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Question Determine the area, in square units, bounded above by f(z) 22 + lOz + 25 and g(z) by the €-axis over the interval [~5,-1].2x 2 and bounded belowGive ...

Question

Question Determine the area, in square units, bounded above by f(z) 22 + lOz + 25 and g(z) by the €-axis over the interval [~5,-1].2x 2 and bounded belowGive an exact fraction, if necessary, for your answer and do not include units.Provide your answer below:FEEDBACKMORE INSTRUCTIONSUBMIT

Question Determine the area, in square units, bounded above by f(z) 22 + lOz + 25 and g(z) by the €-axis over the interval [~5,-1]. 2x 2 and bounded below Give an exact fraction, if necessary, for your answer and do not include units. Provide your answer below: FEEDBACK MORE INSTRUCTION SUBMIT



Answers

Find the total area bounded by the graph of the given function and the $x$ -axis on the indicated interval. $$ y=\sqrt[3]{x} ; \quad[-2,3] $$

Okay, So for this problem were asked to find the area of theory region that is bounded to the left of the vertical line, which is X equals the square root of 2/2 and to the right of or that's inside of X squared plus y squared equals one. So if we look at the graph that I have drawn here we have our X squared plus y squared equals one. And then we have our X equals square to two plus two, and we're looking for the area that's here. So what I want to do first is I won't identify what the's two intersection points are. And we know that on a polar curve that if we happen to have a circle just like we have our polar circle rhetoric circle, we know that the square root of two is going to be at pi over four. And since we're doing within this area, I'm gonna say negative pi over four. So what I want to do is I want to rewrite my two equations as being polar equations. So X squared plus y squared equals one we know is going to be our squared equals one because X squared plus y squared equals r squared. So since R squared equals one, that means r equals one. So this is our 1st 1 and then looking at our X equals square root of 2/2 Well, I happen to know that X equals r co sign of data. So which means that I can substitute my ex for square root of 2/2 equals R co sign of data. And if I were to divide both sides by co sign, we're multiplying by the reciprocal. I know that one over co sign is the same thing. Seek it of data. So this is going to be my second equation. So the next thing I want to do is I want to go back and look at this picture, and it looks like I'm gonna be finding the area from pi over four too negative. A negative report of pi over four. And then I'm gonna be subtracting out this. My ex bite scored a two over to seek it from negative pi over 42 pi over four. So it's gonna be my one squared. It's gonna be my one squared minus my my seek in squared function. So my area function is going to be from a to B of 1/2 and in this case, it's going to be my f if they'd a squared minus GF data squared detail. So if you recall I said that we're gonna be doing our one minus or seek it and we're doing it over Pi over four to negative pi over four or just to make things a little bit simpler, Aiken do things up until zero and do zero to pi over four and then double it. Since it's going to be symmetric, someone had doubled gonna double it. And then 1/2 one squared minus my square root of two over to seek and data squared in terms of detail. Now my two and my 1/2 are going to cancel, and I am going to go ahead and square everything out so it's gonna be one minus. My square root of 2/2 is gonna become to over four or 1/2 seek it squared data and of course, dictator and I'm actually ready to integrate. So the integral of one is gonna be fate up. And then I know that the integral of sequence squared is actually tangent that I'm gonna bring down by 1/2 and then tangent data. And all of this is from pi over 4 to 0. So I'm gonna have pi over four, and then tangent at Pi over four is going to be the square root of two over to divided by the square root of 2/2, Which is going to be what? So I'm gonna have minus 1/2 and then, of course, state of being zero is zero. The tangent of zero is going to be zero. So my final answer is going to be pi over four minus 1/2.

To determine the area bounded between these two functions of this area and black here. We could go ahead and graph each of these and then notice, but the function to is on top of the blue one. And then we want to find our intersection points here. I'm here, and so you could use the graphing utility. But we'll just go ahead and do this step by him. So we'll set these two functions equal, and then we want to solve her isolate for our X here. Okay, so, um, we're gonna do a couple steps at the same time here. We're gonna, um, interchange. We're gonna multiply both sides by two. And then we're going to, uh, multiply that square root over and then square both sides. So it's gonna be this on fourth and then add the X word over. Subtract the 1/4 over. So that will give us 3/4 is equal to X squared. But then again, this will be plus or minus X or another way to say that is does the same thing as negative. Through 3/2, founded on the left and bounded on the rate positive every three over to And so really just concentrating on our set up. Now we're good to go. In terms of the integral, it's gonna be the area from Negative three. Have Route three hands the positive Route three House of the top function to minus the bottom function one over Route one buying That's expert DX. And although this looks, uh, a bit of a mess, the anti derivative of two is just to Lex, the anti derivative of the second part. There is inverse sign, and then we still have the same limits of integration. And so substituting does in and simplifying will end up with a final result of to brew three minus two. Hi, Third's. And so that is the final answer for the area under the curve and black. There, that's it.

We want to calculate the area of the region bounded by Y equals the square root of X y equals two X minus 15. And why go zero if we integrate with respect to X, we'll have to split the region into two sections. One where we integrate from y equals zero two y equals square to Becks and one where we integrate from Mike was two x minus 15 two. White was great backs. Instead, it would be easier to do just one integral in terms of way where we go this direction and we have our top curve for our bottom curve and our top proof. And we integrate from white with zero to whatever this why I value is here at the intersection. So first starts right out our equations in terms of X or in terms of why X equals y squared for white goods. Creative axe and why it was two x minus 15 will become why plus 15 over to equals x so X equals 1/2. Why plus 15/2 I had to find where these curves intersect will just set them equal to each other. So why squared equals 1/2 why Plus 15/2. And if we move everything over to one side, we have y squared minus 1/2. Why? Minus 15/2 equals zero. And then by factoring we get why, minus three times y plus 5/2. And since we're in the first quadrant, we want the positive solution, which will be why equals three. So now we know that were integrating from 0 to 3. So we set up our integral from 0 to 3. And our top curve is two x minus 15. So, in terms of why it waas 1/2 why I plus 15/2 and then our bottom curve was y squared. Now, if we integrate, we get 1/4. Why squared? Plus 15/2 times. Why minus 1/3. Why cute? It was 0 to 3. And if we plug in three, we get 1/4 times three squared plus 15 over. Two times three minus 1/3 times three. Cute. And if we plug in zero, we just gets her How minus zero if you want. And this is 9/4. Put us 45/2 minus. No. And if we give everything a common denominator. We get 9/4 plus 94th minus 36 4th which gives us a final answer of 63 over four, and that is the area of our bounded.

And this is sort of a tricky problem because if you look at the graph shifted up to and then it's a downwards problem like this. So we're looking at why it goes to minus the square root of X. You have to consider the time that you're below the X axis and the time that you're above the X axis and that happens at X equals four Because the square to for us to to -2 would be zero. So what you have to do is set of two. Different integral is the integral from 0-4. Um And I would rewrite that as X to the one half power. So that's good. That will give me the positive answer we want because the area needs to be positive but then I need the absolute value for the integral from forward to nine of really the same function. Uh because we want and bounded from 0-9, I guess I should have clarified that as well. Um Yeah so let's kind of just jump right into this. The integral of both of these are gonna be the same when you add one to the exponent and then multiply by the reciprocal, that new experiments and you'll get the same thing on this one except you just need the absolute value after it's all said and done. So it's the same integral. But what we'll end up happening is you'll get a negative answer. Uh Yeah, so if you plug in 42 towns floors eight. Remember this means the square root of four which is to cube two times two times two is eight times that to be 16. Um And I didn't plug in zero because you're going to get two zeros there. But on this one you need to do the same thing 18 minus the square root of nine is three cubed to be 27 Viking divide by this three as well as with the nine Times that to be 18. That's kind of a chance happening. I didn't expect that Uh minus uh again four times to be eight. And I guess it's all the same work is over here so if I do this correctly uh that looks like it's these are cancelled 8 -16/3. I forgot to put my absolute value eight minus 16 3rd and B. C. I would change that to be 24 3rd. It's equal to eight -1638/3. And then again here you would have a negative but you want to absolute value 8 -16/3 but it's negative and then absolute value it And when you're back to 16/3 so hopefully all my arithmetic is correct. That's the answer I came up with


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