Understanding Area Under Curves with Riemann Sums

In this video, we're going to focus on finding the area under the curve using Riemann sums. Using left endpoints, midpoints, right endpoints, sigma notation and limits, and also by evaluating the definite integral. So let's begin. Let's say if we have a function f of x is equal to, let's say, x squared plus 1. So this is basically a parabola that starts at 1 and opens in the upward direction. But let's focus on the right side of this problem. So let's say it looks something like this. Let's extend it. Now let's calculate the area from x equals, let's say, 0 to x equals 2. So how can we find the area of this shaded region between the curve and the x-axis? So one way in which we can do it, in which we can get the exact answer, we can find the area by finding the definite integral from a to b of f of x dx. Let's go ahead and do it this way so you know what the exact answer is, and then we'll talk about other techniques of getting the answer. So this is going to be the antiderivative from 0 to 2, f of x is x squared plus 1 dx. Now the antiderivative of x squared... is x cubed divided by 3. The way to get that answer is to add 1 to the exponent, 2 plus 1 is 3, and then divide by that number which is 3. Now the derivative of 1 is equal to 1. The antiderivative of 1 is x because the derivative of x is 1, and this is going to be evaluated from 0 to 2. Now this is going to be equal to, we need to plug in 2 and then 0. And I'm running out of space, so let's make some space. this is equal to plug into first the top number into x so it's going to be 2 cube divided by 3 plus 2 and then minus the lower limit of the integral which is 0 so 0 to the third divided by 3 plus 0 2 to the third is 8 And 0 cubed plus 0 is simply 0. So the answer is 8 over 3 plus 2. Which, let's get the decimal value for that. So this is equal to about 4.67. The exact answer is 14 over 3. Now, how can we approximate the area using Riemann sums? Using the left, the right, and the midpoint? Well, let me give you a general idea of what we're going to do. So, here is the function, just the right portion of it. And we want to find the area between x equals 0. and x equals 2. Now the way to do this is to basically find the area of rectangles that are beneath the curve. It can be above or below the curve. Since it's below it, this is an under approximation. So we want to find the area of those rectangles and add them up. That's how we can approximate the area under the curve using Riemann Sum. Now let's focus on an individual rectangle. The width of the rectangle is known as delta x. The height of the rectangle is the y value, or f of x. y is equal to f of x. So that's just the area of one rectangle. Since we have four rectangles, we need to add up the area of each rectangle. So the area under the curve is basically the sum, the sigma notation represents sum, of all the individual areas of each rectangle. Now let's say this is f of x of i, i equals 1, and... it stops at n. So let's use four rectangles to approximate the area. But this is the generalized equation. And I do want to mention one thing. The area of a rectangle is basically the length times the width. The width is delta x. And the length is basically the height of the rectangle, which is f of x. So f of x times delta x gives you the area of one rectangle. And the sigma represents the sum of the area of all of the rectangles. So let's find the area. of the function, or the area between the curve and the x-axis, of the function x squared plus 1, on the interval 0 to 2. Now these are x values. So a is 0, b is 2. Delta x, the width of each rectangle, is the difference between b and a, divided by n. So in this particular problem, b is 2, a is 0, divided by 4. So 2 over 4 is 1 half, or 0.5. That's the width of each rectangle. Now, if you want to find the area using the left endpoints, make a number line from 0 to 2. Now we have 4 intervals, which is basically going to be 5 points. And each interval differs by 0.5. So the first one is going to be 0.5, the second is going to be 1.0, the third one is 1.5. Now, since n is 4, we're only going to use 4 of those 5 points. Even though there's 5 points, there's 4 intervals, or 4 rectangles. So, if you want to find the area using the left endpoints, Choose four points starting from the left. 0, 0.5, 1, and 1.5. If you want to find the area using the right endpoints, choose the four points on the right. 0.5, 1.0, 1.5, and 2. Don't choose the first point on the left. Now, if you want to use the midpoint rule, choose the four points in the middle. 0.25, 0.75, 1.25, and 1.75. Now keep in mind, the exact answer is approximately 4.67. Let's see how close our approximation is going to be to that answer. So let's begin. So let's find the area using the left endpoints. So this is going to be the sum of delta x times f of x. And delta x. We said it's 0.5, but I'm going to write out the function first. So it's delta x times the sum of the height of each rectangle, starting from the left. So we're going to use 0, 0.5, 1, and 1.5, but not 2. So f of 0 plus f of 1.5, or 0.5 plus f of 1 plus f of 1.5. So we know that delta x is 0.5. Now, if we plug in 0 into this function, f of 0 is going to be 0 squared plus 1, which is 1. Now, what about f of 0.5? I'm going to use the calculator for this part. So, 0.5 squared plus 1 is 1.25. Now, f of 1, 1 squared is 1 plus 1 is 2. 1.5 squared is 2.25 plus 1 is 3.25. So let's add the numbers inside the bracket. So 1 plus 1.25 plus 2 plus 3.25 is about 7.5. Half of 7.5 is 3.75. Notice that the answer is not very close to the actual answer, and that's because we have a small number of rectangles, and it's 4. If I were to increase n to 8, for example, this answer would be more accurate. It would be closer to the true answer of 4.67. But notice that it's less than 4.67. That means that it's an under-approximation. The last endpoint. is usually an under approximation whenever the function is increasing. And for an increase in function, the right endpoint is usually an over approximation. The midpoint, using the midpoint rule, The answer is usually more accurate than the left endpoint or the right endpoint. So now, let's try the midpoint. So we can get rid of this. So the area, using the midpoint rule, is going to be delta x times f of, we need to pick a number between 0 and 0.5. That's 0.25 or... 1 4th plus the midpoint between 0.5 and 1 you can draw the interval if it helps this is 0 this is 2 this is 0.5 1 1.5 so we have the points 0.25 0.75 1.25 and 1.75 so this is going to be 0.75 and then 1.25 And finally, 1.75. So let's plug in the values. Delta X is 0.5, and 0.25 squared is about 0.0625 plus 1. So this is 1.0625. 0.75 squared plus 1. is 1.5625 plus 1.25 squared plus 1, that's 2.5625. And the last one, 1.75 squared plus 1, that's 4.0625. So let's add everything inside the brackets. So 1.0625 plus 1.5625 plus 2.5625 plus 4.0625. This is about 9.25, if my math is correct. And half of 9.25 is about 4.625. So as we can see, using the midpoint rule, we're going to get the most accurate answer. Now let's calculate... the area using the right endpoints. We know it's going to be an over approximation for an increase in function, but let's get the answer. So let's draw the number line first. So A is 0, B is 2, and N is 4. So we're going to have 5 points and 4 intervals. So using the right endpoint, we're going to choose 0.5, 1, and 2. So area, we're going to use Using the right endpoints is going to be delta x which is 0.5 times f of 0.5 plus f of 1 plus f of 1.5 plus f of 2. So let's start with 0.5. So 0.5 squared plus 1 is 1.25. 1 squared plus 1 is 2. 1.5 squared plus 1, that's 3.25. And 2 squared plus 1 is 5. So 1.25 plus 2 plus 3.25 plus 5 is about 11.5. And half of that is 5.75. Notice that it's well above 4.67. So let's write the answers that we have. So the area using the left endpoint was 3.25. The area using the midpoint, I forgot what that was. And for the right endpoint, it's 5.75. I believe this was 4.625. Let me double check. And yes, it's 4.625. Now sometimes, if you're given the area of, or using the left endpoints, and the area using the right endpoints, and if you want to approximate the actual answer, average the two areas, the left and the right one, if you add them up and divide by 2, the average is going to be closer to the true answer than these individual values. So, if we... Add 3.25 plus 5.75 and then divided by 2. Notice what's going to happen. 3 plus 3.25 plus 5.75 is 9 divided by 2 is about 4.5 and 4.5 is pretty close to 4.67. At least it's closer than AL and AR. So this is a better approximation than using the left or the right endpoints separately. But if you have the midpoint, that's good too. That's a good approximation. Now let's graph the rectangles and the curve so you can see how it looks like for the left endpoint, the midpoint, and the right point. So let's start with the left endpoints. I'm going to draw a relatively wide graph from 0 to 2. So let's break it into four intervals. This is 0.5, 1, and 1.5. Now, we're going to use the left-hand points. That means we're going to use 0, 0.5, 1, 1.5, but not 2. So, draw a point on the curve at each of those x values. So, at 0, it's going to be right here, and then at 0.5, 1, and 1.5. And then let's draw a rectangle at each of those points. Notice that the rectangles... are below the curve. So that's why we have an under approximation. That explains why the area using the left endpoints was 3.25. It was significantly less than the actual area of 4.67. So that's what we did in the beginning. We found the area of these four rectangles, and then we add them up, and we got 3.25. Now what about the right endpoint rule? How can we graph the rectangles for that situation? So let's begin by making a graph from 0 to 2. And let's make four intervals again. So this time, we're using the right endpoints, 0.5, 1, 1.5, and 2. So let's plot those points on the curve. And this time, instead of drawing the rectangle towards the right, draw it towards the left. So first, let's put a vertical line. Between the point and the x-axis and let's draw each rectangle towards the left So notice that we have an over approximation the area of the rectangle is above the area of the curve So that explains why the area using the right endpoints was a 5.75. That's why it's so much greater than 4.67. As you can see, there is a significant portion of the rectangle that's above the curve. And so that's why the answer is an over-approximation. Now let's graph the rectangles using the midpoint rule. so let's make the four intervals 0.511.5 and 2 so we need four points but using a midpoint so let's put a point on the curve at.25.75 1.25 and 1.75 Now to draw the rectangle, draw a horizontal line that passes through the point. So it's going to look like this. Let's draw another one. And let's draw another one. And one more. So notice that a portion of the rectangle... is above the curve. Also, there's a portion that is below the curve. So the portion that is above the curve and the portion that is below the curve, they sort of balance each other out. And that's why the midpoint rule is a better approximation than the left endpoint or the right endpoint. But that's how you would graph the rectangles if you ever need to. For the sake of practice, let's try another example. So consider the function f of x is equal to x cubed. Now, using four rectangles, estimate the area from 0 to 4 using the left endpoint, the right endpoint, and the midpoint rule. So first, let's find the actual area. It's going to be the antiderivative from 0 to 4, x cubed dx. And the antiderivative of x cubed? Add 1 to the x-pointer, it's going to be x to the 4th, divided by 4, evaluated from 0 to 4. So let's plug in 4 first, and then we'll plug in 0. 4 to the 4th power, that's 256, divided by 4, is 64, minus 0. So the answer that we're looking for is 64. So now let's approximate the area using the left endpoints to begin with. So first we need to calculate delta x, which is b minus a divided by n. a is 0, b is 4. So it's 4 minus 0 divided by 4. Delta x is 1. So let's make the number line. From a to b. 0 to 4 with a width of 1. So the width of each rectangle is 1. So we have 5 points, 4 intervals. So using the left endpoint rule, we need to use the first 4 points on the left, 0, 1, 2, and 3. So the area is going to be delta x, which we know delta x is the width of each rectangle, so that's 1. times f of 0 plus f of 1 plus f of 2 plus f of 3. So f of 0, if we plug in 0 into this function, 0 cubed is simply 0. 1 to the 3rd is 1, 2 to the 3rd is 8, 3 to the 3rd is 27. So 27 plus 8 plus 1. that's 36. So as you can see, this is a bad approximation. 36 is very far from 64, but at least that's what the answer is. Now let's use the right endpoint. Let me just mark this answer down. So al, the area using the left endpoints, is equal to 36. Now let's make another number line. So we could find the area using the right endpoints. So from 0 to 4, we're going to have the points 1, 2, and 3. But we're going to choose the four points on the right. That's 1, 2, 3, and 4, but not 0. So then a of l is going to be delta x, which is 1, times f of 1, plus f of 2, plus f of 3, plus f of 4. So 1 cubed is 1, 2 to the 3rd is 8, 3 to the 3rd is 27, 4 to the 3rd is very big, that's 64. So this is definitely an over approximation. So 1 plus 8 plus 27 plus 64 is 100. So AR is 100. Now let's average 136. So AL plus AR divided by 2. So 36 plus 100 is 136 and half of 136 is 68. So as you can see, if you average the left endpoint and the right endpoint, your answer will be close to the actual approximation. Now, let's approximate the area using the midpoint rule. So first, let's draw the number line. So we need the midpoint between 0 and 1, which is 0.5. We need 1.5, 2.5, and also 3.5. So delta x is 1, and this is going to be times f of 0.5 plus f of 1.5 plus f of 2.5. plus f of 3.5 so.5 raised to the third power is 1 over 8 or simply.125 1.5 to the third power is 3.375 2.5 to the third power is 15.625 and 3.5 to the third power is 42.875 times 1. So let's add the numbers on the inside. So 0.125 plus 3.375 plus 15.625 plus 42.875. So you should get a total of 62, which is very close to 64. Now, we said that if we increase the value of n, if we use more rectangles, the approximation will be more accurate. So let's use the left-hand points, and let's use 8 rectangles instead of 4 rectangles. And let's see if the answer is going to be closer to 64. So right now, it's 36 with 4 rectangles, but let's see what's going to happen if we use 8 rectangles. So let's calculate delta x first, which is b minus a divided by n. So it's going to be 4 minus 0 divided by the 8 rectangles, which is 1 over 2, or 0.5. Now let's draw the number line. So from 0 to 4, let's increase it in increments of 0.5. So 0.51, 1.52, 2.53, and 3.5. So there's 9 points, and we're only going to use 8 of them. We're going to use every point except the last one on the right. So we're going to use the 8 points on the left. So everything from 0 all the way to 3.5. But not the last 0.4. So the area is going to be delta x times f of 0 plus f of 0.5 plus f of 1. plus f of 1.5 plus f of 2 plus f of 2.5 plus f of 3 plus f of 3.5 so Delta X will know to be 0.5 0 to the third power is 0.5 raised to the third power that's a point one two five 1 to the third power is 1. 1.5 to the third power is 3.375. 2 to the third is 8. 2.5 to the third power is 15.625. Plus 3 to the third is 27. Plus 3.5. to the third power. That's 42.875. So now let's add the numbers inside the brackets. You should get 98. So 0.5 times 98 is 49. So notice that this is still an under-approximation, but 49 is closer to 64 than 36. So even though it's not very accurate, it's more accurate than the previous answer of 36. So as you increase n, the approximation is more accurate. It's going to be better. So far, for increasing functions, the area that was obtained using the right endpoints was an over-approximation, and the area that was obtained using the left endpoints was an under-approximation for an increasing function. Now let's see what's going to happen if we use a decreasing function. So let's say that f of x is 1 over x, and let's approximate the area, just the left and the right endpoints, from 1 to 2. And let's use four rectangles. So go ahead and try this example. Now first, let's get the actual answer. So let's find the antiderivative from 1 to 2 of 1 over x dx. The antiderivative of 1 over x is simply ln x, because the derivative of ln x is 1 over x. If you try to use the power rule, if you rewrite it, and if you add 1, you're going to get 0 over 0, which... Doesn't work. So you need to know that the antiderivative of 1 over X is simply ln X. And this is going to be evaluated from 1 to 2. So ln 2 minus ln 1 is going to be ln 2 is about 0.6931. ln 1 is 0. So the area is approximately 0.6931. Now let's start with the left endpoint. Let's calculate delta x. which is b minus a or 2 minus 1 divided by n and n is 4 so 2 minus 1 is 1 so this is 1 4th which is about 0.25 so now let's draw a number line so from 1 to 2 and let's break it into four intervals or five points so it's going to be 1.25 from 1.5 and 1.75 So for the left endpoint rule, we're going to use the four points on the left. And so the area is going to be delta x, which is 0.25, times f of 1, plus f of 1.25, plus f of 1.5, plus f of 1.75. So this is 1 fourth, or 0.25. Now, f of 1, 1 divided by 1 is simply 1. And 1 divided by 1.25 is about 0.8. 1 divided by 1.5 is 2 over 3, or 0.67, and 1 divided by 1.75 is about 0.571. So let's add 1 plus 0.8 plus 0.67 plus 0.571, and let's divide it by 4. So the area using the left endpoints is.76. Notice that it's an over-approximation because it's a decrease in function. AL is going to be an over-approximation for decrease in functions, but it's an under-approximation for increase in functions. Keep that in mind. Now let's approximate the area using the right-hand points. So this time we're going to use 1.25, 1.5, 1.75, and 2. So delta x is still 0.5, and 1 divided by 1.25, we said that's 0.8. 1 divided by 1.5 is 0.67, and 1 divided by 1.75, 0.571. And 1 divided by 2 is 0.5. So notice that the numbers are decreasing as we go from left to right. Therefore, the area using the right-end points should be an under-approximation for decreasing function. Because the numbers to the right should be less than the numbers to the left if the function is decreasing as you go from left to right. Makes sense, right? So now let's calculate the area. So let's add the four numbers inside the brackets. So 0.8 plus 0.67 plus 0.571 plus 0.5. That's about 2.541. And half of that is basically 1.2705. But let's round it and say it's 1.27. For some reason, that just seems too high. And the issue is delta x. Delta x is not 0.5. I made a mistake. It's 1 fourth. It's 0.25. So let's correct that mistake. So 0.25 times 2.541. That's 0.6353 or 0.635. So this answer is not too far away from this answer, but it is an under approximation. Now let's average the area calculated from the left endpoint and from the right endpoint. So this is going to be 0.76 plus 0.635 divided by 2. The average is equal to 0.6975, which is very close to the actual answer. So anytime you average the left and the right endpoints, you're going to get a better approximation. Sometimes you may want to find the area using the definition of a definite integral. The definite integral from a to b of f of x dx is equal to the limit as n approaches infinity Then we're going to have this sigma notation where i is equal to 1, which stops at n, f of x sub i times delta x. Now, as mentioned before, delta x is simply b minus a divided by n on the interval a to b. Now you may wonder what is this X sub i? It really depends on if you want to find the area using the right endpoints or the left endpoints. So using the right endpoints, X sub i is equal to a, that's this a, plus delta x, which is this value, times i. Now if you wish to use the left endpoints, X sub i is equal to a plus delta x times i minus 1. So let me explain to you what this means. Now let's say if we have a function f of x. In the interval 2 to 6, and let's say we want to use 4 rectangles, and we're going to use the right endpoint rule. So let's make the number line from 2 to 6. Now, delta x, which is b minus a divided by n, it's 6 minus 2 divided by 4, which is 4 divided by 4, which is 1. So, the width of each sub-interval is 1. So, we're going to have 4 intervals, 5 points. 2, 3, 4, 5, 6. Now, we want to use the right endpoints. So, we only want to get 3, 4, 5, 6. Now, X sub i is going to tell us these different values. The first point, i, is equal to 1. For the second point, i is 2. For the third point, i is 3. And the fourth point, i is 4. So, if we plug in 1 for i, we should get 3. If we plug in 4 for i, we should get 6. So the general equation X sub i is equal to a plus delta X times i. And keep in mind delta X is the width of the interval, which is 1. And a is basically this number. a is the first point on the interval. a is 2 and b is 6. But we really don't need b for this example. So let's start with x1. Let's replace i with 1. So the first point x sub 1 is going to be a, which is 2, plus delta x, which is 1, times 1 since i is 1. Whatever this subscript is, is the same as this number. So this is basically 2 plus 1, which is 3, and we get this point. Now what about x sub 2? a is 2, delta x is 1, and i is 2. 2 plus 2 is 4, which is the second point. Now what about the last one, x sub 4? So it's 2 plus 1, delta x is still 1, but i is 4. 2 plus 4 is 6, so we get this number. Now, let's try an example using the left endpoints. So let's say if we have a function f of x bounded by the interval 3 to 9, where n is 4, where we're going to have 4 rectangles. and we're going to use the left endpoints. So let's calculate delta x. It's b minus a, which is 9 minus 3, divided by n, and n is 4. So 9 minus 3 is 6, 6 divided by 4 reduces to 3 over 2, which is 1.5. So delta x is 1.5. So now let's create an interval from 3 to 6, I mean 3 to 9. So n is 4, which means we're going to have 4 intervals, or 5 points. To find the next point, let's add 1.5 to 3, that's going to be 4.5. To find the next point, add 1.5 to 4.5, which is 6. 6 plus 4.5 is 7.5. Now, since we want the left endpoints, we want the first four points on the left. Now, let's use the equation to get these points. So, keep in mind, a is the first number, 3. So, for the left endpoints, x sub i is equal to a plus delta x. times i minus 1. So let's calculate x sub 1. a is 3, delta x is 1.5, and i is basically the subscript. Since we're dealing with the first term, i is 1. 1 minus 1 is 0, so we have 3 plus 0, which is 3. So that gives us the first x value. Let's calculate the third x value. So a is 3, delta x is still 1.5. So for the third value, i is 3. history so this is 3 minus 1 3 minus 1 is 2 so we have 1.5 times 2 1.5 times 2 is 3 3 plus 3 is 6 so this gives us the second I mean the third X value so this is X sub 1 this is X sub 2 X sub 3 X sub 4 Now let's calculate x sub 4, which should be 7.5. So the fourth x value is going to be a, which is 3, plus delta x, which is 1.5. And for the fourth value, i is 4. So this is 3 plus 1.5 times 4 minus 1 is 3. And 3 times 1.5 is 4.5, which gives you 7.5. And that's the fourth x value. Now, there are some other equations that you need to know. For example, the sum of the constant c from the first term to the nth term is simply equal to c times n. The sum, let me rewrite that, the sum of i from i equals 1 to the nth term is going to be n times n plus 1 divided by 2. Make sure you write these equations somewhere because we're going to use it for the next two examples. Now there's two more that you need. The next one is I squared, which is going to be equal to n times n plus 1 times 2n plus 1 divided by 6. And there's one more, I to the third. This is equal to... n times n plus 1 divided by 2 squared. Now you might be wondering, okay, what do these equations even mean? So let's say if we want to evaluate this expression. i to the 3rd, starting from 1. and stopping at the fifth term so this is equivalent to 1 to the third plus 2 to the third plus 3 to the third plus 4 to the third plus 5 to the third So you need to plug in 1 and stop at 5 and then add the sum of each of those terms. So 1 cubed is 1, 2 to the 3rd which is 2 times 2 times 2, that's 8. 3 cubed is 27, 4 to the 3rd is 64, and 5 cubed is 125. So let's add these 5 values. So this is equal to 225. So that's one way in which you can calculate the sum for an i to the third expression. The other way is to simply use the equation, because let's say if this was 100. You don't want to add it 100 times. So using the equation, all we need to do is plug in the n value. And n is 5. So then this expression... is equivalent to 5 times 5 plus 1 divided by 2 squared so 5 plus 1 is 6 and 6 divided by 2 is 3 so we have 5 times 3 which is 15 so this is 15 squared 15 times 15 is 225 so we get the same answer Now let's put this information to good use. So let's say if we have a function f of x, which is equal to x squared, and we wish to find the area between the x-axis and the curve from 0 to 3. There's many different ways in which we can do it. We can approximate the area using the left-end points, right-end points, or the midpoint rule. Or we can find the area of the definite integral, or we can use the definition of the definite integral using sigma notation, limits, and all that stuff. So let's start with the definite integral. So the area is going to be the integration from 0 to 3, x squared dx. The general equation is this. The area is the antiderivative from a to b, f of x dx. So the antiderivative of x squared is going to be x to the third divided by 3. So add 1 and then divide by that exponent. And this is going to be from 0 to 3. So now let's plug in 3. This is 3 to the 3rd divided by 3 minus 0 cubed over 3. 3 to the 3rd is 27, and 0 cubed is simply 0. 27 divided by 3 is 9. So the area is equal to 9. Now let's see if we can get that same answer using the other technique, the definition of the definite integral. The other way to do it is using this equation. The area is equal to the limit as n approaches affinity, sigma, starting from the first term to the nth term, f of x sub i times delta x. And let's use the right endpoints instead of the left endpoints because it's simply easier to use it. The math is less complicated. So let's calculate delta X first. Delta X is b minus a over n and we can see that a is 0, b is 3. So it's going to be a 3 minus 0 over n which is simply 3 divided by n. Now X sub i for the right endpoints is a plus delta X times i. And a is 0. So this is going to be 0 plus delta x, which is 3 over n times i. So this is just 3i divided by n. So now let's plug in what we have. So what we now have is the limit as n approaches infinity, sigma. And we said x sub i is 3i divided by n, and delta x is 3 over n. Now f of x is x squared, so f of 3i over n is going to be 3i over n squared, times all of this stuff. So let's make some space. So what we now have is the limit as n approaches infinity. This stuff times 3 squared is 9 times the other 3. That's going to be 27i squared divided by n squared times 1 over n, which is n to the third. Let me do that step by step for those of you who seem unsure about it. So let's just do this part first. Three squared is nine times i squared over n squared and then times three over n. So this is going to be equal to three times nine is twenty-seven. Divided by n to the third now what I'm going to do is separate the I squared from 27 over n to the third So this is the limit as n approaches infinity 27 over n to the third Sigma I Squared now remember there's a formula for the Sigma I square term And that formula, if you recall. It's n times n plus 1 times 2n plus 1 divided by 6. So what we now have is this limit expression times 27 over n cubed. And we can replace this with n times n plus 1 times 2n plus 1 divided by 6. Now at this point, what we need to do is focus on the algebra. So let's distribute n to n plus 1. So that's going to be n squared plus n times everything else. Now let's FOIL these two expressions. So n squared times 2n, that's going to be 2n cubed. And then n squared times 1 is simply n squared. n times 2n is 2n squared. And the last part, which is n times 1, that's plus n. So divided by 6, times everything that is already on the left. So now we can combine like terms. So we have the limit as n approaches infinity, 27 over n to the third times 2n cubed plus 3n squared plus n divided by 6. So what should we do at this point? What would you do at this point? Let's separate this fraction into three smaller fractions. So what I'm going to do is divide each term by 6. So 2n cubed divided by 6 plus 3n squared divided by 6 plus n over 6. Now let's reduce it. 2 over 6 reduces to 1 third. 3 over 6 reduces to 1 half. Or 6 divided by 3, if you divide it backwards, is 2. And 6 divided by 2 backwards is 3. So now let's distribute the 27 over n to the third to these three terms inside. So it's the limit as n approaches infinity. 27 divided by 3 is 9. And notice that the n cubes, they cancel. Now let's put these two together. So that's going to be 27 divided by 2n. You can cancel two of the n values, so leaving one on the bottom. n squared divided by n cubed is 1 over n. Plus 27 over 6 reduces to 9 over 2 if you divide both numbers by 3. So it's going to be 9 over 2n squared. n divided by n to the third is 1 over n squared. Now, whenever you take the limit as x approaches infinity for a constant divided by a variable, this is going to be a constant divided by infinity. Whenever you divide a finite number by an infinite number, it's always equal to 0. So, therefore, this is going to be 9 plus 0 plus 0. So, the final answer is 9. 27 divided by infinity is 0, 9 over infinity is also 0. Now let's try another example. Find the area between the curve and the x-axis for the function f of x minus 2 in the interval 1 to 4. So let's use the definite integral first. So it's going to be the antiderivative from 1 to 4, 5x minus 2 dx. So the antiderivative of 5x is 5x squared divided by 2 and the antiderivative of 2 is simply 2x. So now let's plug in 4. So it's going to be 5 times 4 squared divided by 2 minus 2 times 4 and then minus the lower limit which is 5 times 1 squared divided by 2 minus 2 times 1. 4 squared is 16. 2 times 4 is 8. And then this is going to be 5 over 2 minus 2. 16 divided by 2 is 8. 8 times 5 is 40. And let's distribute the negative signs, so it's negative 5 over 2 plus 2. 40 minus 8 is 32 and 32 plus 2 is 34 so we have 34 minus 5 over 2 so let's get common denominators let's multiply 34 by 2 over 2 34 times 2 is 68 and 68 minus 5 is 63 so the answer is 63 over 2 now let's get the same answer using the definition of the integral So the area is going to be the limit as n approaches infinity, and then we have the sum of the first term to the nth term, f of x sub i, times delta x. so let's find out delta x delta x is a b minus a divided by n so that's 4 minus 1 divided by n which is 3 over n now the next thing that you want to do is find x sub i you can use the left endpoint or the right endpoint but using the right endpoint is much easier than left endpoint so let's do it that way this is going to be a plus the delta x times i, where a is 1. So this is 1 plus delta x, which is 3 over n, times i. So it's 1 plus 3i over n. So now, let's plug in that information. So we have the limit as n approaches infinity. f of 1 plus 3i divided by n times delta x, which is 3 over n. So f of x is 5x minus 2, and we need to replace x with 1 plus 3i over n. So what we now have is the limit as n approaches infinity. this is going to be a 5 now instead of x we're going to replace it with 1 plus 3i divided by n minus 2 and all of that times delta x or 3 divided by n now I'm going to erase everything so just remember that the area is 63 over 2 which is 31.5 so our goal is to get that answer so let's distribute the five to everything inside so this is going to be five plus 15 I divided by n minus 2 now let's combine like terms 5 minus 2 is 3 so we have 3 plus 15 I divided by n times 3 over n This is supposed to be a 1. Now, let's distribute 3 over n to everything inside. so it's gonna be 9 divided by n plus 45 I divided by n squared now what we want to do is we need to separate this into two terms or into two separate parts So this is going to be the limit as n approaches infinity. And then I'm going to separate the n from the 9. So it's going to be 1 over n sigma of the constant 9. And for the last part, I'm going to separate the 45 over n squared from i. So it's going to be 45 divided by n squared, sigma, i. The only reason why I kept the constant is because I didn't have an i term in front of it. But if I had an i term, I would just try to isolate the i term by itself. Now let's review the formulas that we can use at this point. So if we have a constant C, it's going to be C times N. And if it's simply just the variable I, if you recall, it's going to be N times N plus 1 divided by 2. So we can replace this part with 9 times N. and this part with n, n plus 1 over 2. So let's go ahead and do that. So what we now have is the limit as n approaches infinity, 1 over n times 9n, it's c times n, plus 45 over n squared, times n, n plus 1, divided by 2. So this will simplify to this expression 9n divided by n is simply 9. The n variables will cancel. Plus 45 over n squared times, let's distribute n to n plus 1. So it's going to be n squared plus n divided by 2. Now let's separate this fraction into two smaller fractions. so it's 9 plus 45 over n squared and then it's times n squared divided by 2 plus n over 2 now let's take this term and distribute it to both terms on the inside 45 over n squared times n squared over 2, the n squareds will cancel. And it's simply going to be a constant, 45 divided by 2. Now, 45 over n squared times n over 2, n divided by n squared is 1 over n. So it's going to be 45 over 2n. So now, let's apply the limit as n approaches infinity. So 45 over 2n, or 45 over 2 times infinity, will approach 0. So the answer is going to be 9 plus 45 over 2 plus 0. So let's get common denominators. Let's multiply 9 by 2 over 2. 9 times 2 is 18, and 18 plus 45 is 63. So this is the right answer, which is the same as 31.5.

So let's say it looks something like this. Let's extend it. Now let's calculate the area from x equals, let's say, 0 to x equals 2. So how can we find the area of this shaded region between the curve and the x-axis?

So one way in which we can do it, in which we can get the exact answer, we can find the area by finding the definite integral from a to b of f of x dx. Let's go ahead and do it this way so you know what the exact answer is, and then we'll talk about other techniques of getting the answer. So this is going to be the antiderivative from 0 to 2, f of x is x squared plus 1 dx. Now the antiderivative of x squared...

is x cubed divided by 3. The way to get that answer is to add 1 to the exponent, 2 plus 1 is 3, and then divide by that number which is 3. Now the derivative of 1 is equal to 1. The antiderivative of 1 is x because the derivative of x is 1, and this is going to be evaluated from 0 to 2. Now this is going to be equal to, we need to plug in 2 and then 0. And I'm running out of space, so let's make some space. this is equal to plug into first the top number into x so it's going to be 2 cube divided by 3 plus 2 and then minus the lower limit of the integral which is 0 so 0 to the third divided by 3 plus 0 2 to the third is 8 And 0 cubed plus 0 is simply 0. So the answer is 8 over 3 plus 2. Which, let's get the decimal value for that. So this is equal to about 4.67.

The exact answer is 14 over 3. Now, how can we approximate the area using Riemann sums? Using the left, the right, and the midpoint? Well, let me give you a general idea of what we're going to do.

So, here is the function, just the right portion of it. And we want to find the area between x equals 0. and x equals 2. Now the way to do this is to basically find the area of rectangles that are beneath the curve. It can be above or below the curve. Since it's below it, this is an under approximation. So we want to find the area of those rectangles and add them up.

That's how we can approximate the area under the curve using Riemann Sum. Now let's focus on an individual rectangle. The width of the rectangle is known as delta x.

The height of the rectangle is the y value, or f of x. y is equal to f of x. So that's just the area of one rectangle. Since we have four rectangles, we need to add up the area of each rectangle.

So the area under the curve is basically the sum, the sigma notation represents sum, of all the individual areas of each rectangle. Now let's say this is f of x of i, i equals 1, and... it stops at n. So let's use four rectangles to approximate the area. But this is the generalized equation.

And I do want to mention one thing. The area of a rectangle is basically the length times the width. The width is delta x. And the length is basically the height of the rectangle, which is f of x.

So f of x times delta x gives you the area of one rectangle. And the sigma represents the sum of the area of all of the rectangles. So let's find the area. of the function, or the area between the curve and the x-axis, of the function x squared plus 1, on the interval 0 to 2. Now these are x values.

So a is 0, b is 2. Delta x, the width of each rectangle, is the difference between b and a, divided by n. So in this particular problem, b is 2, a is 0, divided by 4. So 2 over 4 is 1 half, or 0.5. That's the width of each rectangle.

Now, if you want to find the area using the left endpoints, make a number line from 0 to 2. Now we have 4 intervals, which is basically going to be 5 points. And each interval differs by 0.5. So the first one is going to be 0.5, the second is going to be 1.0, the third one is 1.5. Now, since n is 4, we're only going to use 4 of those 5 points.

Even though there's 5 points, there's 4 intervals, or 4 rectangles. So, if you want to find the area using the left endpoints, Choose four points starting from the left. 0, 0.5, 1, and 1.5.

If you want to find the area using the right endpoints, choose the four points on the right. 0.5, 1.0, 1.5, and 2. Don't choose the first point on the left. Now, if you want to use the midpoint rule, choose the four points in the middle.

0.25, 0.75, 1.25, and 1.75. Now keep in mind, the exact answer is approximately 4.67. Let's see how close our approximation is going to be to that answer. So let's begin. So let's find the area using the left endpoints.

So this is going to be the sum of delta x times f of x. And delta x. We said it's 0.5, but I'm going to write out the function first.

So it's delta x times the sum of the height of each rectangle, starting from the left. So we're going to use 0, 0.5, 1, and 1.5, but not 2. So f of 0 plus f of 1.5, or 0.5 plus f of 1 plus f of 1.5. So we know that delta x is 0.5. Now, if we plug in 0 into this function, f of 0 is going to be 0 squared plus 1, which is 1. Now, what about f of 0.5? I'm going to use the calculator for this part.

So, 0.5 squared plus 1 is 1.25. Now, f of 1, 1 squared is 1 plus 1 is 2. 1.5 squared is 2.25 plus 1 is 3.25. So let's add the numbers inside the bracket. So 1 plus 1.25 plus 2 plus 3.25 is about 7.5.

Half of 7.5 is 3.75. Notice that the answer is not very close to the actual answer, and that's because we have a small number of rectangles, and it's 4. If I were to increase n to 8, for example, this answer would be more accurate. It would be closer to the true answer of 4.67. But notice that it's less than 4.67. That means that it's an under-approximation.

The last endpoint. is usually an under approximation whenever the function is increasing. And for an increase in function, the right endpoint is usually an over approximation. The midpoint, using the midpoint rule, The answer is usually more accurate than the left endpoint or the right endpoint. So now, let's try the midpoint.

So we can get rid of this. So the area, using the midpoint rule, is going to be delta x times f of, we need to pick a number between 0 and 0.5. That's 0.25 or... 1 4th plus the midpoint between 0.5 and 1 you can draw the interval if it helps this is 0 this is 2 this is 0.5 1 1.5 so we have the points 0.25 0.75 1.25 and 1.75 so this is going to be 0.75 and then 1.25 And finally, 1.75.

So let's plug in the values. Delta X is 0.5, and 0.25 squared is about 0.0625 plus 1. So this is 1.0625. 0.75 squared plus 1. is 1.5625 plus 1.25 squared plus 1, that's 2.5625.

And the last one, 1.75 squared plus 1, that's 4.0625. So let's add everything inside the brackets. So 1.0625 plus 1.5625 plus 2.5625 plus 4.0625. This is about 9.25, if my math is correct. And half of 9.25 is about 4.625.

So as we can see, using the midpoint rule, we're going to get the most accurate answer. Now let's calculate... the area using the right endpoints.

We know it's going to be an over approximation for an increase in function, but let's get the answer. So let's draw the number line first. So A is 0, B is 2, and N is 4. So we're going to have 5 points and 4 intervals.

So using the right endpoint, we're going to choose 0.5, 1, and 2. So area, we're going to use Using the right endpoints is going to be delta x which is 0.5 times f of 0.5 plus f of 1 plus f of 1.5 plus f of 2. So let's start with 0.5. So 0.5 squared plus 1 is 1.25. 1 squared plus 1 is 2. 1.5 squared plus 1, that's 3.25. And 2 squared plus 1 is 5. So 1.25 plus 2 plus 3.25 plus 5 is about 11.5.

And half of that is 5.75. Notice that it's well above 4.67. So let's write the answers that we have. So the area using the left endpoint was 3.25. The area using the midpoint, I forgot what that was.

And for the right endpoint, it's 5.75. I believe this was 4.625. Let me double check. And yes, it's 4.625.

Now sometimes, if you're given the area of, or using the left endpoints, and the area using the right endpoints, and if you want to approximate the actual answer, average the two areas, the left and the right one, if you add them up and divide by 2, the average is going to be closer to the true answer than these individual values. So, if we... Add 3.25 plus 5.75 and then divided by 2. Notice what's going to happen.

3 plus 3.25 plus 5.75 is 9 divided by 2 is about 4.5 and 4.5 is pretty close to 4.67. At least it's closer than AL and AR. So this is a better approximation than using the left or the right endpoints separately.

But if you have the midpoint, that's good too. That's a good approximation. Now let's graph the rectangles and the curve so you can see how it looks like for the left endpoint, the midpoint, and the right point. So let's start with the left endpoints. I'm going to draw a relatively wide graph from 0 to 2. So let's break it into four intervals.

This is 0.5, 1, and 1.5. Now, we're going to use the left-hand points. That means we're going to use 0, 0.5, 1, 1.5, but not 2. So, draw a point on the curve at each of those x values. So, at 0, it's going to be right here, and then at 0.5, 1, and 1.5.

And then let's draw a rectangle at each of those points. Notice that the rectangles... are below the curve.

So that's why we have an under approximation. That explains why the area using the left endpoints was 3.25. It was significantly less than the actual area of 4.67. So that's what we did in the beginning. We found the area of these four rectangles, and then we add them up, and we got 3.25.

Now what about the right endpoint rule? How can we graph the rectangles for that situation? So let's begin by making a graph from 0 to 2. And let's make four intervals again.

So this time, we're using the right endpoints, 0.5, 1, 1.5, and 2. So let's plot those points on the curve. And this time, instead of drawing the rectangle towards the right, draw it towards the left. So first, let's put a vertical line. Between the point and the x-axis and let's draw each rectangle towards the left So notice that we have an over approximation the area of the rectangle is above the area of the curve So that explains why the area using the right endpoints was a 5.75.

That's why it's so much greater than 4.67. As you can see, there is a significant portion of the rectangle that's above the curve. And so that's why the answer is an over-approximation. Now let's graph the rectangles using the midpoint rule.

so let's make the four intervals 0.511.5 and 2 so we need four points but using a midpoint so let's put a point on the curve at.25.75 1.25 and 1.75 Now to draw the rectangle, draw a horizontal line that passes through the point. So it's going to look like this. Let's draw another one.

And let's draw another one. And one more. So notice that a portion of the rectangle... is above the curve.

Also, there's a portion that is below the curve. So the portion that is above the curve and the portion that is below the curve, they sort of balance each other out. And that's why the midpoint rule is a better approximation than the left endpoint or the right endpoint.

But that's how you would graph the rectangles if you ever need to. For the sake of practice, let's try another example. So consider the function f of x is equal to x cubed. Now, using four rectangles, estimate the area from 0 to 4 using the left endpoint, the right endpoint, and the midpoint rule. So first, let's find the actual area.

It's going to be the antiderivative from 0 to 4, x cubed dx. And the antiderivative of x cubed? Add 1 to the x-pointer, it's going to be x to the 4th, divided by 4, evaluated from 0 to 4. So let's plug in 4 first, and then we'll plug in 0. 4 to the 4th power, that's 256, divided by 4, is 64, minus 0. So the answer that we're looking for is 64. So now let's approximate the area using the left endpoints to begin with.

So first we need to calculate delta x, which is b minus a divided by n. a is 0, b is 4. So it's 4 minus 0 divided by 4. Delta x is 1. So let's make the number line. From a to b.

0 to 4 with a width of 1. So the width of each rectangle is 1. So we have 5 points, 4 intervals. So using the left endpoint rule, we need to use the first 4 points on the left, 0, 1, 2, and 3. So the area is going to be delta x, which we know delta x is the width of each rectangle, so that's 1. times f of 0 plus f of 1 plus f of 2 plus f of 3. So f of 0, if we plug in 0 into this function, 0 cubed is simply 0. 1 to the 3rd is 1, 2 to the 3rd is 8, 3 to the 3rd is 27. So 27 plus 8 plus 1. that's 36. So as you can see, this is a bad approximation. 36 is very far from 64, but at least that's what the answer is. Now let's use the right endpoint.

Let me just mark this answer down. So al, the area using the left endpoints, is equal to 36. Now let's make another number line. So we could find the area using the right endpoints. So from 0 to 4, we're going to have the points 1, 2, and 3. But we're going to choose the four points on the right. That's 1, 2, 3, and 4, but not 0. So then a of l is going to be delta x, which is 1, times f of 1, plus f of 2, plus f of 3, plus f of 4. So 1 cubed is 1, 2 to the 3rd is 8, 3 to the 3rd is 27, 4 to the 3rd is very big, that's 64. So this is definitely an over approximation.

So 1 plus 8 plus 27 plus 64 is 100. So AR is 100. Now let's average 136. So AL plus AR divided by 2. So 36 plus 100 is 136 and half of 136 is 68. So as you can see, if you average the left endpoint and the right endpoint, your answer will be close to the actual approximation. Now, let's approximate the area using the midpoint rule. So first, let's draw the number line.

So we need the midpoint between 0 and 1, which is 0.5. We need 1.5, 2.5, and also 3.5. So delta x is 1, and this is going to be times f of 0.5 plus f of 1.5 plus f of 2.5.

plus f of 3.5 so.5 raised to the third power is 1 over 8 or simply.125 1.5 to the third power is 3.375 2.5 to the third power is 15.625 and 3.5 to the third power is 42.875 times 1. So let's add the numbers on the inside. So 0.125 plus 3.375 plus 15.625 plus 42.875. So you should get a total of 62, which is very close to 64. Now, we said that if we increase the value of n, if we use more rectangles, the approximation will be more accurate. So let's use the left-hand points, and let's use 8 rectangles instead of 4 rectangles.

And let's see if the answer is going to be closer to 64. So right now, it's 36 with 4 rectangles, but let's see what's going to happen if we use 8 rectangles. So let's calculate delta x first, which is b minus a divided by n. So it's going to be 4 minus 0 divided by the 8 rectangles, which is 1 over 2, or 0.5.

Now let's draw the number line. So from 0 to 4, let's increase it in increments of 0.5. So 0.51, 1.52, 2.53, and 3.5. So there's 9 points, and we're only going to use 8 of them. We're going to use every point except the last one on the right.

So we're going to use the 8 points on the left. So everything from 0 all the way to 3.5. But not the last 0.4.

So the area is going to be delta x times f of 0 plus f of 0.5 plus f of 1. plus f of 1.5 plus f of 2 plus f of 2.5 plus f of 3 plus f of 3.5 so Delta X will know to be 0.5 0 to the third power is 0.5 raised to the third power that's a point one two five 1 to the third power is 1. 1.5 to the third power is 3.375. 2 to the third is 8. 2.5 to the third power is 15.625. Plus 3 to the third is 27. Plus 3.5. to the third power.

That's 42.875. So now let's add the numbers inside the brackets. You should get 98. So 0.5 times 98 is 49. So notice that this is still an under-approximation, but 49 is closer to 64 than 36. So even though it's not very accurate, it's more accurate than the previous answer of 36. So as you increase n, the approximation is more accurate. It's going to be better.

So far, for increasing functions, the area that was obtained using the right endpoints was an over-approximation, and the area that was obtained using the left endpoints was an under-approximation for an increasing function. Now let's see what's going to happen if we use a decreasing function. So let's say that f of x is 1 over x, and let's approximate the area, just the left and the right endpoints, from 1 to 2. And let's use four rectangles. So go ahead and try this example.

Now first, let's get the actual answer. So let's find the antiderivative from 1 to 2 of 1 over x dx. The antiderivative of 1 over x is simply ln x, because the derivative of ln x is 1 over x.

If you try to use the power rule, if you rewrite it, and if you add 1, you're going to get 0 over 0, which... Doesn't work. So you need to know that the antiderivative of 1 over X is simply ln X. And this is going to be evaluated from 1 to 2. So ln 2 minus ln 1 is going to be ln 2 is about 0.6931.

ln 1 is 0. So the area is approximately 0.6931. Now let's start with the left endpoint. Let's calculate delta x. which is b minus a or 2 minus 1 divided by n and n is 4 so 2 minus 1 is 1 so this is 1 4th which is about 0.25 so now let's draw a number line so from 1 to 2 and let's break it into four intervals or five points so it's going to be 1.25 from 1.5 and 1.75 So for the left endpoint rule, we're going to use the four points on the left.

And so the area is going to be delta x, which is 0.25, times f of 1, plus f of 1.25, plus f of 1.5, plus f of 1.75. So this is 1 fourth, or 0.25. Now, f of 1, 1 divided by 1 is simply 1. And 1 divided by 1.25 is about 0.8.

1 divided by 1.5 is 2 over 3, or 0.67, and 1 divided by 1.75 is about 0.571. So let's add 1 plus 0.8 plus 0.67 plus 0.571, and let's divide it by 4. So the area using the left endpoints is.76. Notice that it's an over-approximation because it's a decrease in function.

AL is going to be an over-approximation for decrease in functions, but it's an under-approximation for increase in functions. Keep that in mind. Now let's approximate the area using the right-hand points. So this time we're going to use 1.25, 1.5, 1.75, and 2. So delta x is still 0.5, and 1 divided by 1.25, we said that's 0.8.

1 divided by 1.5 is 0.67, and 1 divided by 1.75, 0.571. And 1 divided by 2 is 0.5. So notice that the numbers are decreasing as we go from left to right.

Therefore, the area using the right-end points should be an under-approximation for decreasing function. Because the numbers to the right should be less than the numbers to the left if the function is decreasing as you go from left to right. Makes sense, right?

So now let's calculate the area. So let's add the four numbers inside the brackets. So 0.8 plus 0.67 plus 0.571 plus 0.5.

That's about 2.541. And half of that is basically 1.2705. But let's round it and say it's 1.27. For some reason, that just seems too high. And the issue is delta x.

Delta x is not 0.5. I made a mistake. It's 1 fourth.

It's 0.25. So let's correct that mistake. So 0.25 times 2.541.

That's 0.6353 or 0.635. So this answer is not too far away from this answer, but it is an under approximation. Now let's average the area calculated from the left endpoint and from the right endpoint.

So this is going to be 0.76 plus 0.635 divided by 2. The average is equal to 0.6975, which is very close to the actual answer. So anytime you average the left and the right endpoints, you're going to get a better approximation. Sometimes you may want to find the area using the definition of a definite integral.

The definite integral from a to b of f of x dx is equal to the limit as n approaches infinity Then we're going to have this sigma notation where i is equal to 1, which stops at n, f of x sub i times delta x. Now, as mentioned before, delta x is simply b minus a divided by n on the interval a to b. Now you may wonder what is this X sub i? It really depends on if you want to find the area using the right endpoints or the left endpoints. So using the right endpoints, X sub i is equal to a, that's this a, plus delta x, which is this value, times i.

Now if you wish to use the left endpoints, X sub i is equal to a plus delta x times i minus 1. So let me explain to you what this means. Now let's say if we have a function f of x. In the interval 2 to 6, and let's say we want to use 4 rectangles, and we're going to use the right endpoint rule. So let's make the number line from 2 to 6. Now, delta x, which is b minus a divided by n, it's 6 minus 2 divided by 4, which is 4 divided by 4, which is 1. So, the width of each sub-interval is 1. So, we're going to have 4 intervals, 5 points.

2, 3, 4, 5, 6. Now, we want to use the right endpoints. So, we only want to get 3, 4, 5, 6. Now, X sub i is going to tell us these different values. The first point, i, is equal to 1. For the second point, i is 2. For the third point, i is 3. And the fourth point, i is 4. So, if we plug in 1 for i, we should get 3. If we plug in 4 for i, we should get 6. So the general equation X sub i is equal to a plus delta X times i. And keep in mind delta X is the width of the interval, which is 1. And a is basically this number.

a is the first point on the interval. a is 2 and b is 6. But we really don't need b for this example. So let's start with x1.

Let's replace i with 1. So the first point x sub 1 is going to be a, which is 2, plus delta x, which is 1, times 1 since i is 1. Whatever this subscript is, is the same as this number. So this is basically 2 plus 1, which is 3, and we get this point. Now what about x sub 2? a is 2, delta x is 1, and i is 2. 2 plus 2 is 4, which is the second point. Now what about the last one, x sub 4?

So it's 2 plus 1, delta x is still 1, but i is 4. 2 plus 4 is 6, so we get this number. Now, let's try an example using the left endpoints. So let's say if we have a function f of x bounded by the interval 3 to 9, where n is 4, where we're going to have 4 rectangles.

and we're going to use the left endpoints. So let's calculate delta x. It's b minus a, which is 9 minus 3, divided by n, and n is 4. So 9 minus 3 is 6, 6 divided by 4 reduces to 3 over 2, which is 1.5.

So delta x is 1.5. So now let's create an interval from 3 to 6, I mean 3 to 9. So n is 4, which means we're going to have 4 intervals, or 5 points. To find the next point, let's add 1.5 to 3, that's going to be 4.5.

To find the next point, add 1.5 to 4.5, which is 6. 6 plus 4.5 is 7.5. Now, since we want the left endpoints, we want the first four points on the left. Now, let's use the equation to get these points. So, keep in mind, a is the first number, 3. So, for the left endpoints, x sub i is equal to a plus delta x. times i minus 1. So let's calculate x sub 1. a is 3, delta x is 1.5, and i is basically the subscript.

Since we're dealing with the first term, i is 1. 1 minus 1 is 0, so we have 3 plus 0, which is 3. So that gives us the first x value. Let's calculate the third x value. So a is 3, delta x is still 1.5.

So for the third value, i is 3. history so this is 3 minus 1 3 minus 1 is 2 so we have 1.5 times 2 1.5 times 2 is 3 3 plus 3 is 6 so this gives us the second I mean the third X value so this is X sub 1 this is X sub 2 X sub 3 X sub 4 Now let's calculate x sub 4, which should be 7.5. So the fourth x value is going to be a, which is 3, plus delta x, which is 1.5. And for the fourth value, i is 4. So this is 3 plus 1.5 times 4 minus 1 is 3. And 3 times 1.5 is 4.5, which gives you 7.5.

And that's the fourth x value. Now, there are some other equations that you need to know. For example, the sum of the constant c from the first term to the nth term is simply equal to c times n. The sum, let me rewrite that, the sum of i from i equals 1 to the nth term is going to be n times n plus 1 divided by 2. Make sure you write these equations somewhere because we're going to use it for the next two examples. Now there's two more that you need.

The next one is I squared, which is going to be equal to n times n plus 1 times 2n plus 1 divided by 6. And there's one more, I to the third. This is equal to... n times n plus 1 divided by 2 squared.

Now you might be wondering, okay, what do these equations even mean? So let's say if we want to evaluate this expression. i to the 3rd, starting from 1. and stopping at the fifth term so this is equivalent to 1 to the third plus 2 to the third plus 3 to the third plus 4 to the third plus 5 to the third So you need to plug in 1 and stop at 5 and then add the sum of each of those terms. So 1 cubed is 1, 2 to the 3rd which is 2 times 2 times 2, that's 8. 3 cubed is 27, 4 to the 3rd is 64, and 5 cubed is 125. So let's add these 5 values. So this is equal to 225. So that's one way in which you can calculate the sum for an i to the third expression.

The other way is to simply use the equation, because let's say if this was 100. You don't want to add it 100 times. So using the equation, all we need to do is plug in the n value. And n is 5. So then this expression... is equivalent to 5 times 5 plus 1 divided by 2 squared so 5 plus 1 is 6 and 6 divided by 2 is 3 so we have 5 times 3 which is 15 so this is 15 squared 15 times 15 is 225 so we get the same answer Now let's put this information to good use.

So let's say if we have a function f of x, which is equal to x squared, and we wish to find the area between the x-axis and the curve from 0 to 3. There's many different ways in which we can do it. We can approximate the area using the left-end points, right-end points, or the midpoint rule. Or we can find the area of the definite integral, or we can use the definition of the definite integral using sigma notation, limits, and all that stuff. So let's start with the definite integral. So the area is going to be the integration from 0 to 3, x squared dx.

The general equation is this. The area is the antiderivative from a to b, f of x dx. So the antiderivative of x squared is going to be x to the third divided by 3. So add 1 and then divide by that exponent. And this is going to be from 0 to 3. So now let's plug in 3. This is 3 to the 3rd divided by 3 minus 0 cubed over 3. 3 to the 3rd is 27, and 0 cubed is simply 0. 27 divided by 3 is 9. So the area is equal to 9. Now let's see if we can get that same answer using the other technique, the definition of the definite integral. The other way to do it is using this equation.

The area is equal to the limit as n approaches affinity, sigma, starting from the first term to the nth term, f of x sub i times delta x. And let's use the right endpoints instead of the left endpoints because it's simply easier to use it. The math is less complicated. So let's calculate delta X first.

Delta X is b minus a over n and we can see that a is 0, b is 3. So it's going to be a 3 minus 0 over n which is simply 3 divided by n. Now X sub i for the right endpoints is a plus delta X times i. And a is 0. So this is going to be 0 plus delta x, which is 3 over n times i. So this is just 3i divided by n.

So now let's plug in what we have. So what we now have is the limit as n approaches infinity, sigma. And we said x sub i is 3i divided by n, and delta x is 3 over n.

Now f of x is x squared, so f of 3i over n is going to be 3i over n squared, times all of this stuff. So let's make some space. So what we now have is the limit as n approaches infinity. This stuff times 3 squared is 9 times the other 3. That's going to be 27i squared divided by n squared times 1 over n, which is n to the third.

Let me do that step by step for those of you who seem unsure about it. So let's just do this part first. Three squared is nine times i squared over n squared and then times three over n.

So this is going to be equal to three times nine is twenty-seven. Divided by n to the third now what I'm going to do is separate the I squared from 27 over n to the third So this is the limit as n approaches infinity 27 over n to the third Sigma I Squared now remember there's a formula for the Sigma I square term And that formula, if you recall. It's n times n plus 1 times 2n plus 1 divided by 6. So what we now have is this limit expression times 27 over n cubed. And we can replace this with n times n plus 1 times 2n plus 1 divided by 6. Now at this point, what we need to do is focus on the algebra. So let's distribute n to n plus 1. So that's going to be n squared plus n times everything else.

Now let's FOIL these two expressions. So n squared times 2n, that's going to be 2n cubed. And then n squared times 1 is simply n squared.

n times 2n is 2n squared. And the last part, which is n times 1, that's plus n. So divided by 6, times everything that is already on the left. So now we can combine like terms.

So we have the limit as n approaches infinity, 27 over n to the third times 2n cubed plus 3n squared plus n divided by 6. So what should we do at this point? What would you do at this point? Let's separate this fraction into three smaller fractions.

So what I'm going to do is divide each term by 6. So 2n cubed divided by 6 plus 3n squared divided by 6 plus n over 6. Now let's reduce it. 2 over 6 reduces to 1 third. 3 over 6 reduces to 1 half. Or 6 divided by 3, if you divide it backwards, is 2. And 6 divided by 2 backwards is 3. So now let's distribute the 27 over n to the third to these three terms inside.

So it's the limit as n approaches infinity. 27 divided by 3 is 9. And notice that the n cubes, they cancel. Now let's put these two together.

So that's going to be 27 divided by 2n. You can cancel two of the n values, so leaving one on the bottom. n squared divided by n cubed is 1 over n. Plus 27 over 6 reduces to 9 over 2 if you divide both numbers by 3. So it's going to be 9 over 2n squared. n divided by n to the third is 1 over n squared.

Now, whenever you take the limit as x approaches infinity for a constant divided by a variable, this is going to be a constant divided by infinity. Whenever you divide a finite number by an infinite number, it's always equal to 0. So, therefore, this is going to be 9 plus 0 plus 0. So, the final answer is 9. 27 divided by infinity is 0, 9 over infinity is also 0. Now let's try another example. Find the area between the curve and the x-axis for the function f of x minus 2 in the interval 1 to 4. So let's use the definite integral first.

So it's going to be the antiderivative from 1 to 4, 5x minus 2 dx. So the antiderivative of 5x is 5x squared divided by 2 and the antiderivative of 2 is simply 2x. So now let's plug in 4. So it's going to be 5 times 4 squared divided by 2 minus 2 times 4 and then minus the lower limit which is 5 times 1 squared divided by 2 minus 2 times 1. 4 squared is 16. 2 times 4 is 8. And then this is going to be 5 over 2 minus 2. 16 divided by 2 is 8. 8 times 5 is 40. And let's distribute the negative signs, so it's negative 5 over 2 plus 2. 40 minus 8 is 32 and 32 plus 2 is 34 so we have 34 minus 5 over 2 so let's get common denominators let's multiply 34 by 2 over 2 34 times 2 is 68 and 68 minus 5 is 63 so the answer is 63 over 2 now let's get the same answer using the definition of the integral So the area is going to be the limit as n approaches infinity, and then we have the sum of the first term to the nth term, f of x sub i, times delta x. so let's find out delta x delta x is a b minus a divided by n so that's 4 minus 1 divided by n which is 3 over n now the next thing that you want to do is find x sub i you can use the left endpoint or the right endpoint but using the right endpoint is much easier than left endpoint so let's do it that way this is going to be a plus the delta x times i, where a is 1. So this is 1 plus delta x, which is 3 over n, times i. So it's 1 plus 3i over n.

So now, let's plug in that information. So we have the limit as n approaches infinity. f of 1 plus 3i divided by n times delta x, which is 3 over n. So f of x is 5x minus 2, and we need to replace x with 1 plus 3i over n.

So what we now have is the limit as n approaches infinity. this is going to be a 5 now instead of x we're going to replace it with 1 plus 3i divided by n minus 2 and all of that times delta x or 3 divided by n now I'm going to erase everything so just remember that the area is 63 over 2 which is 31.5 so our goal is to get that answer so let's distribute the five to everything inside so this is going to be five plus 15 I divided by n minus 2 now let's combine like terms 5 minus 2 is 3 so we have 3 plus 15 I divided by n times 3 over n This is supposed to be a 1. Now, let's distribute 3 over n to everything inside. so it's gonna be 9 divided by n plus 45 I divided by n squared now what we want to do is we need to separate this into two terms or into two separate parts So this is going to be the limit as n approaches infinity. And then I'm going to separate the n from the 9. So it's going to be 1 over n sigma of the constant 9. And for the last part, I'm going to separate the 45 over n squared from i. So it's going to be 45 divided by n squared, sigma, i.

The only reason why I kept the constant is because I didn't have an i term in front of it. But if I had an i term, I would just try to isolate the i term by itself. Now let's review the formulas that we can use at this point. So if we have a constant C, it's going to be C times N. And if it's simply just the variable I, if you recall, it's going to be N times N plus 1 divided by 2. So we can replace this part with 9 times N.

and this part with n, n plus 1 over 2. So let's go ahead and do that. So what we now have is the limit as n approaches infinity, 1 over n times 9n, it's c times n, plus 45 over n squared, times n, n plus 1, divided by 2. So this will simplify to this expression 9n divided by n is simply 9. The n variables will cancel. Plus 45 over n squared times, let's distribute n to n plus 1. So it's going to be n squared plus n divided by 2. Now let's separate this fraction into two smaller fractions.

so it's 9 plus 45 over n squared and then it's times n squared divided by 2 plus n over 2 now let's take this term and distribute it to both terms on the inside 45 over n squared times n squared over 2, the n squareds will cancel. And it's simply going to be a constant, 45 divided by 2. Now, 45 over n squared times n over 2, n divided by n squared is 1 over n. So it's going to be 45 over 2n.

So now, let's apply the limit as n approaches infinity. So 45 over 2n, or 45 over 2 times infinity, will approach 0. So the answer is going to be 9 plus 45 over 2 plus 0. So let's get common denominators. Let's multiply 9 by 2 over 2. 9 times 2 is 18, and 18 plus 45 is 63. So this is the right answer, which is the same as 31.5.

Transcript for:Understanding Area Under Curves with Riemann Sums

Transcript for:
Understanding Area Under Curves with Riemann Sums