Find the average value of f(x)=25−x^2 on the interval [0,1]

Question

hartnn · Answer

Average value of the function f(x) over an interval (a,b) is 
$\Large \dfrac{1}{(b-a)} \int \limits_a^b f(x)dx$

hartnn · Answer

so, you need to solve,
$\Large \dfrac{1}{(1-0)} \int \limits_0^1 (25-x^2 )dx$

hartnn · Answer

try it out, all you need is
$\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$

hartnn · Answer

why would x be 25 ? x would be x only.

anonymous · Answer

ok im just confused what to put into the equation

hartnn · Answer

have you ever solved an integral before ?

anonymous · Answer

yes

hartnn · Answer

good, so can you solve this, 
$\Large \int \limits_0^1 (25-x^2)dx$
$=\Large \int \limits_0^1 25dx$ $-\Large \int \limits_0^1 x^2dx$

anonymous · Answer

alright 74/3

anonymous · Answer

Then: Find a value c in the interval [0,1] such that f(c) is equal to the average value

hartnn · Answer

so you need f(c) = 74/3

to get f(c), just replace 'x' in f(x) by 'c'

hartnn · Answer

f(x) = 25-x^2
f(c) = 25 -c^2

25-c^2 = 74/3
find c from here

anonymous · Answer

alright thanks

hartnn · Answer

welcome ^_^