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Find the area Of the shaded region_ (Round your answer to 2 decimal places) The [ region bounded by y=x3 2x2-[Sx+ 6 andy = 03+6...

Question

Find the area Of the shaded region_ (Round your answer to 2 decimal places) The [ region bounded by y=x3 2x2-[Sx+ 6 andy = 03+6

Find the area Of the shaded region_ (Round your answer to 2 decimal places) The [ region bounded by y=x3 2x2-[Sx+ 6 andy = 0 3+6



Answers

Find the area of the region described in the following exercises. The region bounded by $y=6 x$ and $y=3 x^{2}-6 x$

We want to calculate uh area of this region that we see right in here. Uh It's the region that is bounded by the functions Y equals six minus X squared y equals x one. X is less than or equal to zero and y equals negative X. When x is greater than or equal to zero. But this is a region in here that we want to integrate. We can see that integrating with X would be best with respect to X. So we're gonna have to still break up the integral into two parts here. Um We're gonna start our integration when X is negative three and go to zero and then we're gonna have another definite integral forms from X zero up to X equals uh positive three. Um So the area we're going to break up the area of this entire region into uh the some of the areas of these two separate regions represented by two different two different definite integral. So the area of this region over here Is going to equal integrating with respect to X. The definite integral from X -3-0. And when you want to find the area bounded between two curves to functions you take the top function the greater and subtract the lesser us. A red minus blue. Red minus Blue is going to be six minus x squared minus six. Now we need to add to that. The area of this region over here on the right side here, we're going to make a definite integral from X 02, X is three. And this time the area between the curves represented by the definite integral from X 02 Xs three. The greater function is the red function. The lesser function is the green function. So the area this region, the area between the two curves is going to be the integral from 0 to 3 of the red function minus two. Green function. Red minus Green six minus X squared minus negative X. Okay area this region in a girl from X 02 X S three read functions of track Green function six minus X squared minus negative X six minus X squared minus negative X. That is minus negative X turns into a plus sacks so we can turn that into a plus. It's attracting a negative vaccine thing is adding X. So once we uh calculate these two separate definite integral where you will have. Okay, total area of this region bounded between these functions. So in a role to elect In a derivative of six is 6. X integral of X squared is executed over three minus integral of X which is X squared over two. We have to calculate that Between -3 and zero. Now we're going to add to that. Uh This definite integral anti derivative of six is six X minus integral of X squared is execute over three plus integral of X. Is x squared overtook. And we have to evaluate this expression Between zero and 3. So the area is going to equal this plus this. Well, this entire expression evaluated at zero is 0. And then we need to subtract this entire expression evaluated at negative three six times X. When X is negative, three is negative, 18 minus Execute over 3 -3 Cube -3 Cubes will be negative 27 Divided by three will be negative nine minus X squared over two. When X is negative three, expert will be positive nine to minus -9 over to. We'll simplify that in a minute. Let's move on and work on this side. This expression evaluated when X is 36 times X. Six times 3 18 Subtract Okay, -3- 00- three. -3- 00- three. Okay, evaluating this expression when X is 36 times X. Six times 3, 18 minus Execute three Cube. Just 27 divided by three years, 9 plus X squared when X is 33 square 99 halfs minus this expression evaluated at X equals zero. While this will all be zero. All right. Uh Let's go ahead and uh try to clean this up a little bit. Alright, 0- All this will be the negative. Okay, zero minus all this will be the negative of negative 18 plus nine minus negative minus plus nine minus nine years. Little easier. Put it in the calculator that way. Okay, so zero minus all this will be the negative of all this. The negative of negative 18 Plus 9 -9 years. And then this side doesn't look so bad face of plus This side just says 18 minus nine plus nine years. My signal zero doesn't change anything. All right. At this point, we can actually drop these square brackets, don't really need them. And if we're careful, you can get your answer in the calculator exactly by entering what you see, including a parenthesis. So, let's go ahead and do that. All right. So, when we enter all of this in the calculator, being real careful, making sure we also enter those parentheses. Uh we get 27. So area of the region bounded between those functions was equal to the sum of two separate definite integral, and the sum of those two separate Definite in the rules came out to be 27. So the area of the enclosed region is 27. This area, this whole area in here is 27

In problems. 65. You want to get the area off the bounded region by these graphs, This line and this line and this graph Let's sketch. This is kids to the graphs toe. Visualize the bounded region we have here. A line well equals X plus six. It's a slow is one. And it shifted up, boy six in the Y direction. 123456 We have here a point, and it has slow with one. Then we have here minus one minus two, minus three minus 45 minus six. This is the point and disappoint. This is why equals explosives. The line y equals minus two x as the boy zero and zero and it has a negative slope equals minus two. Meaning that at X equals one. Yeah. Why equals minus two minus one? We have one equals two. Yeah, this is We have two points. We can draw the line. Finally we have Why equals execute? We have the line by equals. Execute the graph. We have zero and zero point When X equals one y equals one. Another point when X equals minus one, why equals minus one another point and when X equals. Do we have for equals eight something like here and when X equals minus two Why equals minus eight? Something like here This is the graph. Avoid equals, execute. We can see the region here This is the region the boundary division Boy, the graphs here The region is bounded at the top. Bye Ah function that this function is why equals X plus six. The top here is the line y equals X plus six What the bottom function is divided into to function for an interval For this area the bottom function is this line which is why equals minus two x and for the right area this area, the bottom function is why equals execute. Then they calculate the area bounded by these three graphs We need toe divide this area into two areas The first area The total area equals area on plus area too. The first area is bounded at the top by the line y equals X plus six on from the bottom by oy equals minus two x then the first area can be calculated by evaluating the definite integral off the top function minus the bottom function explosive X minus two x the X and we integrate for the bounds off area one we integrate for the bounds of area on. Everyone starts from this point till X equals zero. To get at this point, we need to solve the equation that has the same way. Cordant for the two lines. X plus six and minus two x Let's get the intersection between the two lines. X plus six equals minus two x. We equipped these two white accordance except six equals minus two x. We add two x toe both sides, then three X plus six equals e. We subtract sex from both sides. Then we have three X equals minus six. We divided by three, then X equals minus two. The intersection off the two lines here Heavens at X equals minus two. Then for area one, we integrate from X equals minus two. From here to here, until X equals zero. This is for area. One area to this is area too is bounded boy the boat and the top by y equals X plus six. Then we integrate for example, a six minus the bottom function, which is why equals execute minus execute the X we integrate at the starting point of the region, which at X equals zero until the end off the region, which is here at X equals two. If we want to make sure that X equals two we chick, if this point has the same coordinate in the line and the graph for the line at X equals two, we have to blow six. War equals it. And for this equation, when X equals two, why equals it? They have the same boy, Corden. Then this is the same point They intersect at this point. We integrate from 0 to 2. Let's evaluate this integral the first integral we have X plus six minus toe minus minus two x we have here. We have a sign here. This is minus minus two X then the integration is we have x plus two x three x The integration off three x three x squared Divided by two plus six x Integrate from minus 2 to 0, plus integration of X is X squared divided by two plus six x minus x about four by four Integrate from 0 to 2. We started by the hour bone The service uber X equals zero give zero minus. We substituted by the lower bound. So institute X equals minus two. We have three multiplied by minus two squared. It's just for divided by two plus six multiplied by minus two. Plus we started by the hour bound we substitute X equals to have four. Divide by two plus six multiplied by two minus to to the borrow four divided by four minus. We service uber X equals zero Give zero equals six. Mine. We have six minus 12 which is minus six. And we have a minus here. Six plus to two plus 12 minus. We have four equal stint equals 16 which is the final answer off our problem. And this is the area the total area bounded by these three graphs given in the problem.

In Problem 25. We want to give the area enclosed between the two graphs six. X minus X squared and y equals X. It's it's good to these two graphs to see the area we need X and Y you have here. 123 123 We first graph the light y equals X. It's the line where all X values equals. Why values This is. Why equals X? It's a grave. The parabola. Why equal six x minus X squared? Let's when X equals zero. We have y equals zero. Then we have here a point for the parabola. When X equals warm. We have y equals six minus one by equals five. We have here a point. For example, is five. It's four. When X equals two. We have 12 minus four. Is it something like here it when X equals three. We have 18 minus line equals knowing something like here and we can see that a curve is going to be something like that because it's a problem. You know that something like that it will switch down somewhere it will intersect. At this point, we can get this point. Algae Brickley for equating the two boys here. We quit the two right hand sides to get the point we equate six x minus X squared equals X. We subtract X from both sides. We have five x minus. X squared equals zero. We take X as a common factor. We have X multiply by five minus X equals zero. We have here two solutions for this equation. The first solution is X equals zero. Here. The second solution is X equals five from the second factor which is here. This point and shaded area or the bounded area is this area. We can see that this area is bounded, boy, the graph the parabola at the bottom and the line the the parable at the top and the line in the bottom and using the definition of definite integral we can get the area equals the integration off the top function, which is the problem minus the bottom function, which is the line the X we integrate for the bounds of the area. The area started at X equals zero here until we reach X equals five Here. This is very the bones From 0 to 5. We can simplify this integral from 0 to 56 minutes, 66 months x is five x minus. X squared. The X we integrate five x squared, divided by two minus execute divided by three is up to substitute from 0 to 5. We started by the hour bound we services uber X equals five. We have five Cube divided by two minus five cube divided by three minus. We substituted by the second bone, which gives zero. We have five cube divided by six, which is 100 on the 25 divided by six. And this is the area off the shaded or enclosed area between the two given graphs.

In the problem we have been given that is Why equal -4 upon texas, choir minus six minus six. So first of all this is written as excess choir minus three X Plus two x -6 Or it is X into X -3 Plus two into X -3 This is X-plus two In blocks -3 No we have This as integration -1, 2 to -4 upon X plus two In two x -3 dx Hence it is -4 upon x plus two And works -3 that equals two. Airborne X-plus two plus B upon x minus three mhm So further we have the value of a equal to four upon five and B equals two minus four upon five. So this is integration -1-2, four upon five And works plus two -4 upon five index minus three index So this is regionals four upon 5, Aaron Model X-plus two -4 upon five. Ellen mode x minus three putting the limits minus one and two. So Father, this is written as Put up on five into Ellen would x plus two Upon X -3 The Elements -1 and two. This gives us 4.5. 16 is equal to 4.5 Ln 2 to the power forward. All this is equal to 16.5 Ln two. So overall we have this as the answer to the problem.


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