearing this up for me!
zepdrix
  • zepdrix
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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