MATH 12002 - CALCULUS I 4.4: Average Value of a Function Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.4: Average Value of a Function

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Average Value of a Function

Let y = f (x) be a continuous function on the interval [a, b]. We would like to compute the average y-value of this function. Since there are infinitely many y-values (unless f is a constant function) we cannot simply add them up and divide by the number we have. We will take an approach similar to what we did to compute areas:

1 Compute a finite approximation;

that is, find the average of the y-values at finitely many x-values.

2 Take the limit as the number of x-values approaches infinity.

Since this is very similar to the procedure for computing definite integrals, it should not be surprising that computing average value will involve a definite integral.

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Average Value of a Function

Let y = f (x) be a continuous function on the interval [a, b]. For each positive integer n, partition the interval [a, b] into n equal subintervals, each of length ∆x = b−a

n , with endpoints

a = x0 < x1 < x2 < · · · < xn−1 < xn = b. Compute the average of the y-values at the right endpoints of the subintervals: An = f (x1) + f (x2) + · · · + f (xn−1) + f (xn) n . We define the average value fave of f on [a, b] by fave = lim

n→∞ An.

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Average Value of a Function

To compute the limit of this average value as n → ∞, observe that An = f (x1) + f (x2) + · · · + f (xn−1) + f (xn) n = f (x1) + f (x2) + · · · + f (xn−1) + f (xn) n · b − a b − a = 1 b − a · [f (x1) + f (x2) + · · · + f (xn−1) + f (xn)] · b − a n = 1 b − a · [f (x1)∆x + f (x2)∆x + · · · + f (xn−1)∆x + f (xn)∆x] = 1 b − a · Rn, where Rn is the nth right Riemann sum for the integral b

a f (x) dx.

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Average Value of a Function

Finally, since

1 b−a is a constant, we have that

the average value fave of f (x) on the interval [a, b] is fave = lim

n→∞ An = lim n→∞

1 b − aRn = 1 b − a lim

n→∞ Rn =

1 b − a b

a

f (x) dx.

Average Value

If y = f (x) is a continuous function, then the AVERAGE VALUE

• f f on the interval [a, b] is

fave = 1 b − a b

a

f (x) dx.

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Example

1 Find the average value of f (x) = x2 + 4 on the interval [3, 12].

The average value is 1 12 − 3 12

3

(x2 + 4) dx = 1 9

• 1

3x3 + 4x

• 12

3

• =

1 9 1 3(123) + 4(12)

• −

1 3(33) + 4(3)

• =

1 9 [(576 + 48) − (9 + 12)] = 1 9 [624 − 21] = 603 9 = 67. Note that 12

3 (x2 + 4) dx = 603 is NOT the average value.

The integral must be divided by the length of the interval.

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Average Velocity

Finally, suppose an object is moving along a straight line so that its position at time t is s(t), and so its velocity at time t is v(t) = s′(t). Previously, we discussed average velocity on a time interval t = a to t = b. We now also have the average value of the velocity function on [a, b]. Are they the same thing? Recall that average velocity on the interval t = a to t = b is ∆s ∆t = s(b) − s(a) b − a . Since s(t) is an antiderivative for v(t), we have that the average value of the velocity function on [a, b] is 1 b − a b

a

v(t) dt = 1 b − a · s(t)

• b

a

= 1 b − a [s(b) − s(a)] = s(b) − s(a) b − a . Thus average velocity is equal to the average value of the velocity function.

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