sketch-the-region-enclosed-by-the-curves-and-find-its-area-y2x-1-quad-y-frac15-x7

Question

Sketch the region enclosed by the curves and find its area.
$$
y=2+|x-1|, \quad y=-\frac{1}{5} x+7
$$

Gregory Higby · Accepted Answer

Okay. So, um, the absolute value function that they give you is ah, shifted right. One up. Five. Um, and that should make sense because the when you have X minus one an absolute value and then plus five on the outside. Oh, it&#x27;s well, dyslexia here, I&#x27;ll shifting up to so scratch what I drew earlier. It&#x27;s right one up, too, Ariel right here and some v shape can kind of ignore that. That&#x27;s awful bit, um, and then the other equation is negative. 1/5 X plus seven. Um, so let&#x27;s say that 67 and that&#x27;s down one right, five and then up. One left, five to get a curved look. Something like this. Well, a straight line. So if you want to see where these two curves intersect when I said I&#x27;m equal to each other. So this right equation is basically the line X plus one, which makes sense if you just, you know, enabling. Plus two you get plus one. If you said that equal to *** 1/5 X plus seven and then say you add 1/5 over. So that would be 6/5 x and subtract one over here six and then multiply by the reciprocal to solve for X, you get five. So that&#x27;s gonna be my upper bound right here. Uh, at X equals five Vertical line right there. And then you want to do the same thing with this equation, which is negative. X negative. X plus three can see the Y intercept. There&#x27;s other ways of figuring that out. You want to set that equal to that negative 1/5 X Plus seven Really the same work, but you add 1/5 this way, which make it negative. Four fists Ex Subtract three This way. Seven. Maestri&#x27;s four and then the same thing multiplied by five negative force times five Negative force. These cancel so X equals negative five. So I have two different colors because I wanted to, um, show the green area. Um, being at X equals negative five. So what I want is the in a role of this, which would be from negative five until one. Because that&#x27;s where the equation change of the upper function being that negative 1/5 X plus seven minus a lower function, which is the one I wrote earlier that negative X plus three oops, got ahead of myself and ah, when you distribute that negative in there, it&#x27;s gonna become minus three. So as we combine, if we just simplify that, um, negative 1/5 plus acts will be forfeits. Um, plus four looks very similar to what we had earlier from number 5 to 1 dx weaken. Do the and I deserved of of this, we should be adding one to your exponents. Dividing by two makes it two fists plus four X and that goes from number 5 to 1. Plugging in one into all that you get to fist plus four and then when you plug in negative five into that whole negative five squared is positive. 25 divided by five is just five times two is 10 um, plugging in five or 30 minus 20. So it would get there is probably easiest if I just simplify four minus negative tense. That&#x27;s 14 and two fists. I&#x27;m just gonna leave it like this for right now. So now what I need to dio is this half, which is going from X equals 1 to 5 of a completely different, you know, problem. And then we&#x27;re gonna add those together. So that&#x27;s from 1 to 5. The upper functions still the native 1/5 X plus seven. But now, when I subtract off its ah this equation the X plus one equation that we found earlier, some minus X and it&#x27;s actually minus one because of you have to distribute that minus and so really looks similar to the previous problem. When you combine like terms, Um, from 1 to 5 d x so very similar you add one to your exponents divided by two makes that negative. Three fists, Um, plus six X, and that&#x27;s from 1 to 5 as a plug in five. And for all that, five squared is 25 divided by +55 times negative. Three is negative. 15 plus six times five is 30 and then minus plugging in one. I feel like I&#x27;ve over explain this problem, but stick with me. I am almost there. Um, the best way to do this, though, is to, you know, you get 15. There it becomes minus six. So that&#x27;s nine and three fists. And as I had a plus sign up there somewhere, as I add these together to fist plus three fists is one plus nine is 10 plus 14 is 24 way too much work, but that&#x27;s correct.

Ernest Castorena · Answer

the region, defined by the intersections of two curves Y is equal to X squared minus one, and why is equal to seven minus expert? Okay, let&#x27;s find out when the intersects we&#x27;re gonna set them equal to each other. X squared minus one is equal to seven minus X squared. Let&#x27;s see, bring these over to the other side. Two X squared is equal to eight. Our X is equal to plus or minus two. Okay for those of their two and a section points, let&#x27;s see 111 function be able, the other on their sexually. One is right now. One is going down. It looks like we would always have the negative experience on top. So let&#x27;s compete that area from negative to two. That&#x27;s the whole region of an infection. We&#x27;ll have seven minus X squared, minus six way plus one. So I subtracted the other carefully. Marigot, and take the definite any role. Let&#x27;s see, we&#x27;re going to have sexually to expert a so eight x Len. It&#x27;s 2/3 execute evaluated at negative two and two. Then we&#x27;ll we&#x27;ll get Let&#x27;s see 16 minus 16/3 minus. Let&#x27;s see. Negative 16. Well uh, 16/3. So we&#x27;re gonna end up with, Let&#x27;s see equals two minus 32/3. Well, let&#x27;s see. Multiply by 353 We&#x27;re gonna end up with 64/3, a little over 21.

Gregory Higby · Answer

Okay, so, um, if you were to graph this, um, I believe it&#x27;s to over one minus, uh, one plus x squared, And then the absolute value function looks something like this. Yes, I&#x27;ve come put a rides on this. Um, so that&#x27;s the absolute value of X s. Oh, what I would do because these air even functions. I hope that makes sense to use. I would just find this in a girl and then double your answer. Um, so how do I find that integral is? Ah, it&#x27;s the general from zero. Because that&#x27;s C X and ah, were these two intersect? Um, is that X equals one? Because if you pull you one for X squared is one plus one is two to divide by two is one in the absolute value of one is still want eso The upper function is that to over one plus x squared minus. This value is just, you know, x at this point, um, because of ah, you know, we dropped the absolute value because we&#x27;re only looking at positive values for X. And so then when I need to do as I mentioned earlier, is double that answer. So Now we can go ahead and take the enrolled that so That too, is still in front. So the I hope you&#x27;ve already learned this otherwise it would take me forever to teach that that anti derivative is the inverse tangent of X. You can just google. What is the integral of 1/1 plus x squared? If you didn&#x27;t know that now this too is just a constant, so we can bring it in front. And then this in a girl, you add one to your exponents and then you divide by that new exponents from 0 to 1. So what do we have is now we need to plug in one and for all that. Um So where on the unit circle? Um and then to inverse tangent of zero. I don&#x27;t actually write in zero for this one. Actually not right for that one either. Zero squared is still zero. So I ran out of room. But where in the inner circle, this tangent equal one. Well, that&#x27;s where signing coastline are both the same. That&#x27;s that route to over to route to over to And that&#x27;s the radiant measure because that&#x27;s what this question saying in verse. Tangents. What&#x27;s the radium as your pi? Over four. So what we have is two times to hire or four minus one squared is still one times one half is one half now. Same question. Where is Inverse Tanweer&#x27;s Tanja equals zero. Well, that sign over co sign. And that&#x27;s the radium measure. Zero. So this happens to be zero as well, son. Just not gonna write it. So from here, just distribute that to enable two times two is four. Divided by four is canceled. So I just love to pie in two times. One half is one is our answer on this.

Sky Li · Answer

again. Let&#x27;s draw those two curves first. So say this is our acts My plane in the first Carvey&#x27;s actually, uh, he&#x27;s just a squared X move A shifted to the right by one. So we know, um, a square hotel fax It&#x27;s look like this. So he&#x27;s shifted by once we start from one, and it goes like this. The second is the street line X minus Y equals one. So but why go to zero X equals one? So across this point and x zero by cosa naked across this part our street flying is here. This is our street line. Next, minus y. You close one. Okay. Ah, and then close area is the shaded regions Now, Alice, trading for in the area. Um, you know, to find this outer area, we need to take the integral by the respect, respect to X or respect why it depends on ah how you represent those functions. Say, since I Personally, I don&#x27;t like this square root because if you remember the square root of something, the anti duty off that is it&#x27;s very difficult to remember and looks ugly. So here I&#x27;ll reform the first function. So I take square on both side I&#x27;LL get up Thanks. He closed Tio one plus y square. Hey, And remember over, um our wise quater than zero. The greater the co two zero right, Because the white goes to a square root off something So it always known Active on this since the first function I represented by Why this I also want to represent But why? So I do Here is some I mean why to the other side. So the second function will be activity close to one plus one. Right? Um after this so many in a row with respect. Why? And let&#x27;s find hungry No, a star from zero. And what about this point? The basis our way too. That was so bad off. Why To just say this happens when y plus by square yukos one plus y And we know why too cannot be zero. Which means I lied too. Ecos water because this equation will give us to swoosh is one of them yourself. One and another is zero since white who is not zero. So it has to be one on our boundary with coast from zero to one in the things I the integral. Always red curve minus the left curve or red car abuse of a bus boy. And now our left curve. Here is this one plus life work. Okay, now, Andy. Dear God, this is one half those one canceled. So the anti dirty off. Why is what? Half white square? Minus one cube. My third one like you. Hey, you graduated and in one zero. So Plugin Lycos one. This is what Half minus one third of the cheese. One six in the second term zero. So fix Ivancic will be our answer.

Ernest Castorena · Answer

lost on the area between the two curves. It looks like for 12 years we have let&#x27;s say that&#x27;s one y one y two y ones on the top that&#x27;s in the region from 0 to 2 that anyone and then it reverses in the habit. We want to give you the whole area, but they appeared to be even. Function are on functions. So instead of just putting one over the other, we can just double the inner role of one. So we&#x27;ll have 02 uh, five X over one plus X squared minus the lower curve minus six. The X now I could bring that inside. So we have There are two find two x one plus six for minus two x the eggs. So that&#x27;s computer indefinite. Integral to get area of Peter, evidence will be seen Alan of one plus expert amulet. If I That makes sense. Very general, but I and love minus X squared, evaluated at zero and to see if we evaluated at two. We&#x27;ll get find Helen of by nine. That&#x27;s for subtract five Allen one, which is just zero. My net zero squared to zero. So the area between our career of intersections is five Island five, minus four

Gregory Higby · Answer

Oh, it&#x27;s ah, it&#x27;s problems. Pretty straightforward if you memorize everything. So this is the one over the square root of one minus X squared function. And then why close to is like right here. So the area that we&#x27;re looking for just here. Um, now, if I were doing the problem, I would just do half, because this is an even function. I would just do the integral from zero to, I believe this X coordinate where they intersect. Um, What? You can find it by setting them equal to each other. Because it&#x27;s the equation. Y equals two. I forget to write that down. So you said them equal each other and you square both sides. You have 1/1 minus X squared equals four. Then you could cross multiply, I guess so that before minus four x squared. So subtracting for being negative 33 bores, Um, because you have to divide the four over. So I forgot to mention that so square root that and then when you squared it for you get to, um So my bounds, we&#x27;re gonna be zero to root. 3/2, um, of the upper function being to minus your lower function the the fancy one that they gave you. And don&#x27;t forget, because I almost forgot to hit to write down double that this is will help us find the area. Um ah, yeah. So Ah, this would be a waste of time to rewrite that. I don&#x27;t know. No thinking. Um, So the anti derivative of this, I would be two X and then hopefully you&#x27;ve seen this before. That Andruw of his inverse sign of X. You can just quickly google that if you haven&#x27;t learned it before. Um, and that&#x27;s plugging in from zero to root 3/2. So as we plug this in, you said that to in front. So it&#x27;s double route three over to you can see things will start to cancel minus inverse sign over 3/2 and then minus zero times to still zero. Um, and then they turned this into a plus because it&#x27;s subtracting the negative inverse sign of zero. Um, but if you look on the unit circle, where to sign equal Route 3/2 will signs the light coordinate. That&#x27;s at this order, payer. One half brewed, 3/2. So we want that one. So if you&#x27;re not familiar with the unit circle, that&#x27;s pi or three by over three. And then Warda signing zero. That&#x27;s the Y Coordinate right here in the radiance. Right. There is zero. So this would happen to be zero as well. But you don&#x27;t Never, never heard Siddle check. Um, you might just leave your answer like that. Some teachers prefer it that way. Or distribute that to in because of my right, I only did half on the area. I went from zero to root, 3/2 sided double over here, um, to buy or three. But I like this answer better.

Problem

Sketch the region enclosed by the curves and find…

Problem 17 Medium Difficulty

Sketch the region enclosed by the curves and find its area.
$$
y=2+|x-1|, \quad y=-\frac{1}{5} x+7
$$

Answer

$$
24
$$

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Calculus Early Transcendentals

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Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals