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Find the area of the shaded region.y = Vx Y =X - 2...

Question

Find the area of the shaded region.y = Vx Y =X - 2

Find the area of the shaded region. y = Vx Y =X - 2



Answers

Find the area of the shaded region.
$$x=y^{2}-2$$

For the following problem were given to functions. X equals y squared minus four. Why picturing green X equals two y minus y squared. Pictured in blue, we were asked to calculate the area between the two curves shaded in red. Do this. We're going to take the inner girl with respect Toe. Why? So this will be from why equals zero toe y equals three, since that's where the area shaded in red extends from when going from y equals here to y equals three, the blue function appears over top of the green function. Therefore, we will do the integral from 0 to 3 of two Y minus y squared minus why squared minus four. Wife ash on right here Distributing the negative In combining like terms, we get the integral from 0 to 3 of negative two y squared plus six by D y. When taking the integral, you add one to the exponents and then divide by the new exponents so we'd add two plus one equals three and then we divide by three. So it becomes negative. 2/3. Why cubed plus wide to the first power. So one plus one, then we divide by two so six to buy by. She was three, and there's now three times y squared from 0 to 3, plugging three. And for why we get why cubed times negative. 2/3 which simplifies too negative a team. And we get y squared times three, which is plus 27 minus. And then we put zero in for Why? So it's just gonna be minus zero. So becomes 27 minus 18 to given area in between the two curves of nine.

For this problem were given the following two functions. Why I equals the square root of X Plus two and why equals one over X plus one. We were asked to find the area shaded in red between the two curves. Do this. We're going to take the integral with respect to X and do the top function. Man is the bottom function. The area and bread extends from X equals zero X equals two, so the integral will be from 0 to 2. Function on top is the skirt of X plus two. The function on the bottom is one over X plus one. Therefore, the integral will be the square root of X plus two minus one over X plus one DX. We can separate this into two in a girls with the integral from 0 to 2 of the square root of X plus two DX minus the integral from 0 to 2 of one over X plus one DX. When looking at the integral of the skirt of X plus two, you can look at it as X plus two to the 1/2 power. Therefore, the interval of this is going to be you. Add one to the exponents, so it's three halves and then divide by the exponents so instead of three, have 2/3 times X plus two to the three halves from 0 to 2. Looking at one over X plus one, you should be able to identify that the integral of this is natural log, so the integral be natural Log X plus one from 0 to 2. When plugging in to to 2/3 times X plus two to the three halves, you get 2/3 times two plus two, which is four to the three halfs minus. The new plug in zero zero plus two is too so minus 2/3 times two to the three halfs minus natural log X plus one from 0 to 2. Simplifying this you get four to the three halves times 2/3 which simplifies to 16 3rd minus 2/3 times two to the three halves, which is 4/3 times the square root of two. You then can put two in zero into the other. Integral said becomes natural log two plus one, which is three minus natural log zero plus one, which is one natural log of one is zero so that goes away and the answer become 16 3rd minus 4/3 times the square root of two minus natural log three.

In the problem we have been given Y is equal to X plus one upon X into X -1. So this can written as a upon X plus B upon x minus one which is equal to X plus one upon X & two X -1. Sorry this is x minus one. So no We have to find the value of A&B. So a equal -1 and be equal to two. Therefore integration become 2-5 minus one upon X plus two upon x minus one. The X this is written as minus landmark X plus two. Ellen More X -1, two and 5. Hence it is equal to Ellen marred x minus one whole square upon X two and 5. Therefore this is witness Ln March 16 upon five minus Ellen on half. This equals to Ellen mod 32 upon five or this is written as a landmark 30 to minus Ln More than five which is equal to five. Ellen mud too- Ellen More five Now. Further this equals two 1.8 5 6. So this is the answer.

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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