solve-the-given-problems-by-integration-find-the-first-quadrant-area-bounded-by-yfracln-4-x14-x1-a-2

Question

Solve the given problems by integration.
Find the first-quadrant area bounded by $y=\frac{\ln (4 x+1)}{4 x+1}$ and $x=5$.

Stephen Hobbs · Accepted Answer

given these three functions, the area bounded by all three. And the first quadrant is this shaded black area here, and we&#x27;ll go ahead and set up multiple into girls. First off this intersection, point has an X value of 0.5 and this intersection point as, ah x value of to. And of course, the two of those functions will be origin there. And so the first integral that will set up for the area under the current is from 0 to 0.5 of the top function, which is four x minus bottom function, which is X over four DX, and that will get us the area on the left side here from zero to 0.5. So all that shaded, uh, darker black area And then from here, we don&#x27;t want to add the area under the curve from 0.5 to to and the top function keer now is one over X. Subtract the bottom function, which is X over four, and these will be the wits de exes times, um, everything in parentheses there is the height of all of our rectangles. So there&#x27;s all that. That&#x27;ll add the exact amount of area in this second part and then, in the end, taking on 10 derivatives here, the they&#x27;re all just polynomial so that those steps are fairly straightforward. There we&#x27;ll end up with one point three eight six to nine. The reason why it&#x27;s a little funky is because when we anti drive one over X, we can end up a force with natural log and so that it&#x27;s gonna have some some, uh, irrational numbers. And so this is approximately our final answer for the total area on between all three of those curves.

Gregory Higby · Answer

eso. I just want to look at this graph just to see if it made sense. Thio show the work. Um, and I think it&#x27;s probably best to just look at a graphing calculator and be able to use it. And, uh, if you type these in, you can clearly see that the point of intersections at X equals one Well, at the 0.10 But exit goes, one is what we care about. And then exit goes forward, the upper bound that they give you. Um, but instead of going through the work and showing you the substitution, um, to get the area between the curve, I think it makes perfect sense just to go to a calculator and do the interval from one toe four of your upper function, which is this natural log of X. I&#x27;m just going to copy it down on, then subtract off your lower function again. I&#x27;m just gonna copy it down. Then just tell the reader that&#x27;s in terms of X. And you can definitely tell that this is not a perfect answer. Um, so no matter what, you&#x27;re gonna have to go to a calculator to get the approximate answer so the amount we just use a calculator, get to get it, um, and not waste any time. Um, uh, if if you wanted to, every teacher makes you these. I would do you substitution. Uh, you know where you let u equal a certain value and, you know, split up into two separate into girls, which is perfectly fine. But you still have to go to a calculator anyway. So the answer is 1.491 Using these bounds at X equals one and exit goes for mm.

Julian Wong · Answer

in this problem. We&#x27;re looking from the area bounded by y equals one plus rid X y equals two of a rude X Y equals X over four and the y axis. So I&#x27;ve done the graph, You&#x27;re ready and you&#x27;ll notice that I&#x27;ve actually put another line right down. Why equals one? That&#x27;s because, ah, you know, we&#x27;ve got multiple curves here. Normally we&#x27;re just used to dealing with too. But because we&#x27;ve got three of them, we&#x27;re gonna have to separate this into a couple of different regions. Uh, for the first region, we are going to take the area from 0 to 1 of one plus rejects and minus the red curve. And for the second region from 1 to 4, we&#x27;re going to take the area using the black curve minus the red curve. So we&#x27;ll have to add up the two areas so area is equal to the integral from zero toe once of the blue curve minus the red curves. So one plus rude x minus X over for the ex, plus, the second area is from 1 to 4. Black curve would be to overrule Dex minus X over for the ex. Luckily, for us these air not difficult expressions to take the anti derivative for so we can go straight to it. That will be X. Ah, that&#x27;s X to the power of 1/2. So we know we want extra power three over two and then taking into account the coefficient, that&#x27;ll be two over three X to the three over two. And that&#x27;s one over four X. So that&#x27;s gonna be minus one over eight X squared, evaluated from 0 to 1, and then we gotta add it to anti derivative for the next one. So that&#x27;s two times X to the negative 1/2 which makes it for X to the power of 1/2 minus won over eight X squared, evaluated from 1 to 4. So we&#x27;ve got a little bit to evaluate here, putting one into our first bracket that will be one plus two over three minus one over eight. I&#x27;m gonna skip putting zero in there because that old zero every time add it to the next one. Putting four in there, it&#x27;s gonna give us, um, eat minus two and then putting one in there will give us ah four minus one of rape S O. If we group the first racket together as one fraction, that will be 37 over 24. And if we group everything in the second fraction that that&#x27;s a being 17 over eight, giving them a common denominator and simplifying will actually give us the area of uh 11 over three.

Dillon Huddleston · Answer

um So for this problem, the first we will find the intersection of the well won&#x27;t for the So we can integrate with this back tax since we&#x27;re bounded on the left by the y axis Well integrated, starting from next zero. Now, to find the right expound For a first in trouble, we find the intersection of you to upper curves. Well, I equals one plus shrewd ex Michael too equal to two of the square root backs, so quit. He knows the multiplied by the square backs. Um, we have that, too. Equals on this river backs plus, uh, ex. So we have an equation. Zero is X. Let&#x27;s squared X minus two. I know this could be factored. Zero is squared X plus two squared X minus one s. So then the ribs of interest are rude. AGs you call to minus two and rude X equal to one. I get. Of course. Uh, this word of minus two is on route to eye. It&#x27;s imaginary is not riel, uh, to ignore that value. Some things you have exes discredit more positive screw, which is one. And we could verify that. Why, uh, with X equaled one is 1/4 uh, or for looking at the two upper occurs. Why is one less Route one is to the left hand? And why is to buy Rude Axe is to also over the right hand to verify that X equaled one would be the right boundary for the first in trouble, which is where the two opera curbs me. So the first integral we integrate. Ah, one place through Dax minus expired for So that will give us, um X plus 2/3. Excellent. Three hours. Um, finance for rude axe. And again, this is violated between X equals zero on what? Uh, now this. We had the jiggle, which will be evaluated from one point to be determined from one somebody to be determined. The interview. And in this case, um, will be a salon commodification. Here s 02 of the square root of ax with an anti derivative for rude axe on is an upper bound over them. Um, so into ground in the first thing to go zone, plus food, eggs, mine. It&#x27;s expired. Awful. Which is the lower curved much Hasn&#x27;t a goal, Auntie Derivative minus X squared by neat. And so then our second interval this time of the upper curve is to buy Food X and the lowers X by force. We do may. Too rude Axe two divided by rude act surges forward acts again. Still checking lower curve exit before. So let&#x27;s export read until riveted. No, this is about created from us. One with the two upper curves intersect. I know with they, um the right boundary, right and point to be determined onto that we equate the lower in the upper Kurds so we equate eggs people before with two divided by rude axe I wish is the same as X squared, but by 16 is four by axe, so x cubed is 64 and this will heal the real solution Explore X equal to four and we have two other cubic roots which are not riel. Therefore it&#x27;s moored s. So from this point, it&#x27;s just a matter of verifying d of computing These two and he&#x27;d read this. So the 1st 1 evaluated zero, of course, zero. But you did it when we just technical efficiency. That&#x27;s one US 2/3 is 5/3 less money. Eso that&#x27;s 40 over 24 less three over 24 to 37 over 24. Into that, we add four times before eight less 16 by it is you flat. Six much we subtract for rue one. So four. That&#x27;s that&#x27;s 32 of rage. Uh, it&#x27;s one of rain for 31 of great subtract 31 already from six. That&#x27;s 48 last, 31 in 17 over a. And so we add 17 over 8 to 37 over 24. No, no, that&#x27;s 51 over 24 was 37 is, uh, 88 divided by 24 and you should be 88 24th visible by eights or body. Writer is an operator by a Mediate about my ages. 11 in a 24 but it But it is three. Final answer is 11 3rd and we know that the value. Leavenworth three

Bobby Barnes · Answer

we want to find the area of the region and the first quadrant bounded on the left by the Y access and then also by those three other curves that they give us. So the region that we&#x27;re interested in I&#x27;ll go ahead and color in lack. There&#x27;s gonna be this region right here. So let&#x27;s go ahead and erase all these other lines just so they don&#x27;t distract us. So that&#x27;s the region we care about. Now notice that around here, where the blue and the green lines intersect each other, we have a different upper and lower. So well, we&#x27;re gonna have to do is figure out how to partition these two. So the way I don&#x27;t like to think about finding the area is gonna be the upper minus lower function. So I&#x27;ll call this area one area to sew for area one. This is going to be the integral. So they tell us to start at zero. And then, well, to find that point of intersection, we&#x27;d want toe make the blue and the green lines equal to each other. So it&#x27;s going to do. That would be one plus square root of X secret to over the square root of X. And if you were to go through and solve, we would end up with ex equaling two one. And so that means our upper Balan is gonna be one. And in this region, if we were to pass through it. So he passed through the red line and then the blue. So our upper is going to be one plus root ex and that are lower is going to be X over force. We subtract that all. So that&#x27;s gonna be our first region. And then what about the second region? Well, we already found out that it intersects at one, but now we need to figure out where did the green and the red lines intersect? This point right here. So we&#x27;re going to set those functions ego to each other again. And if you were, go through and solve everything that should give us that X is equal to or so our upper bound is. Or as though the green line in this region is on top of the red, so that should be two over the square root of X minus X over four DX. Now let&#x27;s go ahead and interview each of these air. That should give us our area. That should be a plus. There, no, for each is just going to be power rule. So we should have X now spirit of exes. Ex of the 1/2. So it should be X to the three house power now and then we divide by the new power. So 2/3 and then we should have won eight x squared. And we&#x27;re gonna evaluate this from 0 to 1. Now, for our 2nd 1 again, it&#x27;s just gonna be powerful for each and won over. Expert is X a negative 1/2. So that should integrate to be X to the 1/2. And then we multiply by its new power or multiplied by the reciprocal disc in power, which is gonna be, too. And then that should also give us 1/8 ex squirt. So now we would I wait each of these intervals direction. Let&#x27;s go ahead and simplify this. This should be poor. And now we just plug everything. And so starting with our first general plugging in one should give us one plus 2/3 minus 18 of them. Plug it and zero should just give us sirah. Now plug it in four So X to the 1/2 is really just the square root. So we get two times for eight and then four square to 16 so we&#x27;d subtract off two. And then when we plug in one, that should give us four minus one heat. Now, when we add all of this up, this should give us 11 3rd It looks like so the area between that region that we have shaded in black should be 11 3rd

Problem

Solve the given problems by integration. The gene…

Problem 38 Easy Difficulty

Solve the given problems by integration.
Find the first-quadrant area bounded by $y=\frac{\ln (4 x+1)}{4 x+1}$ and $x=5$.

Answer

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