MATH 12002 - CALCULUS I 4.4: Average Value & Geometry Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.4: Average Value & Geometry

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 12

Average Value

Recall that the average value of a function is found as follows:

Average Value

Let y = f (x) be a continuous function. The average value of f on the interval [a, b] is fave = 1 b − a b

a

f (x) dx. The average value of a positive function can be described in terms of areas.

D.L. White (Kent State University) 2 / 12

Average Value

Suppose y = f (x) is continuous and f (x) 0 for a x b, with average value fave =

1 b−a

b

a f (x) dx on [a, b].

We know that b

a f (x) dx is the area under the graph of f on [a, b].

Since fave =

1 b−a

b

a f (x) dx, this area is also equal to (b − a)fave.

Observe that (b − a)fave is the area of the rectangle of height fave on [a, b].

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 3 / 12

Average Value

Suppose y = f (x) is continuous and f (x) 0 for a x b, with average value fave =

1 b−a

b

a f (x) dx on [a, b].

We know that b

a f (x) dx is the area under the graph of f on [a, b].

Since fave =

1 b−a

b

a f (x) dx, this area is also equal to (b − a)fave.

Observe that (b − a)fave is the area of the rectangle of height fave on [a, b].

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 4 / 12

Average Value

Hence the average value fave of y = f (x) on [a, b] is the number such that the area under the horizontal line y = fave between x = a and x = b is equal to the area under the graph of y = f (x) between x = a and x = b.

f ✲ ✛ ✻ ❄

a b fave

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 5 / 12

Average Value

Notice that the rectangle and the region under the graph of f have the black shaded region in common. Since the total areas are equal, it follows that the areas of the red shaded regions must be equal.

f ✲ ✛ ✻ ❄

a b fave

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 6 / 12

Average Value

If you aren’t convinced by those pictures, you can watch the red shaded region move back and forth:

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 7 / 12

Average Value

If you aren’t convinced by those pictures, you can watch the red shaded region move back and forth:

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 8 / 12

Average Value

We can also interpret the average value fave as the number such that: the area above the horizontal line y = fave and below the graph of y = f (x) between x = a and x = b and the area below the horizontal line y = fave and above the graph of y = f (x) between x = a and x = b are equal.

f ✲ ✛ ✻ ❄

a b fave D.L. White (Kent State University) 9 / 12

Mean Value Theorem For Integrals

By the Extreme Value Theorem, if y = f (x) is continuous on the closed interval [a, b], then f has an absolute minimum m and an absolute maximum M on [a, b]. Thus m f (x) M on [a, b], and so m(b − a) b

a

f (x) dx M(b − a), by a comparison property of definite integrals. Dividing by b − a, we get m 1 b − a b

a

f (x) dx M; that is, m fave M. By the Intermediate Value Theorem, f takes on every y value between m and M, and therefore there is a c with a < c < b such that f (c) = 1 b − a b

a

f (x) dx = fave.

D.L. White (Kent State University) 10 / 12

Mean Value Theorem For Integrals

This is the Mean Value Theorem for Integrals:

Theorem

If y = f (x) is a continuous function on [a, b], then there is a number c with a < c < b such that f (c) = 1 b − a b

a

f (x) dx. The theorem says that a continuous function on a closed interval must take on its average (i.e., mean) value on the interval.

D.L. White (Kent State University) 11 / 12

Mean Value Theorem For Integrals

If y = f (x) is continuous on [a, b] and F is an antiderivative of f , then b

a f (x) dx = F(b) − F(a), by the Fundamental Theorem of Calculus,

and so 1 b − a b

a

f (x) dx = F(b) − F(a) b − a and f (c) = F ′(c). Hence the Mean Value Theorem for Integrals implies F(b) − F(a) b − a = F ′(c) for some c with a < c < b. This is the Mean Value Theorem (for F) that we studied previously.

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