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Problem 12

Evaluate each iterated integral. (Many of these uā€¦

Problem 11

Evaluate each iterated integral. (Many of these use results from Exercises 1-10 ).
$$\int_{1}^{5} \int_{0}^{3}\left(x^{2} y+5 y\right) d x d y$$




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Video Transcript

Okay, so we're getting the double integral, um, and rule from 1 to 2. Um, and then another area or from 0 to 5 of the function Extra fourth times. Y plus why dx dy y eso? This is an integrated inner iterated integral eso We're gonna be working from the inside out. So let's first evaluate the inside and so 0 to 5 x the fourth times. Why? Plus why dx so first. What I'm gonna do is I'm gonna back out this. Why DX And then, since we're integrating respect to X because of this DX here, um, just move the Y to the outside and treat it as a constant because we're we're integrating with respect to X this time. Okay, then we just want to integrate this using the reverse power rule. We just add one to the expert and then divide by that new exponents. Same for one. Basically, we just want toe Think about like what function went derived should give you one. So when you derive X, you should be getting a one because there's implicit power of one here. So you would bring this one down and then subtract by one so would be one x to the zero and then X to deserve power is just one set this with equal one. So we know that this X is the correct anti derivative here. And then we want to integrate this from certified, um, now, just playing it in, Uh, we have five to the fifth over five. So we're playing in Pfeiffer's into X because we're integrating respect to X and then subtract when we plug in zero. So this should just be zero, um, five to the fifth over five. Should just be fights of fourth plus five on this should just be zero. So we're just gonna take that out and then fight to the fourth is 6 25 So we have plus five there on this result in 6 30 Why now? This isn't our final answer. I just calculate the inside integral. So now we got a basically plus this result into the outside, integral. So 1 to 2 of this. So they grow from 1 to 2 of 6 30 Why d y now? This is really simple. Integral. Um, you just want to use the reverse power rule again. So 6 30 Why squared over two evaluate from 1 to 2, but that, um, we just plug in to first. So 6 30 times four because two squared is for but like to then we just want plug in one, but remarked to subtract, um, plugging in one, we get 6 30 times one divided by two. This should result in 6 30 times. Three. Bye bye to because we have 46 thirties and then 16 30. So four minus one is just three. I'm just trying, like, 6 30 as if it were a variable here. And then remind to you too. Bye bye. To because of the common denominator, um, plugging into your calculator, you should be able to get that This should be 9 45 and that is our final answer.

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