When approximating integrals using the trapezium rule, the type of approximation (overestimate or underestimate) depends on the relationship between the trapezoids and the curve. For an even function like sin²x, reflecting the integral bounds (from A to B to -B to -A) produces the same approximation type because the curve shape remains identical. However, reflecting the function itself (changing sin²x to -sin²x) reverses the approximation type, as the trapezoids now lie below the curve instead of above it.
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i didn't know this was on the TMUA spec??Added:
Today, I'm going to be solving a problem from the TMUA. I know a lot of you guys are preparing for the TMUA. This comes from the 2019 paper two. And actually, what I'm going to be doing, I'll give you a little bit of a behind-the-scenes exclusive. Soon, I'm going to be releasing my own papers that I've created for my students, and I'm going to be sharing some of the problems here on this YouTube channel. So, it gives you an extra source of problems, cuz I know that there aren't that many TMUA papers to work from. So, if you do want to see those, make sure you are subscribed and you follow and turn on the notification bells, all that sort of stuff. Anyway, let's look at this one. A student approximates the integral from A to B of sin² x dx using the trapezium rule with four strips.
The resulting approximation is an overestimate. Which of the following is/are necessarily true? If the student approximates the integral from -B to -A of sin² x dx in the same way, the result will be an overestimate. And part two, if the student approximates the integral from A to B of cos² x dx in the same way, the result will be an underestimate. Okay, do pause the video and have a go at this problem yourself.
I'm going to dive right in.
Now, for this question, we don't actually need to use the trapezium rule.
We just need to kind of understand what it does. So, what you don't want to do is try and like use the crazy formula, where you don't even know what n is and all these sorts of things. Well, actually, you do know it's four strips, but you don't know what A is, you don't know what B is. You don't want to write out a whole formula. We just want to think visually about what this is. So, I'm actually going to start with two to begin with, just cuz I think that's easier. So, we've got the integral from A to B of cos² x dx. And the idea is that that's the integral from A to B of 1 sin² x dx.
And now, crucially, the idea is if we're using a trapezium rule, if I split this into two integrals, it might be easier to see. But this one doesn't actually really make a difference. That's just going to correspond to translating the graph of y = -sin² x. This plus one here just corresponds to translating it up or down. So, in terms of the trapezium rule, it has no impact on whether it's an overestimate or underestimate. Um what about this minus sine squared x?
Well, if we know that sine squared x, how would you know, whichever bound will whatever our A and B are whatever sine squared x looks like, which also we should be comfortable with drawing, looks something like this.
Like if we're doing the the trapezium rule with our, you know, strips there and whatever it is and we're doing our four strips, right?
So, I don't know, it might look like this. There's one trapezium. There's two.
There's three and there's four. It's not really equally spread out, but you get the idea.
We work out the area of that, that, that, and that and we get an overestimate, let's say.
Well, the idea here and the thing that we want to think about is well, if we're looking at minus sine squared x, the curve would just be the same, but reflected in the x-axis. So, it's going to look something like this.
Sort of thing. And then these points are still going to be like the same, just the corresponding negatives.
So, it's going to look something I'm drawing this very well.
Something like this.
I haven't drawn that well at all, but you know, you get that trapezium, that trapezium, that trapezium, and that trapezium there.
And the idea here is well, now this is for sure going to be an underestimate.
Because if the red area above was bigger to we know that the red is bigger than the integral of blue, then we know that the green trapeziums, the green trapeziums, just because of area, the area of the green trapeziums will be bigger than the area under the or area bounded by the x-axis and the green curve, but that therefore means that the green area is a more negative number than the area of the green curve.
And so therefore, the the trapezium rule, which is this guy here, would give us an underestimate for the integral of the green area. And so therefore we would get an underestimate. So two is certainly true.
Okay, cool. What about statement one?
Well, we're looking at the integral from minus B to minus A of sin squared X DX.
If I just give myself some space over here, let's move this thing across.
So to look at this thing, again we're just going to think about a sketch. Oops.
So sin squared X very conveniently is an even function. And so if I have A there and I have B there, let's say, and I do the same thing over here. It's not I'm not drawn very well, but minus A will be there and minus B would be there. And you can see that the area is the exact same shape and the curvature between is the exact same. And so if we're using this trapezium rule here to give us an overestimate, whatever the trapeziums look like or trapezias look like, not drawn very well, but we're going to get the exact same but reflected trapezias over here.
So the these trapeziums are supposed to be the same as those trapeziums. Not drawn very well, but they're just reflected. And so if they go above the curve on average over here, they're going to go above the curve on average over here. It's just literally a reflection of the image. And so therefore we're going to get an overestimate here. And so statement one is also true. And therefore the answer here is D. Now, I've solved this problem kind of explaining everything in depth, but strong TMUA students, they will be able to recognize like for example this straight away. And because they see the minus sign and that they've seen this logic before, they'll immediately go great. That's now gone from an overestimate to an underestimate. They can immediately circle two is true.
And similarly with one, they can almost in their heads visualize, okay, cool.
This this reflection or this is just a reflection and thus the areas and nothing actually changes just apart from where I'm drawing these trapezias. And so again, um the this will be at same. If this is an overestimate, this will be an overestimate. So, do lots and lots of these questions, but make sure you're employing these tricks of thinking about what the graphs look like, and thinking about, okay, if it's reflected, then that will change whether it goes from overestimate to an underestimate, or underestimate to overestimate, if it's reflected in the x-axis, that is, or in a horizontal line. But, if it's reflected in the y-axis, that won't change anything. Um anyway, that's how we solve this problem. Stay tuned for my exclusive TMUA style papers, which I'll be sharing very, very soon on this channel.
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