Question

find the average value of the function f(x)=1+x^2 on interval [-1,2]
i need help on last step where you plug the -1, 2 in the soultion

Answer 1

you may try it now.

Answer 2

write down the formula first.

Answer 3

i know but i could not know how mutiply...1/3[x+x^3/3] to the interval -1, 2

Answer 4

you are talking about the average value of the function, use this formula : [f(x+h)-f(x)]/h , wher h=b-a

Answer 5

i think more like 1/b-1 intergral a to b f(x) dx

Answer 6

you are talking then about the mean value theorem by integration

Answer 7

no the average value function...just do the steps for me n illl figure out where im stuck

Answer 8

please

Answer 9

Average value of any function in the given interval is the integral of that function in that given interval divided by the length of that interval.

Answer 10

h=2-(-1) =3 ===> f(x+3) = 1+((x+3)^2 =1+x^2+6x+9 , f(x)=1+x^2

Answer 11

never mind

Answer 12

it depends on the question if it's in the integation or in the deffrentiation

Answer 13

but both way work

Answer 14

intrgation

Answer 15

so just find the definite integration / 3

Answer 16

x^3/3

Answer 17

average value of the function f(x)=1+x^2
= (x+(x^3)/3)/(2-(-1)) with limits -1 to 2
= 2

Answer 18

[x^3/3] (from -1 to 2) = 8/3 -(-1/3) =9/3=3

Answer 19

thank you nagavishnu now i undersatnd thanks thats where i got stuck on

Answer 20

how you mulitply it

Answer 21

I think you missed the integration itself , it should as nagavi posted

Answer 22

\[f_{avg}={1 \over b-a}\int\limits_{a}^{b}f(x)dx={1 \over 2-(-1)}\int\limits_{-1}^{2}(1+x^2)dx={1 \over 3}(x+{x^3 \over 3})_{-1}^{2}\]
\[={1 \over 3}(2+{8 \over 3}+1+{1 \over 3})=4\]
I hope I din't make any mistakes.

Answer 23

you did it as x^3/3 , whearas it is (x+(x^3)/3)

Answer 24

the answer is 2 not 4

Answer 25

it's done now

Answer 26

yes , the final answer = 6/3=2

Answer 27

Just a typo.. It's \(\frac{1}{3}(2+\frac{8}{3}+1+\frac{1}{3})=\frac{6}{3}=2\).