Help mee to solve this?

Question

Find the exact average value of f (x) =3x2 −2x on the interval [1,3].

I think polar bear is right??

Answer 1

first integrate

F(x) = x^3 -x^2 from [1, 3]
= 3^3 -3^2 -1^3 + 1^2
= 27 - 9 -1 + 1
= 18

average = 18/(3-1)
= 18/2
= 9

Answer 2

First off, to calculate the average value of a function from a to b, you have to mind this formula for the average.

Average value of function =

1/(b-a) * Integral(a to b, 3x^2 - 2x)dx

In our case, the interval is 1 to 3. Thus, we replace a and b with those values.

1/(3-1) * Integral (1 to 3, 3x^2 - 2x)dx
1/2 * Integral (1 to 3, 3x^2 - 2x) dx

3
1/2 * (x^3 - x^2)|
1
(Note the floating 3 and 1 is supposed to represent the subscripts for after evaluating the integral).

1/2 * [ (3^3 - 3^2) - (1^3 - 1^2) ]
1/2 ( (27 - 9) - (1 - 1) )
1/2 (18 - 0)
1/2 (18)
9

Answer 3

To find the exact average of "f(x) = 3x2 − 2x" on the interval [1,3], this is what you do:

Find [f(3) - f(1)] / [3 - 1]:

= [21 - 1] / [2]

= 10

* I guess you really DO have to use integration to find the exact value. Sorry for the wrong answer.

* To puggy: Thanks for explaining it so well. I didn't know how to do it because i'm in calculus 1, but now I know. Thanks again.

Answer 4

average value = 1/(3-1) integral from 1 to 3 of 3x^2 -2x dx
= (1/2) (3x^3 /3 -2x^2 /2 evaluated from 1 to 3)
=(1/2) ( (3)^3 -3^2 -(1^3 -1^2))
= (1/2) (27 -9 )=18/2 = 9
.
.