Do Now - #22 and 24 on p.275 Graph the function over the interval. Then (a) integrate the function over the interval and (b) find the area of the region.

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Presentation transcript:

Do Now - #22 and 24 on p.275 Graph the function over the interval. Then (a) integrate the function over the interval and (b) find the area of the region between the graph and the x-axis. Possible graph window: [0, 2] by [–5, 3] An antiderivative of the function: (a)

Do Now - #22 and 24 on p.275 Graph the function over the interval. Then (a) integrate the function over the interval and (b) find the area of the region between the graph and the x-axis. Possible graph window: [0, 2] by [–5, 3] (b) Area =

Do Now - #22 and 24 on p.275 Graph the function over the interval. Then (a) integrate the function over the interval and (b) find the area of the region between the graph and the x-axis. Possible graph window: [0, 3] by [–5, 5] An antiderivative of the function: (a)

Do Now - #22 and 24 on p.275 Graph the function over the interval. Then (a) integrate the function over the interval and (b) find the area of the region between the graph and the x-axis. Possible graph window: [0, 3] by [–5, 5] (b) Area =

Average Value of Functions Section 5.3b

First, remind me… How do you find the average of n numbers?  Sum the numbers and divide by n !!! Our New Challenge: How do we find the average value of an arbitrary function f over a closed interval [a, b]???

Does this look familiar??? We take a large sample of n numbers from regular subintervals of the interval [a, b], taking number from each subinterval of length The average value of the n sampled values is Does this look familiar???

So what happens as n approached infinity? We take a large sample of n numbers from regular subintervals of the interval [a, b], taking number from each subinterval of length The average value of the n sampled values is Can you say RIEMANN SUM??? So what happens as n approached infinity?

Definition: Average (Mean) Value If is integrable on [a, b], its average (mean) value on [a, b] is

The Mean Value Theorem for Definite Integrals If f is continuous on [a, b], then at some point c in [a, b], The value f (c) in the MVT is, in a sense, the average (or mean) height of f on [a, b]. When f > 0, the area of the shaded rectangle a c b is the area under the graph of f from a to b. b – a

The average value of the Find the average value of the given function on [0, 3]. Does the function actually take on this value at some point on the interval? Use NINT!!! The average value of the function on [0, 3] is 1. The function assumes this value when Since this x value lies in the interval [0, 3], the function does assume its average value in the given interval.

The rectangle with base [0, 3] and with height equal to 1 (the average Find the average value of the given function on [0, 3]. Does the function actually take on this value at some point on the interval? The average value of the function on [0, 3] is 1. The function assumes this value when 4 The rectangle with base [0, 3] and with height equal to 1 (the average value of the function) has area equal to the net area between f and the x-axis from 0 to 3. 4 –5

Quality Practice Problems Find the average value of the given function on the given interval. At what point(s) in the interval does the function assume its average value? Find the point: In the given interval:

Quality Practice Problems Find the average value of the given function on the given interval. At what point(s) in the interval does the function assume its average value? Find the point: Both are in the given interval!