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Find the area bounded by the graphs of the following equations over the given interval. Compute answer to three decimal places: y=X 6x; y = 0; -1<x<2...

Question

Find the area bounded by the graphs of the following equations over the given interval. Compute answer to three decimal places: y=X 6x; y = 0; -1<x<2

Find the area bounded by the graphs of the following equations over the given interval. Compute answer to three decimal places: y=X 6x; y = 0; -1<x<2



Answers

Find the area bounded by the graphs of the indicated equations over the given interval. Compute answers to three decimal places. $$ y=-2 x^{2} ; y=0 ;-6 \leq x \leq 0 $$

So if we want to find the area between these two curves, the first thing I would do is go ahead and just sketch what the graph looks like. That way you can go ahead and figure out what functions on top in which functions on bottom. So I'm first going to factor on an X from this. Actually, I'll do the same for Marine. So this first one is going to be X Times X squared minus six X plus three. And then this here is going to have a double zero, um, at three. Or I should say, this is a perfect square because of the X Times X minus three squared. So if I were to come over here to try to grab this so at zero, we have a zero, and then at three, we also have a zero. So this function has a lien coefficient of one. It's an odd polynomial, so it should look something kind of like this. So it's going to pass through this era that is going to come back up at some point. Uh, actually not like that. It's going to come down and then it's actually just going to touch here because it's a double multiplicity, and then it'll come up and then it will have to come back down. It looks something kind of like that. Uh huh. Now we can go ahead and graph. Why is it would axe? So we have us here, and I believe it would look something like this. Okay, so notice are three points of intersection are going to be those three points right there. Actually, let me do this in red. These three points here. So we need to figure out what is a what is b and then what is C and so this is not drawn to scale or anything. Um, but just to kind of give us idea, I think you can see how, in this first region, the green is on top, but in the second region over here, the green is on the model, Uh, and so once we figure those out that we can use that to help us set up our interval. But let's go to figure out what those are. So we're gonna just set those equal to each other. So we execute minus six. X word plus seven X is good. X so I'm just subtract X over. Say is excused minus six. Expert. Yeah, plus eight X is equal to zero. So now we can go ahead and factor out an X like we did before. And now we ask ourselves. Okay. What numbers will, um, multiply to eight, but add to negative six. Uh, and if I'm not mistaken, it should be X minus four and X minus two. So then that would give where X is equal to 02 and four. So that means a is zero b is two, and then see is four. So now, um, this first region here goes from 0 to 2. So actually, let me write this down here. So this is going to be the integral from 0 to 2. And then we take our top line, which is the grain. So the top minus our bottom, which is blue, Mhm. And so that will be our integral there. And then for our second region over here, this area is going to be well, we're integrating from 2 to 4 because those are points of intersection there. And then we have the blue line on top and then the green line on the bottom. And now we can go ahead and just plug these and especially me under wide this in green here. So I don't forget now what you're saying we just plug everything in and integrate. So this is going to be the integral from 0 to 2 of So the top in this case is degree and should be execute minus six X squared plus 15 x and then minus just x so that gives us integral from 0 to 2. Oh, x cubed minus six x squared plus eight x. And to integrate this, we can go ahead and use powerful to integrate. So it would be, um, X to the fourth over four minus six X cubed over three and then plus eight X where over two. And then we evaluate from zero to, uh, let me go ahead and simplify this because that should just be too. And that should be four. Let me bring that up. So x to the fourth over. Four minus two x cubed, uh, plus or x squared evaluate from 0 to 2. So we go ahead and first plug in to so we'd have 16/4, which would be four. Uh, then we have, uh, minus 16. And then we would have plus 16 and then minus. Well, if we plug in zero, we just get zero. So these 16 just counts out, and then we'd be left with four. Okay, so the area that first region is four. Now, we come out here and do the same thing for this one. Um, but since this is just the opposite ordering of that, we can just take what we have down here and then flip all the signs. So this is going to now be the integral from 2 to 4 of negative execute plus six x squared minus eight x dx. And if you were to plug those that you would see would get that same thing, um, and then integrating it It would be pretty much the same idea where we just get to this down here. But all the signs are fluid. So just to kind of save some time. So this would be negative X to the fourth over. Four plus two x cubed and then minus four x squared or evaluating from 2 to 4. So we go ahead and first plug in four. Um, so four to the fourth, divided by 4. 64. So this is gonna be negative. 64 plus for cube times too. So would be plus 1 28. And then that would just be four cubed for the last ones of minus 64. And then we're gonna subtract off when we plug in four. Which would I'm sorry when we pulled into which should give us, so it would be so again, Two to the fourth. Um, actually, we already plugged to it up here, so it was four. Minus 16 plus 16. But remember, we switched the order, so this would be negative for here. Okay, so notice this negative 1. 61 20 negative 60. All these council, and then the negatives here council. And we're just going to be left with four again. So our area, it's going to be the first area plus the second area, which is just four plus four, which gives us eight. So our total area overall is eight

Here. The area can be fined by this formula. So I'm just writing in this formula minus three, that is lower limit toe a parliament zero minus x square, minus 20 BX going ahead and solving it further So I can I devalue edge, Just look at it carefully. Minus act to the power to plus one by two, plus one minus 20 x lower limited minus three and upper limited zero. I usually formula here, which is given by acts to depart and equal toe act to the part and plus one by and plus one. This is the formula I used here to integrate the value going forward and just integrating it. So I get the value edge minus. Ask you by three minus 20 X and lower Limited minus three apart. Limited zero. Solving it further. I get the value edge minus zero cube by three minus 20 multiplication zero minus minus. Trick you by three minus 20 multiplication minus three. Going forward and solving it further. I get the value edge minus zero minus minus 27 by three plus 60. So on solving it further, I get the value edge. So just look at it carefully, minus zero minus 51. And on solving it, I get the final area value age 51 for this question.

The required area for this problem can be calculated by this formula minus integration off minus one says zero. Act one minus x bx Going forward and solving it I can I devalue age minus 10 minus X plus X is squired bx So on integration I get the value like this. So just look at it carefully. I get minus X to the power one plus one by one plus one plus x to the power to plus one by two, plus one minus one and he is zero. I use your formula which I'm writing it here so that you can learn the concept X to the power and the X is equal to X to the power and plus one by n plus one. This is the formula used here now on integration. I get the value like this which I am writing here minus X squared by two plus X cubed by three and the lower and upper limit ese minus one and zero. Solving it further. I get the value like this on putting the a parental world limit minus zero square by two plus zero cube by three minus minus minus one is square by toe, plus minus one cube by three. So on solving it further, I gave the final value edge minus minus five by six, which is equal to five by six. So five by six is our answer. If you want to change it into dismal, you will get 0.8333 This is our answer for this question.

The area for this question can be calculated by the formula. So I'm just writing that formula a year. This is the formula and we have to calculate this value. So on integration, I get the Value Edge Act to the power to plus one by two plus one plus two x and putting the lower limited zero and a par limiters three as we know that X to the power and is equal toe X to the power and plus one by and plus one. This is the popular I used here, going ahead and solving it further. I get the value edge X cubed by three plus two eggs, 0 to 3. So solving it further I get the value edge trick you by three plus two multiplication three minus zero cube by three plus two multiplication zero and on solving it further, I get the value as nine plus six minus zero. So on solving it, I get the final answer, which is equal to 15


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