More on Volumes & Average Function Value

Chapter 6.5

March 1, 2007

Average

On the last test (2), the average of the test over 2 classes was:

=  TestScores

i=1

142

∑1428.1

FYI - there were 71 who scored a 9 or 10, which means the median was a 9.

Chapter 6.5

March 1, 2007

Average Function Value of f(x) on the interval [a,b]

We can divide the interval [a,b] into n subintervals

and average the selected function values.

Δx =b − a

n

f (x1) + f(x2 ) + ...+ f(xn)n

=1

nf (xi )

i=1

n

∑

or  1

n=

Δxb−a

⎛⎝⎜

⎞⎠⎟

=Δxb − a

f (xi )i=1

n

∑

=1

b − af (xi )

i=1

n

∑ Δx

Average function value

If we let then number of points selected go to infinity

We arrive at the Definite Integral!

Ave f ≈1

b−af(xi )

i=1

n

∑ Δx

limn→ ∞

1b−a

f(xi )i=1

n

∑ Δx

Ave f =1

b−af(xi )a

b

∫ dx

Find the average function value ofover the interval

f (x) =x2 −1

Ave f =13

x2 −1( )0

3

∫ dx

0, 3⎡⎣ ⎤⎦

Find the average function value ofover the interval

f (x) = x

Ave f =14

x0

4

∫ dx

0, 4[ ]

Average function value

If we let multiply both sides of the formula, we get:

Thinking area: The area of the rectangle (b-a) by Avef has the same area as the area under the curve as seen…..

Ave f =1

b−af (xi )

a

b

∫ dx

(b−a)Avef = f(xi )a

b

∫ dx

Find the average function value ofover the interval

f (x) = x

Ave f =14

x0

4

∫ dx

        =43

0, 4[ ]

x =43

 x=164

Solving for x, we get:

The area of the green rectangle = the area under over the interval

f (x) = x0, 4[ ]

Let R be the region in the x-y plane bounded by

Set up the integral for the volume of the solid obtained by rotating R about the line y = 3,

a) Integrating with respect to x.

b) Integrating with respect to y.

x =y2   and  x=4

Integrating with Respect to x:

Outside Radius ( R ):

Inside Radius ( r ):

Area:

Volume:

x =y2   and  x=4

y =± x  and  x=4

Integrating with Respect to y:

Length of Slice:

Inside Radius ( r ):

Area:

Volume:

x =y2   and  x=4