Want this question answered?
why doesn't wiki allow punctuation??? Now prove that if the definite integral of f(x) dx is continuous on the interval [a,b] then it is integrable over [a,b]. Another answer: I suspect that the question should be: Prove that if f(x) is continuous on the interval [a,b] then the definite integral of f(x) dx over the interval [a,b] exists. The proof can be found in reasonable calculus texts. On the way you need to know that a function f(x) that is continuous on a closed interval [a,b] is uniformlycontinuous on that interval. Then you take partitions P of the interval [a,b] and look at the upper sum U[P] and lower sum L[P] of f with respect to the partition. Because the function is uniformly continuous on [a,b], you can find partitions P such that U[P] and L[P] are arbitrarily close together, and that in turn tells you that the (Riemann) integral of f over [a,b] exists. This is a somewhat advanced topic.
It can be expected to change gradually over time, but the difference from one year to the next, or even in thousands of years, will be insignificant.
Why does the price of a bond change over its lifetime?
A counter accumulates an unknown quantity of external events over a known interval of time.The measurement of interest is typically frequency when the events are periodic. If the events are random, the measurement involves event density over time.A timer accumulates a series events of a known interval over an interval that is being measured.The measurement of interest is typically the time elapsed between two events. If the start and stop events recur periodically, multiple measurements can be made and averaged, allowing for increased resolution.Counter/timers in MPU's are typically just counters that count external events in counter mode and processor cycles in timer mode.
Slave codes I think
The linear function changes by an amount which is directly proportional to the size of the interval. The exponential changes by an amount which is proportional to the area underneath the curve. In the latter case, the change is approximately equal to the size of the interval multiplied by the average value of the function over the interval.
Average velocity is change in position (displacement) divided by the interval.
There have to be two (or more) ordered pairs for an average rate of change to make any sense. Your question does not.
measure of the average responsiveness of quantity to price over an interval of the demand curve. = change in quantity/ Quantity ___________________________ change in price/ Price
The mean value theorem for differentiation guarantees the existing of a number c in an interval (a,b) where a function f is continuous such that the derivative at c (the instantiuous rate of change at c) equals the average rate of change over that interval. mean value theorem of integration guarantees the existing of a number c in an interval (a,b)where a function f is continuous such that the (value of the function at c) multiplied by the length of the interval (b-a) equals the value of a the definite integral from a to b. In other words, it guarantees the existing of a rectangle (whose base is the length of the interval b-a that has exactly the same area of the region under the graph of the function f (betweeen a and b).
Acceleration is the rate of change of velocity - in symbols, a = dv/dt. Or for average acceleration over a finite time: a(average) = delta v / delta twhere delta v is the change in velocity, and delta t is the time interval.
The range of a function is the interval (or intervals) over which the independent variable is valid, i.e. results in a valid value of the function.
V = d / tVelocity is the change in distance over an interval of time.
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
The average velocity over an time interval is the average of the instantaneous velocities for all instants over that period. Conversely, as the time interval is reduced, the average velocity comes closer and closer to the instantaneous velocity.
yes, aka rise over run.
In general, the acceleration during that time interval could vary considerably. However, we can calculate the average acceleration during the interval. The change in speed is 20 meters per second - 5 meters per second = 15 meters per second, and this change in speed occurs over a 3 second interval. Thus the average change in speed over this interval is 15 meters per second/ 3 seconds = 5 meters per second per second = 5 meters/second2