Approximate the area under the curve f ( x ) = 2 3 x + 5 f\left(x\right)=\frac{2}{3x+5} f ( x ) = 3 x + 5 2 over the interval [ 4 , 10 ] \left\lbrack4,10\right\rbrack [ 4 , 10 ] using the Midpoint Riemann sum with n = 3 n=3 n = 3 subintervals.

This problem we're asked to approximate the area under the curve of this function f of x is equal to two over three x plus five on the interval from four to 10 using the midpoint Riemann sum with n equals three subintervals. Okay. Now how do we go about doing this? Well the thing you need to understand is there is some function that we're dealing with, this function right here, and we're trying to find the area underneath whatever curve this function makes, but we're approximating this. So if you imagine we have some kind of graph, and I don't know exactly what this function's going to look like. So what I'm going to do is just kinda write some arbitrary curve on my graph here, and then say this is the function that we're dealing with. Obviously, that's not this function, but we're drawing some curve here. And we're trying to find the area from four to 10 on the on the x axis. This is the interval. So we go from four all the way over to 10, and we're trying to find it this area approximately underneath this curve. Now to go about solving a problem like this, what we would we've done in the past is taken our curve and filled it with a bunch of rectangles. But doing these Riemann sums allows us to now use an equation to figure out what the approximate area is going to be. Now we're specifically asked to use the midpoint rule, and for the midpoint rule, the equation looks like this. We're going to have the midpoint m of n is equal to the sum that's gonna go from k equals one to n. That's going to be of f, and then we're going to have x of k minus one representing the left side, plus x of k representing the right side, all divided by two. So if we take the left side plus the right side on the x axis, and then we kinda like average this, that's gonna give us that midpoint, and then we're going to multiply that by delta x the width. So let's go ahead and see if we can use this equation. Now first off, I'm gonna deal with this delta x, because this one's usually the most straightforward. The equation is b minus a over n defined delta x, b being the high value of 10, a being the low value of four, and then that's all divided by the number of subintervals, which I can see is three. So we have 10 minus four over three, which all comes out to two. Meaning two is going to be the consistent width for this midpoint Riemann sum. So we can take this delta x as two and bring it outside the integral. So we're going to have that the midpoint Riemann sum is equal to two multiplied by this summation. So we'll have the sum from k equals one. And then n is the number of subintervals three, and we're gonna have f of x of k minus one. So we're just gonna rewrite this x of k minus one plus x of k all divided by two. Now what I can do is make this a bit more specific by taking this portion of my function and plugging it in to the function we have in this problem since that's what we're doing. And the way that I'm going to do this is by taking every place I see x and replacing it with this whole thing that we see in here. So we're gonna have two times the sum from k equals one to three, and then this whole thing right now is gonna be two over, and then we won't have three, and that's going to be times, and then this whole portion. So we won't have x of k minus one, plus x of k, all over two, is going to be plus five. So this is what we ultimately get when we write out the summation, and something else I can do is combine the two constants that I see here. See that we have a two and a two there, I can bring the that two up to the front, make it a four, multiplying this all out. So what I would do is rewrite this whole thing right here, and I'm just gonna go ahead and copy this, so I don't have to rewrite this twice. So I'll just put it down right about here. So going back to this sum, we're just gonna go ahead and rewrite this, And this time I'm going to take this two, knock it off the top, giving us a four out front, and then just a one on top of this fraction. And now we have things about as simplified as we can get it before actually evaluating this sum. So let's go ahead and now evaluate the sum since that would be our next step here. So doing this, well, we're going to have this whole thing is equal to four, and then we're going to multiply out the function values. Now the way that I can do this is by taking a look at the interval that we have. This goes from four to 10. And notice that our width is two. Now since I'm dealing with a midpoint here, that's gonna be the midpoint on the x axis between two points. Now going over to this kind of mock up graph that we drew over here, notice that we'd be at four. And if we went a delta x or two units, we would be over here at six. Now what would be the midpoint between four and six? Well, four plus six over two would give us five. That would just be averaging those. Right? So that would give us this midpoint of five. This would be the midpoint and the rectangle that we would look at if we were doing some rectangle approximation. So what we're doing is we're going to a midpoint of five between four and six as the interval starts at four as we saw up here. So going down here, what we would have is this sum is gonna start as being one, and it's gonna be over three times, and then we have this midpoint of five, and that's gonna be plus and and then it's gonna be what we have here. So we have one over three times five plus five. Right? Because I just wrote up this whole thing, three times five plus five, and then we're going to go ahead and add on the next piece of this. So the next piece is gonna be one, and then we're going to have over three, and then we need to go two units up. So we started with five, and then we're going to go to seven. It's gonna be plus five, then we'll have plus one over three times two units up. It's gonna be nine plus five. And the reason why that I'm always going two units up here is because notice that our delta x was two. So that's how much I increment up by when I do this. Now notice that I have one, two, three pieces here, and we were going from one to three. So this all looks right. So all you need to do is evaluate these numbers. And if you put this into your calculator and you get a decimal approximation, you should get the midpoint is approximately equal to 0.48 as the approximate area using a midpoint. So that is how we can do a midpoint Riemann sum, and that would be the solution to this problem. So hope you found this video helpful. This is the general process for midpoint Riemann sums. Let me know if you have any questions, and I will see you in the next video.

8. Definite Integrals

Riemann Sums

Business Calculus

8. Definite Integrals

Riemann Sums

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객관식

Approximate the area under the curve $f$\left$(x$\right$)=$\frac{2}{3x+5}$$ over the interval $\left\[\lbrack\)4,10\(\right\]\rbrack$ using the Midpoint Riemann sum with $n=3$ subintervals.

0.55

0.48

0.43

0.61

검증된 단계별 안내

Step 1: Divide the interval [4, 10] into n=3 subintervals. To do this, calculate the width of each subinterval, Δx, using the formula Δx = (b - a) / n, where a=4, b=10, and n=3.

Step 2: Determine the midpoints of each subinterval. For each subinterval, the midpoint is calculated as the average of the left and right endpoints of the subinterval. For example, the midpoint of the first subinterval is (x₁ + x₂) / 2.

Step 3: Evaluate the function f(x) = 2 / (3x + 5) at each of the midpoints found in Step 2. This will give the function values at the midpoints.

Step 4: Multiply each function value from Step 3 by the width of the subinterval, Δx, to approximate the area of each rectangle in the Midpoint Riemann sum.

Step 5: Add up all the areas of the rectangles from Step 4 to approximate the total area under the curve over the interval [4, 10]. This sum is the Midpoint Riemann sum approximation.

4:22m

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Approximate the area under the curve f(x)=23x+5f\(\left\)(x\(\right\))=\(\frac{2}{3x+5}\)f(x)=3x+52 over the interval [4,10]\(\left\[\lbrack\)4,10\(\right\]\rbrack\)[4,10] using the Midpoint Riemann sum with n=3n=3n=3 subintervals.

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Approximate the area under the curve $f\(\left\)(x\(\right\))=\(\frac{2}{3x+5}\)$ over the interval $\left\[\lbrack\)4,10\(\right\]\rbrack$ using the Midpoint Riemann sum with $n=3$ subintervals.