The iterated integral for Jo reversed is given bydydr with the order ofintegrationSelect one:Gl ev dxdySs e" dxdy No correct answerJl, e" dxdySSSS" e...

Evaluate the iterated integral by first changing the order of integration. $$\int_{0}^{2} \int_{x}^{2} 2 e^{y^{2}} d y d x$$

In this problem we wish to sketch a given region and reverse the order of integration. For integral 0 to 1 integral one to eat at the Y. F. DX. DY. This question challenges our understanding of multiple controls and how to evaluate them. Remember that evaluate the integral requires us to utilize a single variable integration techniques and steps, whoever were not evaluated integral, we're just reversing the order integration. And to do that we need to look at the region are to understand what the new limits will be. So the region R is given here. It's given as X between one and why between 10 and one above the curve of Mexico's Either Y. We can rewrite that Mexico either Y as Y equals X. Thus we see that we have in terms of why L. A. Next one and in terms of X. One to eat because we integrate from above the curve and then from over these X values. Thus we can identify our double integral as integral one to eat integral. Next one. F D I C S.

In this problem, you have to evaluate the created integral by first changing the order of integration. Now we have given that double integral from 0 to 1 and why one three eggs e to the power x Q dx dy via a force will draw the graph. So in the graph, this one here is why exists This one here is X axis on this region. Here is the region off integration. Now we'll write double integral from 0 to 1 and by 13 eggs into the power X cubed DX dy equals two different angles from 0 to 1 and from zero to eggs three x into the power exp you de vie de X now evaluating the limit will God 0 to 1 three x y into the power X cube from why's equals to zero toe wise equals toe eggs D x now signifying this we got intake the 0 to 13 X square into the power x q d X. Evaluating the limit into the Power X Cube From 0 to 1, we got the final answer as e minus Once that's the solution

So for this problem we're given this double integral and note that it doesn't give us the function. Just the bounds and the order of integration. And it's asking us to essentially switch the order of our bounds. So to do that let's sketch out what the base of this integration looks like. And so I know that on my inside into girl here my why is going from X. 21 So if I write that as an inequality, I know my Y. Is greater than or equal to X and is less than or equal to one. When I look at my outside into girl, my variable of integration is X. And the X is going from 0 to 1. So writing that as an inequality my ex is greater than or equal to zero and less than or equal to one. So as we sketch out what this base looks like, we know we're really only needing to go up to one in either the X or the Y. Direction. And we know that the relationship between X and Y is that Y is greater than or equal to X. So if we take this line here, that's the line Y equals X. And we know that why is also less than or equal to one. So it's below this line right here. And so we can see that the area that we're looking for is this triangle. This is where Y is bigger than X, but less than one. And our exes are running from 0 to 1. So our exes go all the way across. So as we switch our bounds of integration, we just want to make sure that we're covering this exact same area on our grass. So we know that we're still going to be evaluating the integral of that same function. Instead of dy dx we want dx d Y. So X is on the inside, we need to evaluate our integral with respect to X. First. So my ex is here that's left to right. We can see they start at zero and we'll go to the right until we hit this line. And that line is where X is equal to why. So are bound will go to why? So from X equals zero, two, X equals Y in that horizontal direction. And then we evaluate and we need just integer values for the outside are integral with respect to Y. And we need all of these values from when Y is zero all the way up until why is one. So those bounds go from 0 to 1. This is our new integral with those flipped bounds.

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