Question

Calculus.

Answer 1

Use the Mean Value Theorem to determine the `average` of f(x) on the interval [0,2].

Is this maybe what they meant to ask?
Because area would simply be the integral from 0 to 2...
It wouldn't require MVT at all.

Answer 2

if i remember right,

if you take an interval [a,b] and calculate the secant slope for line AB

the point on the curve inside the inverval c, where the slope of the tangent is the same as the secant AB

that is the average value , mean value over the interval

Answer 3

that value at x=c, will be the height of a rectangle....

Answer 4

a rectangle over the same interval , with a height of f(x) at that x=c value, will have the same area as the function does over the interval

Answer 5

Right.... Can you set it up for me?

Answer 6

So the definate integral for f(x) over the interval
=
(b - a) * f(c)

Answer 7

2x^2+3 **

Answer 8

(b-a)*f(c) is a rectangle base (b-a) and height f(c), that mean value

Answer 9

?

Answer 10

\[\int\limits_{0}^{2}(2x^2 + 3)dx = f(c)*(2 - 0)\]

Answer 11

34/3 ?

Answer 12

solving that for f(c)

you get the height of a rectangle that has the same area as the given function

34/3 = f(c) * 2

f(c) = 34/6 = 17/3
and the c value for x
c = 2root(3)/3

Answer 13

not sure what way the book says the formula, probably similar, i just remember for the areas, that middle c value will give you the height , so a rectangle with that height has the same area as the function, over the interval

Answer 14

definite integral = area of rectangle f(c)*(b-a)

Answer 15

alright thanks. here is another one.

Answer 16

yeah a similar version of this mean average value is seen with slopes of lines too, not only the areas

you remember the average value of a function over an interval [a,b]
is the slope of the line from each endpoint value... secant slope?

Answer 17

\[\frac{ f(b) - f(a) }{ b-a }\]

Answer 18

so we can just put the lower and upper bounds and integrate right?

Answer 19

\[\int\limits_{0}^{2} 3x^2-2 dx \]

Answer 20

-2-36-2/-2 the way you did it?

Answer 21

-12 not -36

Answer 22

Please exclose the integrand in parentheses:\[\int\limits\limits_{0}^{2} (3x^2-2) dx\]

Answer 23

when i do it dan's way i get 8

Answer 24

sorry had to step away a min


the slope of a secant line through the interval end ponts (ave function value)
is seen also at a tangent to the function at a point (c,f(c)) in the interval

Answer 25

right. When i plug in f(b) and f(a) into the equation you gave me I got 8

Answer 26

That's ok :)

Answer 27

similar to the last way to solve the area prob above, the secant slope is this over interval [a,b]

\[\frac{ F(b) - F(a) }{ b-a}=f'(c)\]

notice that is also the definite integral value in the numerator from a to b

Answer 28

secant slope = tangent slope at c

you can use the definate integral to get the F(b) - F(a)


rearranging the formula, it is the same as the area prob above

Answer 29

alright i got 8 for the left side how do i solve to get the right side?

Answer 30

\[\frac{ 1 }{ b-a } \int\limits_{a}^{b}f(x)dx=f '(c)\]

Answer 31

here the idea is slope of a tangent at (c,f(c)) as the average value of the curve

last time, it is the height of a rectangle having same area as the curve

Answer 32

so f'(c)=8 so 8 is our answer?

Answer 33

same fundemental idea, in a dff way

Answer 34

so 8 is our answer?

Answer 35

i forgot what the actual prob was..

Answer 36

here u go

Answer 37

This one is 30)

f(x) = 3x^2 - 2 [0,2)

Answer 38

3(0)^2-2-3(2)^2-2/0-2=8

Answer 39

\[\frac{ f(b) - f(a) }{ b-a}=f'(c)\]
\[\frac{ 1 }{ b-a } \int\limits\limits_{a}^{b}f(x)dx=f (c)\]


sorry i forgot, when you change f(b)-f(a) in the slope for the secant line to the definite integral value F(b)-F(a), the right needs to be integrated also, f ' (c) goes to f(c)

Answer 40

\[\large \frac{ 1 }{ 2-0 } \int\limits\limits\limits\limits_{0}^{2}(3x^2-2)dx=f (c)\]

= (1/2) * 4


the integral is 4, so you get

f(c) = 2

Answer 41

and so the c value is

c=2root(3)/3

Answer 42

this is for #30?

Answer 43

yeah , f(x)=3x^2 - 2 ; [0,2]

Answer 44

the definite integral for that is 4, and over the interval length [2-0], it becomes 2

that is the value , f(c) = 2

Answer 45

so c is
2 = 3*c^2 - 2

gives you c = 2*root(3)/3