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Point) Consider the area shown below: The top curve (in blue) is _ for Ay_Vx, and the bottom curve (in red) isy = x and we have used the notation Dy(Click on the fi...

Question

Point) Consider the area shown below: The top curve (in blue) is _ for Ay_Vx, and the bottom curve (in red) isy = x and we have used the notation Dy(Click on the figure for a larger version )Write Riemann sum for the area; using the strip shown: Riemann sum = 2Now write an integral that gives this area fz where aand bFinally; calculate the exact area of the region, using your integral area

point) Consider the area shown below: The top curve (in blue) is _ for Ay_ Vx, and the bottom curve (in red) isy = x and we have used the notation Dy (Click on the figure for a larger version ) Write Riemann sum for the area; using the strip shown: Riemann sum = 2 Now write an integral that gives this area fz where a and b Finally; calculate the exact area of the region, using your integral area



Answers

Write a Riemann sum and then a definite integral representing the area of the region, using the strip shown. Evaluate the integral exactly.

The question gives us this triangle and ask us to find the area using a remand some with vertical rectangles. So first we're going to start by putting this on a coordinate plane just being ry access. This is our origin and this is our X axis. So now we need to find the function that will give us this line here in order to find the height of our rectangle. So we can see that it's just a linear function. So we can find the slope by using the rise of a run. Our rise is negative three and the run is six. Source slope is equal to negative one half and we can see that the X. Or the y intercept is at three. So the function is shifted up three. So our function is going to look like negative one half X plus three. And so now that is going to give us the height of this rectangle. So we can put that into our remained some, which is first going to look like the limit as and approaches infinity. The some from I equals one two. And so now again we need the length times width of the strike tangle and our length is going to be negative one half X plus three. And our with it's going to be delta X. So this is our remit some. And now in order to put this into a definite integral, we can do the definite integral from 0 to 6 on the X. Axis of negative 1/2 x plus three delta X. And now thanks to the integration some and subtraction rules. We know that we can find the anti derivative of these two terms separately and put them together in our and a girl. So We will have the anti derivative of negative 1/2 x. Which is equal to negative X squared over four plus The anti derivative of three which is three X. And this is from 0-6. And now we can plug in and solve so we'll get negative six squared over four plus three times six and that's all minus negative zero squared over four plus three times zero. And so this right side is going to cancel out since we have all zeros. And so we're left to solve the left, which is going to be negative 36 over four Plus three times 6 is 18. So we have negative nine plus 18 Which gives us our answer nine. And this is the area of the triangle.

We want to construct a remand some using horizontal rectangles in order to solve for the area of this triangle. So we'll start by labeling the length of our strip as W. And filling in this missing length here From W. to the top of the Triangle as five -H. And so now we can imagine that W. Is the base of a second triangle that is similar to the original triangle Since it has the three angles in common. And so since we have a second, similar triangle, we know that the proportion of the basis of these similar triangles will be equal to the proportion of the heights of the similar triangles. So in writing that we would write W over three is equal to five minus H. The height of the smaller triangle To the height of the larger triangle five. And so now if we multiply both sides by three, we get an equation that will give us the length W for some height H. So rewriting this, we get 15 -3 h over five. And that leaves us with 3 -3/5 H. So this is the equation for the width of our horizontal strip and when we put it in our remote and some will multiply it. Bye the hi of our strip in order to get the total area of the strip which is given over here as delta H. And putting that into the form of agreements. Um We have the limit as N approaches infinity Of the some from I equals 1 to end. And so this is our remained some. And now constructing a definite integral, we'll have the definite integral From 0 to 5 using the total height And we will drop down the 3 -3 5th H delta H. And now we can solve this integral. We'll get The anti derivative of three is 3 H. And the anti derivative of 3/5 is three H squared Over five times 2. This is from 05, so we get three H minus three H, squared over 10, From 0 to 5. And now we can plug in are bound zero and 5. So we get three times five minus three times 5 squared over 10 minus three times 0 zero three times 0 squared over 10. And now these terms are going to cancel out since they're multiplied by zero. So we're left with 15 minus three times 5 squared is 75/10 And we're left with 75/10 or 15 halves. And this is the area of the triangle.

Okay, This question wants us to evaluate this integral using geometry. So we're just going to graph the function and find the area under the curve. So we're just graphing the absolute value function, but it's shifted and reflected, so we start at positive one, and it's just gonna be a V opening downward and are integral goes from zero to to. So now we can see we have two different areas that we have to deal with. We have the green triangle, and then we have the Red Triangle. So we just need to find the area of each of these. And then we can add them together, making sure to subtract the Red Triangle because it's negative. So we need the base and the height. So it's good to know what point this is, because that's where we change from positive negative. So we're looking for where one minus the absolute value of X is equal to zero, and that happens at the point were negative. Absolute value of X is equal to negative one, so the absolute value of ax equals one. So X is just plus or minus one. And if we needed to this points negative one. So now let's just find the heights. So one height is just the Y intercept at 01 and the other height, we just plug in that just plug in to to get one minus the absolute value of two. So negative one. And now we can just plug in so 1/2 base times height for each triangle to find this. So our area is just gonna be the difference of these two triangle areas which we confined because the 1st 1 has a base which runs from 01 So it has a base one, and its height is the Y. Intercept at one. Then we're subtracting base to which runs from 1 to 2. So it has based one, and then the height is down one so also won. And we already accounted for the subtraction. So we just have 1/2 times zero or zero. So we have zero net area, and that makes sense looking at the graph because coming down here, the Red Triangle is

Okay, This question wants us to evaluate this integral using geometry. So we're just gonna find the area under the graph between negative 12 So let's just draw some axes here. And first we know what the absolute value graph looks like. Well, we have a negative sign in front, so it's just Z. But pointing out downward and are integral is going from negative one positive too. I'll move this over a little bit for scale. So if we want the area under the curve here we have this first area between the origin and then similarly, we have another area for the positive X values, but both of them are below the axis, so we're gonna have just adding to negative areas together. And it's not too difficult to figure out this area because each of these are just triangles. So are integral is just equal to the some of these two triangle areas. And we're gonna call the heights negative to get our area signed to match and then all factor 1/2. So for the first base, our triangle goes from negative 120 So it has a base equal to one, and then the height. Well, if we plug in negative one into our function, we get negative one. So the height of the triangles, negative one, then for our second triangle were running from 0 to 2. So the basis to in the height is the ab negative. Absolute value of negative, too. So negative, too, because we just evaluated in the graph to find these points. So again, we know this point is too negative too. And this point is negative. One negative one. So we can just find the area is 1/2 times negative. One minus four or negative five halves.


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