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Problem 38 Easy Difficulty

Although it is often true that a double integral can be evaluated by using either $d x$ or $d y$ first, sometimes one choice over the other makes the work easier. Evaluate the double integrals in Exercises 37 and 38 in the easiest way possible.
$\iint_{R} 2 x^{3} e^{x^{2} y} d x d y ; \quad 0 \leq x \leq 1,0 \leq y \leq 1$




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Video Transcript

All right, so we want to evaluate this double Integral over our region are where the region are is defined by both of our variables X and y lying between zero and one. So access between zero and one and why is between zero and one onda function that we're integrating over this region is two x cubed e to the expert one. Now, the hint that they give us for this problem is if we're able to do so, it might benefit us to change the order of integration. We can't always change the order of integration, but in this case, it turns out that we are able to do that simply because the region that we're integrating over is a rectangle. We know that for integrating over a rectangle, weaken, integrate in any order we like. So, um, it's up to us whether or not we want to keep it. How how it is and integrate with respect to X first and then why or change it so that we're integrating with respect to y first and then X on the way that I would decide is I look at the Inter grand and I think about is that Instagram's simpler in terms of why or is it simpler in terms of X? So in terms of X, we have a cubic of x times an exponential within X squared in the exponents, so quite complicated. But in terms of why, why only appears once it appears in the exponents of that exponential? And that's it. So because it's simpler in terms of why I would choose to integrate with us back to why first. So let me write this down. I'm going to write the balance on the integral so our outermost integral is gonna be in terms of X. So the bounds or X equals zero and X equals one. And the innermost integral is in terms of why which also goes between your own one. Do I d. X. So now we want to evaluate this innermost integral with respect to why so the first thing I would like to do is factor or anything that doesn't have ah wide term in it. And I'm gonna leave off the X equals and why equals on the balance, because it should be clear. So I'm gonna factor out this two x cubed from the first or the Inter most integral. And so the innermost integral we're left with e to the X squared y do I the X. So now we would like to evaluate the innermost integral. And I think we could do that. The function that we're dealing with is eating the expert Why and again, with respect to why we're gonna treat any terms with X as just a constant. So this into grand is really e to some constant times. Why? And we know the anti derivative of that and that is just e to the X squared y divided by expert. We're evaluating that from zero toe one. I'm sorry. That should be why and after we evaluate that, we're gonna get E to the X squared over X squared minus you to the zero, which is one over X squared GXE and even mawr. We can distribute this two x cubed in and a next squared will cancel. So this becomes two X each of the X squared minus two x, the X and so now we're into grand is actually the sum of two functions in terms of X, the first being two x e to the X squared second being minus two X So we can write this as the sum of two intervals. So our first integral is just two x e to the X squared minus the integral of two X. And this second, integral is very straightforward. The anti derivative of two X is X squared. So this evaluation is X squared from 0 to 1, which is one squared minus zero squared, which is just one So that second intervals just one. And this first integral. Well, I don't know, the anti derivative off the top of my head s What? I don't know. The anti derivative. The first thing I would like to do is try and make a use substitution. So when I look for a u substitution, I look for a function of X within my into grand, and I also look for the derivative of that function, lying somewhere also in my instagram or some constant times that function. So here it seems natural. If we choose you to be x squared, the derivative is two x dx which lies in our inte grand right. So, making this substitution, we're gonna get e to the you do you minus one and here we have to be careful about our bounds because this integral is now in terms of you. And since you is equal to X squared and X is between zero and one, you is also gonna be between zero and one, and the anti derivative of each of you is to the U. So this becomes each of you evaluated from 0 to 1 minus one, which is each of the one minus each of the zero minus one, which is e minus two. And that's our answer.