5. (a) Estmate the area under the graph of f(x)=1+x^2 from x=-1 to x=2 using three rectangles an

5. (a) Estmate the area under the graph of f(x)=1+x^2 from...

Key Concepts

Definite Integral

A definite integral quantitatively represents the area under a curve between two points on the x-axis. This concept is fundamental in calculus for finding accumulated quantities such as area, displacement, and total accumulation, and it forms the basis of numerical integration methods when explicit antiderivatives are difficult or impossible to obtain.

Riemann Sums

Riemann sums are a primary method of approximating the area under a curve by dividing the interval into equal or unequal subintervals and summing the areas of rectangles whose heights are determined by the function's value at selected sample points within these subintervals. As the number of subintervals increases, the approximation typically converges to the true value of the definite integral.

Right Endpoint Method

The right endpoint method is a specific type of Riemann sum where the function value at the right endpoint of each subinterval is used as the height of the approximating rectangle. This approach may systematically overestimate or underestimate the area, depending on whether the function is increasing or decreasing over the interval.

Left Endpoint Method

Analogous to the right endpoint method, the left endpoint method uses the function value at the left endpoint of each subinterval to determine the rectangle's height. Like the right endpoint, the accuracy of this approximation depends on the function's behavior along the interval, and errors tend to decrease as the number of subintervals increases.

Midpoint Method

The midpoint method uses the function value at the center of each subinterval as the height of the approximating rectangle. This method often provides a more accurate estimation of the integral compared to using the left or right endpoints, particularly when the function is smooth, because it tends to balance over- and underestimations within each subinterval.

Numerical Accuracy and Convergence

Increasing the number of rectangles (or subintervals) in any Riemann sum generally improves the approximation's accuracy because it more closely captures the function's variation over the interval. This principle is key in studying the convergence of numerical methods to the exact value of the definite integral as the partition becomes finer.

Transcript

00:01 Good afternoon, numerade students, or evening, whichever time of day it is for you.

00:06 So today, our question will be about finding integrals under a curve using rectangles.

00:11 So i've already drawn here what i think the graph looks like for the question that we'll be dealing with.

00:16 And so when it comes down to solving integrals using rectangles, it basically comes down to this.

00:22 Trying to find all of this area is pretty challenging.

00:25 But if we simply drew rectangles under the curve that kind of followed the shape of the shape of, the graph, you should get an okay approximation for the area under the curve.

00:37 So part a asks us to use, i believe it was, three right endpoints.

00:43 Is that correct? yes, three rectangles and right endpoints.

00:47 So what they mean by right endpoints is they're referring to, where do we stop with our rectangle? here, when i did my sample drawing here, i went up here on the left and went to the right.

00:58 That's what we would call using a left endpoint.

01:02 For a right endpoint, we would want to start here, go up to the curve, and then go left.

01:10 And that's why it's the right endpoint because the right part of the rectangle ends on the curve.

01:18 So first off, we know that the interval wants us to go from 1 minus 1 to all the way up to 2.

01:26 And we can see that we could form.

01:28 Let me just draw a number line.

01:29 I'll keep coming back to this.

01:31 For a little bit.

01:32 So we have minus one, one, nope, that's next to zero, one, and then two.

01:40 I'll make these a little more clear.

01:42 And so if we drew three right end point rectangles, we see that we have this interval, we have this interval, this interval, and that's all three of our intervals.

01:54 And we can see that if we want to use the right end point, we should use 0, 1, and 2.

02:00 And so going back up to our graph here, we can see that when i say zero, it means that we have to find the function at zero and then find the area of this rectangle.

02:10 And using the formula for base times height, we can say that the height of the rectangle is the function at that point, in this case it's f of zero.

02:22 And the width is this interval right here, which as we can see here from zero to minus one, the interval is one.

02:29 So for part a using three endpoints, we're going to want to find, i'll say, r3 equals 1 times f of 0 plus 1 times f of 1.

02:46 Whoops, let me rewrite that real quick.

02:49 One times f of 1 plus 1 times f of 2.

02:55 Because we're using these three trying rectangles right here on the right endpoints.

03:00 This one, this one, and this one.

03:06 And if you were to calculate all that, you would find that it should equal eight.

03:11 And i won't go into the specifics of that, because this should be relatively simple, i feel.

03:17 And if it's not, then by all means, take a little bit of time to try and solve it on your own, where i just simply omitted substituting in these values right here.

03:29 So next, it asks us to do six endpoints.

03:31 Well, in order to do six, we could just try and divide these ones in half.

03:36 So we have 1 .5, 0 .5, and minus 0 .5.

03:43 And that gets us this interval plus this interval, this one, this one, this one, and this one.

03:51 And that's six intervals, which means that we'll have six rectangles.

03:55 And i'll try to draw them here.

03:56 One, two, three, four, five, six.

04:04 So all of your six rectangles should look something like that.

04:07 And i'll try to hit undo, oops, whoops, whoops.

04:11 I'm going to hit undo too much.

04:14 And so for our six, as i'll call it, we'll notice that our width has actually changed a little bit.

04:20 Instead of being from zero to one with width one, in this case we went from zero to point five, which means our width is only 0 .5.

04:29 So we're going to have 0 .5 times f of something.

04:33 What's the first f of something? last time we did f of zero because our first interval was this one right here, which went from minus 1 to 0.

04:41 But in this case, our first interval is minus 1 to minus 0 .5.

04:44 And since we're wanting to do the right value, that's why we have this r here, we'll use the value on the right, which is minus 0 .5.

04:53 So we have 0 .5 times f of minus 0 .5.

04:58 Plus 0 .5 times f.

05:02 Now what's the next one going to be? well, we just go over one more, and it's 0.

05:07 So you keep on doing that all of these until you end up with f of 0 .5 times f of 2, because that's our final thing.

05:18 And if you want to take a moment to pause this video and try and calculate all that on your own, by all means go for it.

05:26 My calculations got me this to being 6 .875, which is a little smaller than r3.

05:34 And we can see that sort of makes sense because, for example, if we went over here, we didn't use this rectangle in our calculations, but just as an example, you see here that we have all of this extra space here that we technically don't want.

05:51 But if we did two rectangles for right here and right here, we'll see that we only had this much extra space.

05:58 So that's why even though we have more rectangles, our answer is actually smaller than before.

06:04 So now we're going to change colors.

06:07 We're going to do l3 this time.

06:09 And the process is pretty similar.

06:11 Just notice right here, instead of doing the value on the right side, we'll want to do the left side.

06:17 So with this, we want to start with minus 1, and then we want to do 0, and then 1.

06:22 And for our, actually, i'll go ahead and write that down here.

06:26 We got l3 equals 1 times f of minus 1, plus 1 times f of 0, plus 1 times f of 1.

06:40 And putting all that together, my answer for this is 5.

06:46 And if we use the 6, l6, we'll have minus 1, minus 1, 0 .5, 1, and 1 .5.

06:56 Because these are our intervals, and we want to use the left side this time.

07:00 So let's see if i can undo the correct amount.

07:03 Hopefully that was it.

07:04 Looks like it...

Please give Ace some feedback