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# Sketch the graph of a continuous function on $[0, 2]$ for which the right endpoint approximation with $n = 2$ is more accurate than Simpson's Rule.

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Integration Techniques

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Okay, so this question wants us to find an approximation such that the error from the Simpson approximation is greater in the area of the Wraith approximation, which usually isn't the case. So to do this, let's just construct a graph. We don't need it necessarily define a rule for it. We just want something that messes with Simpsons Rule and really favors the right end point rule. So let's say that this is zero right here and this is one and this is 0.5. So let's say we have a function that's flat up until it reaches one, and then it's sort of spikes upto one, and then it stays there. So just by looking at this, we should see that the area the area is approximately 1.5. It's gonna be very close. So area should be close to 1.5 just a little bit less. But to find our approximations, let's do the right approximation first, which is Delta X Times effort two plus f of one. Remember, we skip the left point in a right some So Delta X is one f of two is one, and death of one is on half so the right approximation This 1.5, which is almost exactly what we think are real area is. But let's find the Simpson approximation. So the Simpson approximation is 1/3 time's half of zero plus four times after one plus effort, to which is 1/3 time's well f of zero, this 0.5 four times F of one, which is again 0.5 plus half of point to which again, we said, is one. So it's equal to 1/3 time's one 1.5 plus two. She's equaled a 3.5 over three, which is equal to about 1.16 So we see that are too is 1.5, and as to is 1.16 so and our actual is approximately 1.5. So are right. Good point is the better approximation.

University of Michigan - Ann Arbor

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Integration Techniques

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