solve-the-given-problems-find-the-area-of-the-parallelogram-with-vertices-at-0020-2-1-and-4-1-by-i-2

Question

Solve the given problems.
Find the area of the parallelogram with vertices at (0,0),(2,0)
(2, 1), and (4, 1) by integration. Show any integrals you set up.

Stanley Enemuo · Accepted Answer

Okay. Uh huh. So we want to find the area between the curve Why is equal to X cubed -2 x squared minus x. Close to and why is equal to X squared minus one. With the conditions that access Less than or equal to one. So the area the area is X cubed -3 x squared minus explosive story.

Kenneth Kobos · Answer

So we have these three points which are going to form a triangle, and we want to determine the area of the enclosed triangle. So let&#x27;s put the points 20 02 minus 11 No, we collect, We connect them to find the triangle and we want to determine the area of this enclosed region here. To do that, we&#x27;re going to need the equations of the line of the lines. So this one over here is why equals the slope is one. So why equals X and the Y intercept is too. This bottom one here, why is evil to the slope? Here is negative one so negative x and again the Y intercept is too. And this last one is why is equal to the slope? Here is negative one over three x plus some B values. So let&#x27;s figure out to be valued. We know why equals negative 1/3 x plus be We can plug in any point on this line. So, for example, we have this 0.20 so we can just plug in zero equals negative one over three times two plus b and then what we get is 2/3 equals B. Okay, so negative. 1/3 X plus two over three. Great Know what we want to do is integrate. To figure out the area area is equal to integral. So first word we&#x27;re going from X equals negative. One toe X equals zero. It&#x27;s going to be top function minus bottom function. The top function is X plus two X plus two minus the bottom function, which is negative. 1/3 X plus two or three negative 1/3 X plus two over three d X plus. Now we&#x27;re integrating from zero two x equals two. It&#x27;s going to be the top function, minus the bottom function. Top function is negative. X plus two. Uh, over here, that&#x27;s negative. X Plus two negative X plus two and then we&#x27;re subtracting the bottom function, Which again is the line negative. 1/3 X plus two over three d x. Okay, once we calculate, all of this will have the area. So let&#x27;s just simplify things a little bit First. Integral Negative 1 to 0. Here we have an okay, let&#x27;s use a different color. Here we have X plus, wondered X. So that works out the 4/3 x then we have a plus two and then we have a minus 2/3 s. So that works out to plus four over three DX plus integral from 0 to 2. Let&#x27;s simplify this a little bit. We have a negative X plus a 1/3 ex that works out to negative 2/3 X. Next, we have a plus to minus 2/3 which is plus for over three. And in all of this DX. So now we just carry out the intercourse. The anti derivative of the first integral is for over six x squared, plus four over three acts going from negative 1 to 0. That&#x27;s zero plus, the anti derivative here isn&#x27;t negative to over six X squared plus four over three tonnes X and this is going from 0 to 2. Now we just plug in the numbers so plugging and zero here just gives us a zero No plugging in a minus one. Here we have a four over six, which is 2/3 uh, plus for over three times minus one. Next we plug in to two. Squared is four. That&#x27;s an eight over six. Ah, plus, he re plug in to hear we have eight over three, and then we subtract and we plug in zero, which gives us a zero. Okay, so the zeros air gone. And now, after we simplify everything we end up with. So this 1st 1 is going to give us A to over three. And we have also plus four over three, which works out to six over three, which is to so the area then closed region is too.

Robert Daugherty · Answer

Section 76. Problem number 44. So here I am asked to find the area between a curve which you see given here, one over the square root of X square monies to X plus two area between that curve, the X axis when x goes from 0 to 3. So let&#x27;s take a look at it first. So what you see here is the graph of the the initial curve. What you see here, I&#x27;ve got X goes from zero 23 And so I&#x27;m being asked to find this shaded area, which means, basically, I need to find the integral from 0 to 3 for this function, and it turns out to be about 2.3 to 5. So I&#x27;m gonna get an exact value, and it&#x27;s approximation should be very close to ah, that number. So, in this case, what I&#x27;m doing, I&#x27;m using, um, integral look up tables. And so let&#x27;s just take a look. I&#x27;ve been using the table that you see here that referenced, um, and so I go and I look one up and this is illustrates. You got a sort of be on top of your game here. There&#x27;s a typo in this particular problem. So you see this Actually, what this one is is this is the integral of the one that I&#x27;m looking for One over a x weird plus b x plus c d x. They forgot to put the square root sign. And this works when a is greater than zero. So this is actually what I&#x27;m looking forward. So all the key reason that you&#x27;re gonna have to really know how to integrate you can&#x27;t rely always on integral table toe work it for you. So let&#x27;s go back and let&#x27;s just write down What? That solution Ah, would be, um so according to our interval look up table, I know that the indefinite integral one over square root X squared plus B x plus See the X is going to be equal to one over the square root of a natural log to a X plus B plus two square root of a and in the square root of a X squared plus BX plus c And then all of this, plus a constant. And this is when a is greater than zero. So in my case, I need to do this particular problem where a coefficient of the quadratic term is one be coefficient of the linear term is negative. Two. And see the constant term is positive, too, so I would need to substitute those values in. So just using this formula, you&#x27;ll have one over the square root of one natural log two times one times acts plus negative, too, plus two times one, um, and then the square root of X squared, minus two X plus two. And then all of this is evaluated from 0 to 3. So let&#x27;s clean this up just a little bit. This is the natural log of two X minus two plus two square root of X squared on its two X plus two. Evaluate from 0 to 3. So let&#x27;s do this evaluation. Calculus is over. We&#x27;re into arithmetic now, so substituted three into that expression so you&#x27;ll have the natural log. And when you substitute a three, you have six minus two plus two. You substituted three. You&#x27;re gonna X square. That&#x27;s 99 minus six is 33 plus two is five that when you substitute a zero, you&#x27;ll have natural log zero minus two plus two, and then the square root of to. So what we have now we&#x27;ve got the natural log of four plus two, Route five minus the natural log. And I&#x27;ve got to root two minus two. If I put all of these together, this is the natural log four plus two, Route five divided by 22 minus two. In fact, or two out of the numerator and denominator, You&#x27;re gonna get to two plus Route five over to, um Tu minus one. These 2/2. That&#x27;s going to be one. So this final answer is going to be the natural log two, plus Route five over route to minus one. Substitute that into your calculator. That is about 2.32 50 And then how does that compare? You see, 2.3 to 50 So we were correct in making that work. So again, the key here Waas. I needed to graph it to actually see what&#x27;s happening between zero and three to tell me that the integration will indeed find the area under the curve. Then I write down this integral and I go to my integral table. And in this case, there was an error in my integral tables and hopefully years will not have that. Um, correct that Arab used the formula substitute, do the arithmetic and get down to the final answer here.

Kenneth Kobos · Answer

So we have these three points and they&#x27;re going to form a triangle, and we want to find the area of the enclosed triangle. So let&#x27;s just quickly plucked. Explains zero zeroes at the origin. 31 is around there, and 12 is up here. So we see that we have this triangle. If we connect these points and we want to determine the area of this triangle Ah, using calculus. So we&#x27;re going to be integrating. Of course. We&#x27;re going to need to know, uh, so going from over here to hear, we&#x27;re going to need to know the top function. We&#x27;re going to need to know this function. So this line and we&#x27;re also going to want to know this line. So determining these equations of these lines is should be quite simple. Uh, because they&#x27;re all in the form, like was an explosive. Be so here. The slope is too two acts, and we&#x27;re touching the origin s. So that&#x27;s that equation here. The slope is we&#x27;re going from zero 23 So here this slope is 1/3 X and there&#x27;s no be value. Since it&#x27;s touching the origin here, the slope is negative. Half y equals negative half X. But of course, there is this be value, since it&#x27;s not touching the origin. But I&#x27;ll owe let you figure out how we get that plus meet. It works out to be plus five over too. Yeah, we saw for that because we know it&#x27;s why equals negative Half X plus some B. And then we know that on this line we have, for example, we could take this 0.1, too, so we can just plug in the values two negative half one plus B. This lets us suffer B equals five over, too. That&#x27;s how we get the the five over to here. Five over two. Okay, so now we have the equations of our lines. Uh, we have this one, this one and this one and then So what we need to figure out our, um, is the area. So let&#x27;s do that area equals integral. So we&#x27;re going to break up the integral here at X equals one. So the top line overhears two x and the bottom is 1/3 ex. So our first integral is going to be top two x minus 1/3 x on our limits of integration is We&#x27;re going from X equals 02 X equals one. So we need to put those limits there. This is D X plus into go Now. We&#x27;re going from one too. I believe it was three yet 123 over here. And our top function is negative. Half X plus five over too. So negative half X plus five over too. And then we&#x27;re subtracting the bottom function, which is 1/3 ex. So we&#x27;re going to subtract one over three x d. X now, before we carry out, these inter girls were just going to simplify them a little bit. Uh, integral 0 to 1 two x minus. Wondered X two is six over three. So we have five over three X, the ex plus integral from 1 to 3. Oh, here we have this five over, too, and then we have a negative 1/3 and then we have a negative half. So those workout too? Um, 1/3 plus 1/2. So to over six plus one over. Plus three over six. That works out to five over six x dx. Okay. And now let&#x27;s just carry these intervals. So the anti derivative here is five over six X squared. We&#x27;re going from 0 to 1. Then we add the anti derivative. Here is five over to X minus five over 12 X squared. And this is going from 1 to 3. Now we just plug in our numbers. So here we have five over six, minus zero plus plugging in three. We have three times five is 15 over too. Minus plugging in three again. We have 88 times five is 40 40 over 12 and then we plug in one. So five over too. Um, minus five over 12. So after we simplify all of this, we end up with the value of five over too. So five over two is the enclosed area.

Amanda Stone · Answer

in this problem were given the coordinates negative to zero negative to negative four, 40 and seven negative four and were asked to find the central an area of this figure. So if we graft this, we get a trapezoid that looks like this. And to find this century, we take the average of the X values an average of the Y values. So in this case, are X values are negative to negative 24 and seven. So if we plug this in, we get native to minus two plus four plus seven, divided by four. And if we do the same thing for y values, we get our Y values as zero negative 40 and negative four. So we just have zero minus four plus zero minus four, divided by four. So if we reduce this a little bit negative two minus two is negative. Four four plus seven is 11 s. We have negative four plus 11. Which leaves us with 7/4 and for why Value? We just have negative for minus four, which is negative h over four. So if we reduce this, we can&#x27;t reduce our X value 7/4. So we just leave that, as is negative 8/4 also equals negative too. And this is the central oId. So if we go back to our diagram and we go to 7/4 and go down negative too, it&#x27;s gonna be about right there, and that is going to be the center of our diagram. To find the area, we&#x27;re just going to use the equation for the area of a trapezoid, which is a plus B over two times H where a is this hop portion of our trapezoid. B is the bottom, and H is our height. In this case, A is equal to six. B is equal to nine and H is equal to four. So if we plug these values for A B and H into our equation, we get a equals six plus nine over two times four, which equals 15/2 times 4/1. So we can cancel out to from these two terms, and we&#x27;re left with 15 times to which equals 30 as our area, and that&#x27;s it

Wen Zheng · Answer

the problem is used, given transformation. She already begin to grow. You are is a parole. Aground with braces two one three one nine Q three three Nicked one on one off the transformation. This that&#x27;s going to work, Force House likes you. Asked we Why is able to our force Tom&#x27;s remind us three years? The first the region are It&#x27;s funny to buy. Why you going to axe us war? Why he could to Pax plus two why he could to connective reacts. Why could connective reacts half eight? By the transformation we have the region as its equal to you&#x27;re mean you is between active or connect you one connected to these between zero and it on then partially exp lai. Why should you be if they want to my force So we have to grow r Well, ax Plus it. Why? Hey, it&#x27;s the control integral from zero to eight into profound negative. Or make two one negative too. And full ax. Ask it why you seek or two three minus for you, Holmes, about a horse you you feel which is controlled by force into girl found zero to eight six re plus already. I mean and the answer is one o it

Problem

Solve the given problems. Set up the integrals (d…

Problem 35 Easy Difficulty

Solve the given problems.
Find the area of the parallelogram with vertices at (0,0),(2,0)
(2, 1), and (4, 1) by integration. Show any integrals you set up.

Answer

2

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