Approximating areas with Riemann sums
For fun, let's try to approximate the area under the curve y is equal to the square root, the principal root, of x minus 1, between x is equal to 1 and x is equal to 6. So I want to find this entire area. Or I want to at least approximate this entire area. And the way I'll do it is by setting up five trapezoids of equal width. So this will be the left boundary of the first trapezoid. This will be its right boundary, which will also be the left boundary of the second trapezoid. This will be the right boundary of the second trapezoid. This is the right boundary of the third trapezoid. This'll be the right boundary of the fourth trapezoid. And then finally, this will be the right boundary of the fifth trapezoid. And since we're going from 1 to 6, so we're traveling 6 minus 1 in the x direction, and I want to split it into five sections, the width of each trapezoid is just going to be equal to 1. And so if we say that the width of a trapezoid is delta x, we just can now say the delta x is equal to 1. So let's set up our trapezoids. So the first trapezoid is going to look like that. It's going to go like that. Actually, it's going to be a triangle, not really a trapezoid. Then the second trapezoid is going to look like this. I guess you could say a trapezoid where one of the sides has length 0 turns into a triangle. And then the third trapezoid is going to look like this. And then the fourth trapezoid is going to look like that. And then finally, you have the fifth trapezoid. So let's calculate the area of each of these, and then we will have our approximation for the area under the curve. So let's do trapezoid-- or I really should say triangle-- this first shape, whatever you want to call it. What is the area of that going to be? Well, the area of a trapezoid-- and you'll see this will just turn into the area of a triangle-- it's the average of the heights of the two sides of the trapezoid, the way we've looked at it-- or you could say the average of the heights of the two parallel sides, I guess is the best way to say it. So f of 1-- that's the height here-- plus f of 2, all of that over 2, and then we're going to multiply it times our delta x. Actually, let me do that in that same red color to show you that this is the area of that first trapezoid, so times delta x. And as you see right over here, if you look at it, the f of 1 is just going to be 0. So you're going to have f of 2 times-- so it's going to be this height times this base times 1/2, which is just the area of a triangle. Let's look at the second trapezoid, trapezoid two right over here. What is its area going to be? Well, it's going to be f of 2 plus f of 3. f of 2 is this height. f of 3 is this height. So we're taking the average of those two heights-- divided by 2; that's the average of those two heights-- times the base, times delta x. And then let's do trapezoid three. I think you're getting the idea here. Trapezoid three is going to be f of 3 plus f of 4 divided by 2 times delta x. And then-- let's see. I'm running out of colors. This is trapezoid four right over here. So plus f of 4 plus f of 5, all of that over 2, times delta x. And then we have our last trapezoid, which I will do in yellow. So this is trapezoid number five. I'll scroll down a little bit, get some more real estate. So it's going to be plus-- I'll just write the plus over here-- plus f of 5 plus f of 6 over 2 times our delta x. So let's see how we can simplify this a little bit. All of these terms, we have a 1/2 delta x, so let's actually factor that out. So remember, this is our approximation of our area. So area is approximately-- remember, this is just a rough approximation. It's very clear-- actually, you might say, this is pretty good using the trapezoids, but it is pretty clear that we are letting go of some of the area. We're letting go of that area. We're letting go of some of this right over here. You can barely see it. Some of this right over here, you can barely see it. But we are-- this is going to be, it looks like, an underestimate, but it is a decent approximation. But let's see if we can simplify this expression. So it's approximately going to be equal to, I'm going to factor out a delta x over 2. And then what I'm left with-- and I will switch to a neutral color-- if I factor out a delta x over 2, then I have just an f of 1. And then I have two f of 2's, so plus 2 times f of 2. And I'm doing this because you might see a formula that looks something like this in your calculus book, and it's not some mysterious thing. They just really summed up the areas of the trapezoids. And then we're going to have two f of 3's-- plus 2 times f of 3's. Plus we're going to have two f of 4's-- plus 2 times f of 4. And then we're going to have two f of 5's-- plus 2 times f of 5. And then finally, we're going to have one f of 6-- plus f of 6. If you were to generalize it, you have one of the first endpoint, the function evaluated at the first endpoint, one of it at the very last endpoint, and then two of all of the rest of them. But this is just the area of trapezoids. I'm not actually a big fan of when textbooks write this. Because when you see this, it's hard to visualize the trapezoids. When you see this, it's much clearer how you might visualize that. But with that out of the way, let's actually evaluate this. Lucky for us, the math is simple. Our delta x is just 1. And then we just have to evaluate all of this business. f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1, so that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1, so this is just going to be 2. Actually, let me do it in that same-- well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully, you realized that. I was just sticking with that pen color. Then f of 3. 3 minus 1 is 2-- square root of 2. So the function evaluated at 3 is the square root of 2. So this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate it at 4, you get the square root of 3. So this is going to be 2 times the square root of 3. And then you get 2 times the square root of 4-- 5 minus 1 is 4. 2 times the square root of 4 is just four. And then finally, you get f of 6 is square root of 6 minus 1, is the square root of 5. And I think we're now ready to evaluate. So let me get my handy TI-85 out and calculate this. So it's going to be-- well I'm just going to calculate-- well, I'll just multiply. So 0.5 times open parentheses-- well, it's a 0. I'll just write it, just so you know what I'm doing. 0 plus 2 plus-- whoops. Lost my calculator. Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4-- I'm almost done-- plus the square root of 5-- so let me write that-- gives me-- now we are ready for our drum roll-- it gives me-- and I'll just round it-- 7.26. So the area is approximately equal to 7.26 under the curve y is equal to the square root of x minus 1 between x equals 1 and x equals 6. And we did this using trapezoids.
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