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### Evaluate each iterated integral. (Many of these u…

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Rutgers, The State University of New Jersey
Problem 19

## Discussion

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## Video Transcript

All right, So we have the integral 2 to 4 of the integral from 3 to 5 of our function x over y plus y over three dx dy y. So let's start off by evaluate the inside any role. So we have to go from 3 to 5 of X over y plus y over three DX. Um, okay, so because we're integrators, but two X first, you want to think of all of the labels as constants, namely, wise are all the constants in this case? So, basically, if I were to rewrite this thinking of them as constantly have won over, why as our constant Times X and then why over three another constant TX and then anti driving everything, Um, since one over wise are constantly just pulled that through. And then we integrate respect to X. So we have X squared over two plus, since whatever three is a constant, we just slop that fat Exxon there, um, which is just the reverse with power rule. If you try to drive this respect to, actually, we just get the wiper three. So it's the correct, um, anti derivative. And then we evaluate this from 3 to 5 since we're okay. So when I was first learning this, I got a little confused about, like, what Variable I should be plugging in these limits three and five and a simple trick where it's just like, what I do is I just look at the bearable right here that I'm integrated spent two. I'm like, Oh, okay, so I don't have to put these limits back into X, okay? And it gets a little tricky sometimes because you're anti decorative can get a little tricky. But if you just follow that like just looking back and seeing which variable years integrated respect to that's available you want to plug into. So we have one over. Why times we plug in five into X squared. So five squared over two and then playing into acts again. We have five. Why over three? We subtract this by the value or the value we get from playing in three. So we have went over. Why times 9/2. Because three squared is nine, um, plus three. Why, over three this should be equal to 25/2 y plus five. Why over three minus nine over to why and then minus again. Just why? Because three divided three is one and then some flying. This we get that, uh, we get 16. Over to. Why? Because 25 minus nine is 16. So 16 over to why? And then this should just be 2/3. Why? So this is the, uh or actually, let me justify a little bit more eight over. Why? Finest 2/3. Why? Uh, this is the value we get from the inside in general. And now we have to plug it back into the outside game rules. So look at the outside. Negro. We have and grow from 2 to 4 of a over. Why, my 2/3? Why de y This should be pretty simple. Um, so the first component, we just have eight Ellen, the absolute value of y. And then the second component should just be 2/3. Why squared? And then two by two by two, then evaluated from 24 Um, OK, so let's just supply a little bit first before we pull you in. It's going to take this to basically canceled with the top two. So we have widespread over 342 are 2 to 4 and then we plug in for first. So we have eight Ln absolute value of four minus 16/3 and then subtract this entire quantity by what we are what we get from 1,000,000,000 to. So we have eight Ellen of absolute value of two, um, minus 4/3. There should result in a Ln of four minus eight Ellen of 2 16 Thursday. I'm just grouping up similar numbers. So I just put the Ellen together so and that minus 16 3rd plus 4/3. And this should result in, um what can I do to the first component or the 1st 2 numbers is just factor out that eight city of Eleanor for my cell in two minus 12/3. This is equal Teoh by the rules of lager thumbs. When you have the same base right here of e implicitly and you're subtracting them, basically you can put the inside numbers as so. So when you're subtracting, this becomes the numerary. This becomes the denominator, and that they log on the outside remains the same. So we have Ln four divide by two and then minus four. And this is finally eight Ellen of two plus four. And that is our final answer