Question

Definite Integral problem...
Calculate the Average value of the given function of the interval provided.
\[f(x) = \frac{ 1 }{ (3 + \sqrt{x})^{4} }~~,~~[4, 9]\]

Answer 1

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

Answer 2

do you know the formula?

Answer 3

hehe ^ like mentioned above :D

Answer 4

\[\frac{ 1 }{ (3+\sqrt{x})^4 } = (3+\sqrt{x})^{-4}\]
use the anti-power rule and the formula ofc.

Answer 5

idk anti power rule, but I do understand the equation stuff you posted...
erm...
I did u = 3 + rootx
\[\int\limits_{5}^{6} 2u ^{-3}-6u ^{-4} du\]???

Answer 6

you cant use u-sub for this..

Answer 7

:0 why not?

Answer 8

anti-power rule..
is
\[\int\limits_{}^{} x^n = \frac{ x^{n+1} }{ n+1 }\]
\[av(f) = \frac{ 1 }{ 5 } \int\limits_{4}^{9} (3+\sqrt{x})^{-4}\]
\[= \frac{ 1 }{ 5 } \left[ -\frac{ 1 }{ 3(3+\sqrt{x})^3 } \right]^9_4\]

Answer 9

just evaluate that using the fundamental theorem of calc now. and simplify.

Answer 10

nuh uh me didn't learn dat

Answer 11

and you're doing u-sub??

Answer 12

yeah...

Answer 13

techniques of integration are AFTER the basic rules -.-

Answer 14

meh teacher dumbdumb

Answer 15

or teacher went over anti power rule, but not tell us name...
could you give simple example of anti power rule?

Answer 16

wut happened to chain rule in the thing you gave?

Answer 17

\[\int\limits_{4}^{9} \frac{ 1 }{ (3+\sqrt{x})^4 } dx\]
even if you let
\[\ u = 3 + \sqrt{x} \]
your
\[\ du = \frac {1}{2\sqrt{x}} dx \]
\[\ 2~ du = \frac {dx}{\sqrt{x}}\]
you can't do u-sub cause you don't have the 1/(sqrt{x}) needed...

Answer 18

put du
so you have 2rootx to even it out
then rootx = u-3

Answer 19

i'd rather you try to learn this stuff from the beginning starting with the basic rules of integration..
i don't know why you're doing attempting u-sub without prior knowledge.

Answer 20

and that won't work.

Answer 21

http://www.mathsisfun.com/calculus/integration-rules.html
there's a list of the rules.

Answer 22

you can try u = sqrt(x)
x = u^2
dx = 2u du