Question

find the values of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval

f(x)=x^3 i am lost, i am trying to take the integral, then solving it but i have i gotten the wrong answer right , i have been using the mean vaule theorem but it is not qorking please help

Answer 1

on what interval?

Answer 2

[0,3]

Answer 3

what does the theorem say?

Answer 4

it says there exist a number c in [0,3] such that
\[f(c)(b-a) = \int\limits_{a}^{b} f(x) dx\]

Answer 5

a = 0
b = 3
f(x) = x^3

solve for c

Answer 6

if f is integrable [a,b] then the average value is the function on [a,b] os

avg. 1/b-a \[\int\limits_{a}^{b}f(x)dx\]

Answer 7

the average value is f(c). In other words, it's the y value, not x. c is an x value in [0,3]

Answer 8

you're supposed to find c. Not f(c)

Answer 9

i have been trying to use the theorem, like i see in my notes i get something else then the answer in the back which is 3\[3\sqrt[3]{2}/2\]

Answer 10

answer should be 2 (2/3)^(1/3)

Answer 11

you probably looked at the solution to a different problem. Either that or, you copied the wrong question

Answer 12

no its the same quesition and answer

Answer 13

maybe i looks at the wrong formula hang on
\[\int\limits_{a}^{b}f(x)dx=f(c)-[b-a]\]

Answer 14

i made a mistake. But the correct answer is still different from yours
https://www.wolframalpha.com/input/?i=c%5E3+%283-0%29+%3D+%281%2F4%29+%283%29%5E4

Answer 15

yeah, your formula is wrong

Answer 16

it's f(c)(b-a). not f(c) - (b-a)

Answer 17

ok thats where i am going wrong thank you