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Find the Area of Region Bounded Above by Two Different Functions QuestionConsider the region, R, bounded above by f(z) 22 62 + 9 and g(x) over the interval [3,9]. F...

Question

Find the Area of Region Bounded Above by Two Different Functions QuestionConsider the region, R, bounded above by f(z) 22 62 + 9 and g(x) over the interval [3,9]. Find the area of R:3r + 27 and bounded below by the €-axisGive an exact fraction, if necessary, for your answer and do not include units_

Find the Area of Region Bounded Above by Two Different Functions Question Consider the region, R, bounded above by f(z) 22 62 + 9 and g(x) over the interval [3,9]. Find the area of R: 3r + 27 and bounded below by the €-axis Give an exact fraction, if necessary, for your answer and do not include units_



Answers

Sketch the region $R$ bounded by $y=x+6, y=x^{3},$ and $2 y+x=0 .$ Then find its area. Hint: Divide $R$ into two pieces.

So here we have a region bounded by why equals X minus one quantity squared and why equals seven X minus 19. Now, as you can see on the graph I've generated, I stretched out the X axis a whole bunch, and it still shows up as a teeny tiny sliver right in here. So are our intersection points or somewhere around here and here ish. So we know our I mean, our general form in a row from a to B of F of X minus dfx DX. If you were to draw a vertical line in here, you'd have to zoom way. You would find that the line is the higher is the larger function. So our line is our f of X, and our parabola is our g of X. Um and then we need to find our intersection points. So we're gonna set this. These two equations equal to each other equals seven X minus 19. If you expand out the X minus one squared into a polynomial and then combine all the terms on one side, you end up with X squared minus nine X plus 20 equals zero. This factors into X minus four Times X plus five getting us an ex of four and five. So we have all of our pieces for our integral. So now we can fill it in, so we're gonna have the integral from 4 to 5 of seven X minus 19 minus X minus one. Squared DX. Now, if we expand the X minus one squared and distribute the negative one that's out front, it'll get us a polynomial that is easier to anti derive. So we're gonna do that. So we have the integral from 4 to 5 of negative X squared plus nine X minus 20 DX. Now we can take the any derivative which is going to get us a negative. X cubed over three plus nine x squared over to minus 20 X from 4 to 5. When we perform our substitution ins, I'm gonna bring it over here for space. We end up with negative 1 25/3 plus to 25/2 minus 100 minus negative. 64/3, 1 44 over to minus Haiti getting common denominators and combining everything together. We end up with an overall area of this little sliver being 1/6 and you're done

For this problem, we are asked to sketch the region bounded by the graphs of the functions and find the area of the region where we have F of Y equals Y times two minus Y and G of Y equals negative Y. Now plotted over here, I have the region where the horizontal axis, I have said as Y and then the vertical axis we could call X of Y or whatever we want. Um so we can see that we'll be are the region here is going to be for X between zero and 3. So the way that we actually find the area is by integrating, Sorry, I said X between zero and three, I mean why between zero and three? So we'll be integrating from 0-3. Ah We'll also note that this black line is the white times two minus Y. It's the greater value. So that's the one that we will have a front. We have Y times two minus y minus negative Y. Which obviously is going to turn into white times two minus Y plus Y dy. And then we can turn that y times two minutes. Why into two y minus y squared? So then we have two y minus Y squared plus Y. Which we can reduce further down to just three y minus y squared Dy. Then upon integration we'll get daisy will get three. Yeah man there we go. Okay three we'll have three Y squared over two minus y cubed over three Evaluated from zero up to three. Which is going to end up just being three times three squared or three cubed over to -3 Cubed Over three. So we can put everything to the same common denominator. We'll have three to the power of four Over 6 -2 times three. Cute over six. In which we can then factor out the three cubes. So we'll have three cubed over six outfront times three minus two, Sometimes just one. We're left with three cubed over six and now three square is going to be 99 times three is 27. We have 27/6 and 27 is itself nine times 36 is two times three. So we're left with 9/2 as a result.

Here is the area that we want to calculate the integral while I was still pulled, divided by three East is between nine for tickled to If I did why right across to 19 plus on minus on. The boy equals Stroup Plus turned minus Read the Koran clear to speaks minus to peace. So it is integral begins from mind Ah, minus nine. Credible to divided boy. Really quite 17 to nine. Radical too divided by radicalised 17 Integral begins from Linus. Ready corned 16 36 minus two East tube to four. Divided by three weeks. Re boy, please. It was two in three. Club begins minus nine. Credit card. Who do you want it by reading? Look 17 to nine. Medical too. Do you really did 20 but 17 So over three eggs Los. Read the Koran 36 minus two x the You can preserve us 4.3 east to dividers between minus nine. Reddick Want to Did this really radical? 17 and nine Radical. Untrue. Divided by reading incident plus breath control in drink or the glove. 18 minus two. Thanks his sport. When calculating it comes a little. So the child with the other. Of course. The cheese ready called to divide and rule. Who wanting to buy Reddick on 18 minus X tube, plus one other school. What's a good way? 18. The news in bears off Eakes they avoided worry. Three. Read the country when they can build a thesis. Parts these parts again we put the figure on it becomes a little So have one Divided by two want have been quite 18 issues. Mine. Nine In theirs. 09 Ready called to divided for radical 17 divided by three A girl Morning This who's in verse? Oh minus nine. Radical too. Give leaders were defense. 17. The wind that blurry three replicant mish becomes nine Scenes in theirs. Three Be widened 23. The constant in game minus seems in verse minus three being like a boy.

For this problem, we are asked to sketch the region bounded by the graphs of the functions and find the area of the region where our functions are f of Y equals y squared, and G of Y equals Y plus two. Now, for the sketch or for the plot that I have here, I did set X to be the function of Y. So we have y squared or x equals y squared and X equals Y plus two. You don't necessarily have to do that, but we can see that we're going from negative one is when they first intersect or negative one wise when they first intersect and positive to wise when they next intersect. So to find the area will take the integral from negative one up to positive too of y plus two minus y squared dy. So that will come out to y squared over two plus two, Y minus y cubed over three, evaluated from negative one up to two. All that's left is to plug in those values and calculate so one second. The result you arrive at should be nine over to


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