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In the following exercises, approximate the average value using Riemann sums L 100 and R 100 . How does your answer compare with the exact given answer?
[T] over the interval the exact solution is
[T] over the interval the exact solution is
the exact average is between these values.
[T] over the interval the exact solution is
In the following exercises, compute the average value using the left Riemann sums L N for How does the accuracy compare with the given exact value?
[T] over the interval the exact solution is
[T] over the interval the exact solution is
The exact answer so L 100 is not accurate.
[T] over the interval the exact solution is
[T] over the interval the exact solution is
The exact answer so L 100 is not accurate to first decimal.
Suppose that and Show that and
Show that the average value of over is equal to 1/2 Without further calculation, determine whether the average value of over is also equal to 1/2.
Show that the average value of over is equal to Without further calculation, determine whether the average value of over is also equal to
so divide by the length 2 π of the interval. has period π , so yes, it is true.
Explain why the graphs of a quadratic function (parabola) and a linear function can intersect in at most two points. Suppose that and and that Explain why whenever
Suppose that parabola opens downward and has a vertex of For which interval is as large as possible?
The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, and
Suppose can be subdivided into subintervals such that either over or over Set
Suppose f and g are continuous functions such that for every subinterval of Explain why for all values of x .
If for some then since is continuous, there is an interval containing t 0 such that over the interval and then over this interval.
Suppose the average value of f over is 1 and the average value of f over is 1 where Show that the average value of f over is also 1.
Suppose that can be partitioned. taking such that the average value of f over each subinterval is equal to 1 for each Explain why the average value of f over is also equal to 1.
The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, Dividing through by gives the desired identity.
Suppose that for each i such that one has Show that
[T] Compute the left and right Riemann sums L 10 and R 10 and their average for over Given that to how many decimal places is accurate?
[T] Compute the left and right Riemann sums, L 10 and R 10 , and their average for over Given that to how many decimal places is accurate?
so the estimate is accurate to two decimal places.
If what is
Estimate using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value
The average is which is equal to the integral in this case.
Estimate by comparison with the area of a single rectangle with height equal to the value of t at the midpoint How does this midpoint estimate compare with the actual value
From the graph of shown:
a. The graph is antisymmetric with respect to over so the average value is zero. b. For any value of a , the graph between is a shift of the graph over so the net areas above and below the axis do not change and the average remains zero.
If f is 1-periodic odd, and integrable over is it always true that
If f is 1-periodic and is it necessarily true that for all A ?
Yes, the integral over any interval of length 1 is the same.
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