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Find the area of the region enclosed by the curves y =4 cos _ and y =4 cos Zx for 0<XsI:24 cos *cos 2XSet up the integral(s) that will give the area of the regio...

Question

Find the area of the region enclosed by the curves y =4 cos _ and y =4 cos Zx for 0<XsI:24 cos *cos 2XSet up the integral(s) that will give the area of the region. Select the correct choice below and fill in any answer box(es) to complete your choice

Find the area of the region enclosed by the curves y =4 cos _ and y =4 cos Zx for 0<XsI: 24 cos * cos 2X Set up the integral(s) that will give the area of the region. Select the correct choice below and fill in any answer box(es) to complete your choice



Answers

Sketch the region bounded by the curves and calculate the area of the region.$$y=\sin ^{4} x \cos x, \quad x \in[0, \pi / 2].$$

So you need to sketch the region close by. Those foolery in curves r say this is our x y plain. And so the first function is cosign packs that in some ex ico So wan Whose empire? Ecos huan Negative one so and we're accident post tow her half cose AIPAC seacoast Terrible. The maxi code zero coasts and zero is pie. Couldn't Ciro is one all right? People of this function should look like and this is a union function. So is symmetric by y axis look like this. It's actually cause one. This is not your war. Excuse me. This is our Mexican snack you want and Lycos connected in the second on the second curve. It's a problem. No X equals zero. This curve across a bike was naked one the waxy Sarah Why post make it one across this point? And then why could zero no ax eco's to a positive and that give by half So I call it suppose to quite this make you happy This problem back ten okay. And then close regions in between this shaded area? No. So you orderto finance area. We write on the intercom role for area ico swindle grow with respect, Teo Truce acts because those functions can be represented by acts explicitly. So this is a coast too integral with respect of the base, Back to backs and X goes from negative or half to part with him. In here you are. It's the upper curve minus the Lord coz I pax miners blacks were right, That's what. So it will be minus wax square plus four All right, next to the value of this indefinite things grow thiss definitely grow. So it's fine. And he knew it. The first term will be over. Pi sign packs in the second term will be, um for over three. Execute a third Who just x Yeah, you value? Uh huh. No. So the first time that calm Craxi coast, where half the no sign pi over two is one nice world Were pi miners for three plus one Miners plugging because inactive half So as you can see those are odd function Therefore everything flip their signs So when you hear is the first time we'LL be inactive What? Have, uh thank you for our pie But sorry been mistake We're plugging X equals one half here So this is not forward three times what this is for over three times one over eight. Now here is just And plus we're half right in the second term. Just sleep the sign. So the second This is Manas. Where? Over pi and plus for three times. Well, eight. Linus, where Right. And since we have this negative sign for bringing in our thing, we'LL flip again. Therefore, we'LL get two times the first first herbs. We only need to calculate the first term. And what flight by two. Right, So here will be over six. Plus my half That's our three. So he's just too or pie last to third.

What I like about this problem is usually cosign starts up one But if you're multiplying by two it's got to go up to two and if you look at the bounds that they give you uh from negative pi over to deposit prior to co sign of those two values are zero. So even if you consider the other function which is negative equals negative cosign lex. Um they never cross each other. So all you have to do is evaluate the integral from negative pi rubber tube to positive pi over two of the upper function which is that to co sign banks minus the lower function which is negative. So it turns into plus cuts and nicks G. X. So what I just think that this way is the derivative of sinus positive co signs. So the integral of co sign must be positive sign. And then to just goes along for the ride. Same thing with that co sign but there's no coefficient in front or along from negative pirate too to positive pira tube. Everything about the unit circle that. Yeah and they have much stomach scroll. Sino para two is one. And sign of negative pikachu is negatively. So plugging empire or two you get two times one plus one and then you have to subtract off two times negative one plus negative. Well that's going to give you three minus negative three which would turn your answer in 26

So in this problem, we have to sketch the region's bounded by the curves coastline events and sign into X between X equals zero and X equals pi over two. And we have to find the area of these two regions. Yeah, So, um, the first thing is, look at let's look at co sign of X X equals zero and X equals pi over two. So if I plug in co sign of zero, I'm gonna get one. And if I get co sign and then for the X equals pi over two, I'll get co sign of pi over two, and that's equal to zero. We also note that CO sign of X is positive in this region, so it doesn't intersect the X axis. It just goes from one 20 Um, so it's like a sign of two vets on the same um on the bounds of the same interval. So we're gonna look at sign of two times zero, which is a sign of zero that's equal to zero. And then we have signed of two times pi over two. That's going to equal sine of pi, and that is also going to equal zero now it just doesn't go from 0 to 0, right? It has to do something in between that, Um but we note that sign of, um sign of two x is positive in this region. So let's find out when Sign of two of X is equal to one. So I'm gonna say sign. So let me put it right here. Sign of to vex is equal to one. Yeah, And that happens when to vex is equal to pi over two or when X is equal to pi over four. So at pi over four, sign of two of X is gonna equal one. So let's use this information to draw our graph, so I'm going to have so I'm gonna have a pie or two here. Pi over four. I'm gonna have zero. This is my X axis is my y axis, and I'm going to have one right here. And so co sign of X is just going to go and let me draw one a little higher here. So I'm gonna draw up here. So one coastline of X is going to go from one to pi over 120 during zero to pi over two. So it's going to look something like this. Yeah. Sign of two vets is gonna hit one. So let me draw you not draw a wine. We just draw a kind of a dashed line here. So science to Vex is gonna hit one at Pi over four, and it goes from 0 to 0 during at the bounds of zero to pi over two. So it's gonna look something like this. So notice we have two regions here that these two curves create. We have regional one and region to. So now, since we have the graph of these regions, we now have to find out what separates this region right here. Right, Because we have this. We have this point right here that we're not sure what it is yet, and we have to figure out this point before we can take the area of these two regions. So what I'm gonna do is I'm going to, um I'm going to set co sign of X equal to sign a tube X. This is an intersection point, so I can do so. This is the way to find the intersection point between co sign of X assigned to vex in order to get sign of two vets into a co sign function. What I can do is I can do co Sign of X is equal to co sign the pi over two minus two bets. So now, since the truth functions are the same, I can set the arguments equal to each other. Now, if X is equal to pi over two minus two X. And when you saw this, you'll end up getting X is equal to pi over six. So this is this point right here. It's five or six. This is the intersection point. Okay, so now let's go ahead and find the area of these two curves. So are the region bounded by these two curves, so I'm gonna first, right. Um, some area is equal to one plus two right in the first. So the first region is gonna be the integral from pi over six. I'm sorry. From zero two pi over six, and we're gonna put the top function first. So we're gonna put, um, co sign of X first, and then we subtract sign of two X from that function, and then we're gonna have plus the integral from five or six to want a pi over two. And this time we're gonna switch the order of this would have sign of two X first, and I'm gonna subtract co sign of X from it. So this area, I'm gonna kind of doing this this part right here. So the area. So I'm gonna take the integral of these two functions or yeah, integrate these two functions and I'm get I'm going to get, um, sine of X minus. Um, actually, it won't be minus. It will be plus co sign of to Vex over to. Right. So this is my first, uh, first region, and it goes from 0 to 5 or six. And then we're going to add the second region, which is going to be, um, after integrating those two functions will get negative co sign of to vex over to minus sine of X and that that goes from five or six two pi over two. So now, after I plug in these bounds of the integral, I'm gonna get sign pi over six plus co sign pi over three over to I'm gonna put this in parentheses minus sign of zero plus co sign of 0/2, and I'm going to add, um I'm going to add, So I'm gonna plug in pi over. Tune in this in this interval or this function right here, and I'm gonna get negative co sign a pie over to minus, sign the pi over two, and then it's minus. So then I'm gonna plug in pi over six, so I'll get negative co sign Pi over three over to. If I can fit this in here, I'm gonna get minus sign of paper. Six. Yeah. So let's, um let's simplify this a little bit. So I forgot to move the thing up. So I plugged in five or six. So I plug in five or six and zero into this first, uh, in these first two in this first region right here. And I got sine of pi over six plus co sign pi over 3/2 minus sine of zero plus co sign of 0/2. So this is the first region right here. The second region is plus minus coastline pi over two minus. Sign pi pi over three and then minus, um, ways of pi over three. I think it's pi over two. I'm sorry. Since Pi over two right here and then minus negative coastline pi over 3/2, minus sign five or six. So all this is going to equal is science. So we have signed five or six. This is going to be equal to one half. So one half plus co sign Pi over three is one half as well, so it's gonna be plus 1/4 minus. Um um and then we have signed zero. That's gonna be zero co signing. 0/2 is one half, and then we have plus, um, negative co sign Pi over two. It's going to be negative. It's gonna be one half minus. Sign pi over two is gonna be one. So one half minus one, and then we're gonna have minus. And then we put it kind of under here. You're gonna have co signed pi over three, which is one half, so it's gonna be minus 1/4 and then, um, minus sign a pi over six, which is gonna be one half. So after simplifying these, uh, these values were going to get, um so notice one half this one half right here cancels with this one half and then we have negative one half in here. And then we have, um so this one half is going to so this one half is going So let's let's do it this way. So let's go. Um, one half So we'll have 1/4 over here, plus negative one half minus, um, negative 3/4 and then adding all this together, we'll end up getting one half. So after all that, we find out that the area of these two regions between the regions bounded by coastlines of exercise to vex of one and two. There's nothing but one half.

For this problem were given to functions y equals co Sign X, shown here in blue and y equals Sine X, shown here in green and were asked to sketch the region between the curbs and X equals zero and X equals pi of to So from X equals zero X equals pi of two. They're two different regions in like for area between the curves. For the first region, it's from X equals zero. So this point right here the second reason region is from this point, it's a pi over two. So to find this intersection point so we create these can create these two separate enter girls to estimate the area, we set the two functions equal to each other to find the intersection point. So we set co sign next equal to sign two x and find the X equals pi over sex and pi over do as well as further multiples and derivatives of it. So therefore, the first integral to estimate the area we from zero toe high over six of the function on top so co sign X minus the function on the bottom signed two X and then plus the next year of it, which will be from PI over +62 pi over two. The function on top signed two x minus The function on the bottom Co Cenex that's written out here The integral of co sign X is sine X and the derivative I mean the the integral of sine two x is negative. 1/2 co signed two x, which to figure that out you can kind of work backwards by thinking, Well, what's the derivative of like co sign two x and working that way to figure that out? Then the integral of sine two x is negative. 1/2 coastline two x in the integral of co sign. Next is Cenex. So we plug in pi over six for X from the first in a girl and the sign sign Pi over six is 1/2 plus. Then it would be co sign of two times pi over six, a co sign of pi over three, which is 1/2 then 1/2 times 1/2 is 1/4 minus parentheses and then we plug zero in for X. So the sign of zero is zero and then the coastline of zero is one time for 1/2 is 1/2. Then we add the next integral to it. So you plug in pi over two for X for the second. Integral. So the co sign of two times pi over two is just the co sign of pie, which is negative one times negative, 1/2 assed positive, 1/2 minus the sign of pi over two, which is one so minus one minus parentheses that we plug pi over six in for X. So be co sign of two pi over six or co sign of pi over three, which is 1/2 times negative. 1/2 this negative 1/4 minus and then sign of pi over six, which is 1/2 adding. All this together we get the area to be 1/2.


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