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Sketch the graph of a continuous function on $ [0, 2] $ for which the Trapezoidal Rule with $ n = 2 $ is more accurate than the Midpoint Rule.
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Calculus 2 / BC
Techniques of Integration
Oregon State University
Harvey Mudd College
In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative.
The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis.
The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
Sketch the graph of a cont…
Use (a) the Trapezoidal Ru…
Okay, This question wants us to find a function. Such the trap is I'd rule is a better approximation in the mid point, which usually isn't the case sort of do this. Let's find a function that totally messes with the midpoint approximation. So to do that, we gotta find a function that does it not represent itself at the mid points, which means acts totally different at the mid points. And it does everywhere else, but it still has to be continuous. So let's say we have something like this. So it starts it one, then it hit zero at the midpoint goes back up. Then it goes back down a hits 1.5 again. And then you see the pattern so we can see for sure that area from midpoint, it's just gonna be Delta X plus the sum of f of mid points, which is just Delta X Times zero, which equals zero. Because the only points were using the sample from are the zeros right here. And if you're wondering what this function is, I just picked absolute value of co sign of Pi X just to get the period and positive definite nous that we wanted. So for the trap is oId. On the other hand, we get 1/2 times f of zero plus two F of one plus f of two, which is 1/2 times one plus one plus one. So area from the trap is oId is too, and actual area. Oh, and just for reference, we should write again. The area from midpoint is equal to zero, so we know the actual area is positive. So we should already have an intuition that the midpoint is a terrible approximation. But if we actually calculate this integral out, it's approximately 1.27 So the area of the trap is oId is closer.
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