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Find the area bounded by y = 2 ; y = √x + 2 and y = 2 - x...

Question

Find the area bounded by y = 2 ; y = √x + 2 and y = 2 - x

Find the area bounded by y = 2 ; y = √x + 2 and y = 2 - x



Answers

Find the area bounded by $y=\frac{2}{\sqrt{64-4 x^{2}}}, x=0, y=0,$ and $x=2$.

In problem 28. We want to calculate the area enclosed between these two graphs. The first graph is a parabola y equals X squared these X And here is why this is a problem you have here. Zero one on one minus one and one. We have here one minus 11 When is this to this four and so on? This is a problem. Why equals X squared? That's a graph y equals square root of x square root of X starting boy When X equals zero by equals zero when X equals one y equals one when X equals four y equals two Then we have this point this point and at four x equals to this point something like that. The enclosed area is this area this area has at the top Why equals square root of X and at the bottom it has Why equals X a square. Using the definition off the definite integral the area includes between two functions equals the definite integral off the first function which year is root x the top function, the definite integral of the to function, minus the definite integral off the bottom function which year is X squared tabloid by the X, and we integrate from zero toe one integrate from zero to one, which is the bounds off the enclosed area from 0 to 1. Let's integrate integration. Off the square root of X square root of X is something like square root of X is something like X the war off half integration off extra The bar of half we add one to the bar and divide by the new book we have X. It was about off 1.5, divided by 1.5 minus X cube divided by three. We evaluate this integral from 0 to 1. We started by the hour bound we have one divided by 1.5 minus one, divided by three minus. We substitute by the lower bound the service. Uber X equals zero equals zero. We have one divided by 1.5 inches, two thirds minus one third equals one third, which is the final answer off our problem and the area enclosed by the two functions given in the problem

In this part we can draw. The graph is something like this. So ah, burning wounded part will be like this. And it is here x equal toe to park because if illiquid, why won't acquired to try to all temple gate cans Also x equal to two whereby one is take personal first cup and why do it a second cup? So finally area will be Why one minus jailer due to DX No, it is did little to x minus two by x x squared minus two x plus two DX difficult to X minus two or DX excess choir minus two x Bless Toby X So it is If we consider that square managed to express to as dirty for two x minus two dx will be ditty later on. We'll change the limit. X equal to zero equal to do X equal to two for six minus four. Answer to seen. So if the limit our scene Ifill port deal a report Execute Kojiro thought it Final answer. Maybe possibility, because limiter here always seem so. The final lots of baby occurred to Chico

Embroiling 23. We want to calculate the area Enclosed boy. This line and this graph Let's sketch of them if we have your ex envoy One, 23 I have 123 That's through the line. We have y equals X plus two. It's when X equals zero. You know that. Why will be to you Have a point here and the slope is one. This means we have here a point. And here a point and so on. And you're a point. This is the line. Why equals explodes too. The second graph is y equals X Quit is something like that. We have here one on one and at minus one. We have a buoyant here and we have zero and zero Those when X equals two. We have y equals four here. This point is shared between the two functions. Then we have a point here. The point here point here. A point here and so on. We have this graph is boy equals x squared and the area in between the area enclosed is this area. We can get this area using the definition of different integral the integration off the top graph the equation off the top graph minus the equation. Off the bottom graph, we have the equation off. The top graph is why equals X plus two minus the equation off. The bottom graph is y equals X squared. We have DX here, and the bones is the starting point of the bounded area here and the starting point of the bounded area. Here we have the bounce from minus 1 to 2 to make sure that the two points intersects at X equals minus one and X equals two. We can substitute here by X equals minus one. It gives why equals minus one plus two equals one. And when we substitute by X here equals minus one, it gives why equals one. This means they intersect at X equals minus one. To make sure the intercept at X equals two. Substitute by X equals two in the line to two plus two gives four. And when substitute in the function here by X equals two, we get to power. Two equals four and this means they intersect. They gave us the same point, minus one on one, two and four, meaning that thes bones are through our correct Let's integrate Integration of X is X squared, divided by two. Integration of two is two x minus. The creation of X squared is execute. Divided by three we integrate from minus 1 to 2. We started by the upper bound we substitute by X equals two to the bar of two, divided by two plus to multiply it by two minus two cube divided by three minus. We substitute by the lower bound X squared divided by two equals half minus xmas to block by minus one is to plus one third. Let's evaluate these two terms. The first term is to plus four equal six minus eight, divided by three equals 13th minus half minus to plus one third equals minus and we have another minus 7 12 by six equals nine, divided by 24.5, which is area off the shaded region and enclosed by the line and the curve, which is a final answer off. Our problem


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