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Find the area of the region under the given curve from 1 to 2.3x...

Question

Find the area of the region under the given curve from 1 to 2.3x

Find the area of the region under the given curve from 1 to 2. 3x



Answers

Find the area of the region under the given curve from 1 to 2.

$ y = \dfrac{1}{x^3 + x} $

This question tests you and your ability to find the area between two curves. So the two curves given to us. This question are why equals x squared minus one and y equals three. So our first step is to determine where the two curves intersect as those are going to be the bounds for our integral. So we can do this by sitting 22 equations equal to each other and solving out for X. So we have exposed to I'm sex mice too, the U minus. So our two bounds are gonna be X plus two and x minus two. Our next step is to set up our integral. Be used by visualizing graph to see which the equation is on top, which equations on the bottom. So why it was three. It's just a horizontal line. So and then X squared minus one is upward facing. Proble with a y intercept of negative one. So it looks something like this where these two points are thanks negative two X positive too. So you can see that this is the area we want in our uh for our integral. So because why equals three is on top, We're gonna and subtract that from exporting minus one. There are so attracted by expert minus one. And this we have to forget forget R. D X. And this becomes four minus X squared D. X. And if you solve this integral out, it should be equal to 32/3, which is our final answer.

This problem is asking for the integral from one to two of this function. And let me rewrite this denominator by factoring chips just three minus X. That's the area under the curve from one to two. So here the numerator and denominator of the same degree. They're both degree too. So we should along the vision here. So go to the side and do long division because we always should do this before a partial fractions. It's minus three X and then we have three. X Plus one is our remainder. So using this long division, I can rewrite this integral negative one and then three X plus one x three minus x t x And then we'LL go out and do partial fraction to composition on this fraction here using what the author calls Case one. So here we have a three minus X. So it's good and multiply both sides of this by distant dominator here on the left that on the right, so much factor on the X and we see that b minus a must be three and three a must be won. So we get that moves. So here we have one equals three soft ray and then used in this equation. Here we have B minus a third is three or weaken right that is nine over three. So soft for be there. Just add that one over three over. So we have our A and B. Now let's just go ahead and plug those in Now. We could replace this fraction here with the partial fraction. And let's write that on the next page. Once a two, we have the negative one from a long division and then for a partial fractions. That's our area. Now let's go ahead and evaluate these inner rules. If this last one. If this is bothering you with the three minus X, feel free to do it. Use up here. You could do U equals three minus six and to uni equals negative DX. That should work. So let's integrate these three. The first one just is minus X for the next one. Natural log. Absolute value X and for the last one, ten over three with a minus natural log. Absolute value three minus X, and you could see the negative is coming from the use of. And don't forget our UN points wanted to it's good and plug those in one at a time. Ellen. One there. We know that zero Ellen one. That's also zero minus ten over three. Ellen, too, And then just go out and simplify this eleven over three. Natural onto. That's our answer.

Find the area of the region between these curves Ah, using, um an integral with respect to X and another in hieroglyphics wide. So let's do with respect to X. And I think it be easier if we graft this to show it looks like they have life was negative one. And we have why equal through my sex squared? It's from this point to this point and we find these boundaries by setting these two questions equal to each other. So you negative one equals three minus x squared, which leaves us with X squared equals four X equals closer minus two. So the bounds over interval or negative 2 1 3 4 2 2 to 2 and we have our top equation minus or bottom equation. We have three minus X squared minus minus one, and that just gives us four minus x squared. The ex we integrate. We have four x minus X cubed over three. Plug in our and points and we get first we get eight minus a over three. Then you get minus all of negative eight ah, plus eight over three. And if we simplify that, we get 32 over three. So solving with respect to access pretty easy. And now we're going to do it with respect to why and I'm going to write this again. So why equals three minus X squared? But instead I'm going to write it as X equals. If we work the other side, the sponsor minus square of three, minus why and then we have y equals negative one. And so who still drying out like those native one and we'll have, Why calls through my square. But instead, we're going to treat it, of course, with X equals, and we now have to find our boundary points on in terms of why which are negative one and three. So we have sounds and they give 1 1 3 1 to 3. And we have our top equation and our bottom equation. Or we're talking about a question which is top and bottom, which is just going to be square of three minus y, minus the negative square of three months. Why, as that's our bottom and top equation B y. So instead we go inside, we have to root 31 swine. And if we integrate this, we get three minus. Why to his 1/2 We just add one to the exponents. Three over too. 2/3. Don't forget the other two that we had there originally. And we also remember that because it's negative. We have to have a native so that if we take the derivative, we see that this negative from the general counsel's out. Those negative. And this is what we get for when we integrate, we have free and negative one as our bounds. So let's plug that in Pleven three. Of course, we just get zero. So we have zero minus. If we take negative one and plug it in and we get four in here for two. The three over to power is just four cubes or 60 for a squirt of that eight. Eight times negative two times to over three. Which is this times This times this We get negative 16 times, too, which is negative. 32 or 30 minus that. So they get 32 over three. So either way, taking the derivative respect to or taking interval with respect to X and y kisses The same answer for the area

So the purpose of this problem seems to be to give you an example where you can set up an integral to find an area of a region by integrating with respect to X and then with why and you should get the same answer. So rather than evaluating the integral Sze, what's just show how we can set set them both up. So first, let's just try to sketch this region in question. So we have This is not drawn super to scale. We have this upside down parable. Oh, that shifted up three units. Okay, then we have the line. Why equals negative one? So let's see how he would set up the integral with respect. Tow X First Sweden. We need to find these two intersection points and then we know how to stop the integral. So let's set them equal. We would have three minus X squared is equal to negative one, which is equivalent to X squared, is equal to four. And so we have excess plus or minus two. So the area would be the integral from minus 2 to 2 of the upper function, minus the lower function. So you have three minus X squared, minus a negative one. Okay, so that if we evaluate this, that will give us the area. Let's see also how we should be able to set it up by evaluating integrating with respect to x. Sorry, Integrating with respect the wine we already showed out integrate with respect to X, so you'd have to look at this green function. And so as for being a function of X in terms of why, if you tilt your head to the right by 90 degrees or see this doesn't really pass the vertical line test. It's not really a true function, but we can write it as two separate functions by Let's just solve for X here in terms of in terms of why, so you'd have X squared is equal to three minus y, which would give us now if we take square roots, X would be plus or minus the square root of three minus why it's so to the right of the Y axis that's square root of three minus y and to the left, we have minus square a three minus y, so we can also integrate like this integrate. Why, from here we would be going from one from minus one, 23 and our upper function in terms of why would be squared of three minus y minus negative squared of three minus y. Okay, so we should be able to also evaluate this integral and get the same answer is this one, but we'll just show how to set it up and leave it there.


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