∫(0,2)∫(y,4) sin(x)√xdxdy
what are you trying to do here?
evaluate the double integral
But I have to switch the order and i dont know how to do that.
is this like this?? \[\int\limits_{0}^{2}\int\limits_{y}^{4}\sin(x)\sqrt{x}dxdy\]
yes but the integral for dx goes from y^2 to 4 sorry
its equal to sin4 - 4cos4 i can show steps but it won't be till later today.
could you just tell me the integral then after the order has been switched!!
@EEmarr Sorry but I have trouble recalling the methodology used here, do you have any idea how to proceed? maybe then I can contribute something
I don't think EEmarr is online. But my teacher said this integral would be impossible if i dont switch the order of integration.
switching the order has to do with looking at where the graph lays and the limits that are already being used. you then have to interchange the x's and y's. i haven't worked these problems in over a year so it will take me a while to look over the material.
ok so the limits should change to \[\int\limits_{0}^{4}\int\limits_{0}^{\sqrt{x}}\] look at this video is explains it faster than i can. also on the first one it may be from 0 to 2 i can't tell without drawing a very detailed graph.
by first one, do you mean the integral for dx or dy?
I will check out the video, thanks!
so when you change the order from dxdy. it will then be dydx so i mean dy
Ok, by the way would i still be integrating sqrt(x)sinx
All right, I got the answer. Thank You very much!!!!!
No problem
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