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[-/1 Points]DETAILSSCALCCC4 12.2.019.Calculate the double integral_3r sin(r + y)dA, R = [o5] * [o;]The value of integral is...

Question

[-/1 Points]DETAILSSCALCCC4 12.2.019.Calculate the double integral_3r sin(r + y)dA, R = [o5] * [o;]The value of integral is

[-/1 Points] DETAILS SCALCCC4 12.2.019. Calculate the double integral_ 3r sin(r + y)dA, R = [o5] * [o;] The value of integral is



Answers

$1-80$ Evaluate the integral.
$$\int_{1}^{3} r^{4} \ln r d r$$

Section 6.1 problem 19 We're dealing with a problem where we need to do integration by parts. So you've got calculus and by very nature, you got calculus, algebra, the fact that is a definite integral. You're gonna have arithmetic to in this problem. So in this case, we normally let you be the polynomial. The exception is typically, when you have a longer rhythm term, you're gonna let you be that log a rhythm term, because when you different change the law algorithm function, the law goes away and you have something that is easier to work with a rational function. Then we can let Devi equal r cubed de are So when we integrate that, um by integration will get V is equal to our to the fourth over four. So that tells us that let's work on the indefinite interval. First of to integrate our cube natural log of RDR is going to be UV. So art of the fourth natural log of our over four minus the integral of VD You so that's gonna be 1/4 and you're gonna have our to the fourth of r R D R said. This is our to the fourth natural log of our over four minus 1/4 the integral of our cubed de are So this is art of the fourth natural one of our over four and then minus 1/16 are to the fourth plus a constant of integration. So that is the indefinite integral. So that tells us that if I'm asked to evaluate this from 0 to 3, shoot me from 1 to 3. So the interval from 1 to 3 of our cube natural long of RDR is going to be equal to our to the fourth. The art of the fourth natural long of our over four minus 1/16 are to the fourth and this gets evaluated from X equal one two x equal three So now comes just the arithmetic part. So I like to sort of do it like this to make sure I don't make mistakes with my signs of the 1st 1 This is plugging in X equal three. You get three to the fourth and 81 natural log of 3/4 minus and you get 81/16. So that's what I get by substituting in the value of three when I substitute the value of one. Um, the natural log of one is zero. So you're gonna have zero minus 1/16 times one. And so if I look at what I have here, I've got, um 81/4 natural log of three, minus 81/16. That's 1/16. So that is what minus 80/16 minus 80/16 inches minus five. So this turns into 81/4 natural lago three minus five, and that is our final answer. So, again, the key here was integration by parts letting you be the law algorithm term. That made it very easy because it turned that second integration just into polynomial integration, which was much simper when we had before. And then it was just a matter of keeping track of the arithmetic to get to a final answer.

Okay, So to begin this into girl, I'm going to find the anti derivative of our with respect to Z. So this gives me integral 0 to 2 in a girl, 0 to 2 pi. So the anti derivative of our with respect to Z is gonna be our Z and then going to evaluate this from our toe one. So then here, we're gonna have deep data d r so continuing here. So if we evaluate this, we're going to get integral 0 to 2 integral 0 to 2 pi. Then we're gonna go ahead and plug in one, and then we're gonna plug in our and we're going to subtract that. So with that, we're going to get ar minus horror squared di fada, d r And then from here, we're going to now take the anti derivative of ar minus R squared with respect to data. So that means I'm just gonna basically have a data outside. So we have in there, girl 0 to 2 in a girl, 0 to 2 pi, and then also we're integrating this. So this will give us data ar minus r squared. And then we're going to evaluate this from 0 to 2 pi d r So now if we go ahead and plug into pie and then subtract what we get when we plug in zero and for data but zero just gonna be zero. So we're just gonna basically plug into pie so we have integral 0 to 2 two pi ar minus r squared d r So again here to pie is a constant so we can bring it outside two pi times the integral of ar minus R squared so 0 to 2 of or times d r So now if we just take the anti derivative of ar minus R squared, this will give us two pi times are squared over two minus R cubed over three. Evaluate this from 0 to 2. So if we continue here, we'll have to pie Times two squared over two minus two. Cute over three. So we plug in zero, we'll just end up with zero so we can just go ahead and simplify this. So this is two pi times for over two, which is two and then minus 8/3. And if we simplify, this will get a result of negative four high over three. That's gonna be our solution for this problem.

In this problem we wish to evaluate the following integral the integral of r e t R over to D R. This question is challenging understanding of integration techniques in particular challenge your understanding of how to apply the technique known as integration by parts. So remember that the integration by parts formula says that integral. You D V equals UV minus integral. VD you. Thus you if we can find U and DV for argument integral, we can decompose and use integration by parts. So we have that U S r d u S t R. Thus tv must be er over to D r V S two E V R over to the US by integration by parts. We have to R E. There are over two minus two integral easier or to D R as our integral. The exponential integral on the second. It simplifies as to our zero to minus 40. There are over two plus constant integration. See

For the given definite and they first when it continued it to the world for our they are now taking. This is the first function on. This is the second function we get Ellen R. And Terrace. Enough with the par four year minus until there's enough data on the other and I'm not okay, that's enough. Nothing apart for India born into the air so that in the quarter and then tow happened about five. My five minus one by part to not talk about 555 meeting is, um, in the area, so that will be a portal. Well, what a fight. Fight fight, Magnus. Until there's enough part of the power for my five, they This gives us then art. And after five by five, when I start off, our five like got a different from the limits. 123 So did it work? Bluntly. Trick. It's about five by five minus. Take the quantified by 25. It's minus and then one and wanted the par 55 minus one like 25. So when you saw this, we can't answer it. 43 my friend, Elementary minus to quantity. My quantified


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