## xartaan Group Title Check my work on changing the order of integration? $\int\limits_{0}^{1} \int\limits_{0}^{x^3} e^{x}\sin(y)dydx$ one year ago one year ago

1. xartaan Group Title

$\int\limits_{0}^{1} \int\limits_{0}^{y^{1/3}} e^{x}\sin(y)dxdy$

2. xartaan Group Title

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 Group Title

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 Group Title

|dw:1362707277183:dw|

5. xartaan Group Title

Ok cool, that is what my sketch looked like as well. So I did it right?

6. zepdrix Group Title

$\huge \large \int\limits_{y=0}^1\quad\int\limits_{x=y^{1/3}}^1\quad e^x \;\sin y \;dx\;dy$ Hmm like this I think?

7. xartaan Group Title

Isn't x going from 0 to y^1/3 not from 1? I guess I am missing something

8. zepdrix Group Title

|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 Group Title

yep

10. zepdrix Group Title

We're switching the limits, so now our X will relate to the function.|dw:1362707796144:dw|

11. xartaan Group Title

oooooooh, I think I see! so x goes from the curve, to the line x=1

12. zepdrix Group Title

ya c:

13. xartaan Group Title

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 Group Title

Wait, smaller value would put it on the bottom.

15. zepdrix Group Title

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 Group Title

<3 Thank you so much for clearing this up for me!

17. zepdrix Group Title

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.