Question
Asked Apr 29, 2019
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We can improve any Riemann approximation by
increasing n, the number of rectangles used in our
approximation. In the previous approximations we used
n- 3, so now we will try n 6
25
20
For this value of n you will only need to calculate one
Riemann sum - either the left, right, or midpoint
Riemann sum. Make your selection below, then move
the points on the graph to the left so that the image
matches your choice
15
10
Left Riemann sum
Right Riemann sunm
Midpoint Riemann sum
2 -10
help_outline

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We can improve any Riemann approximation by increasing n, the number of rectangles used in our approximation. In the previous approximations we used n- 3, so now we will try n 6 25 20 For this value of n you will only need to calculate one Riemann sum - either the left, right, or midpoint Riemann sum. Make your selection below, then move the points on the graph to the left so that the image matches your choice 15 10 Left Riemann sum Right Riemann sunm Midpoint Riemann sum 2 -10

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Expert Answer

Explanation:

Yes. Functions that increase on the interval [a,b] will be underestimated by left-hand Riemann sums and overestimated by right-hand Riemann sums. Decreasing functions have the reverse as true.

The midpoint Riemann sums is an attempt to balance these two extremes, so generally it is more accurate.

The Mean Value Theorem for Integrals guarantees (for appropriate functions f) that a point c exists in [a,b] such that the area under the curve is equal to the area f(c)⋅(b−a).

We can extend this idea to each subinterval in a Riemann sum. Each subinterval ii has a point c_i where the rectangular area matches the area under the function on the subinterval.

There likely isn't a nice pattern for identifying these points, however. The midpoint Riemann sum is usually the best easy approximation for these points.

Final Statement:

The option midpoint riemann sum is better approach.

...

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MathCalculus

Integration

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