# Average Value of a Number

• May 5th 2010, 11:41 PM
JollyJolly15
Average Value of a Number
Can someone please help me out with this problem? It is the last one I have for my calculus assignment due tomorrow! We haven't done anything like this in class, so I have NO idea what to do.

Here is a link to a picture of the problem because I couldn't type it on here.

http://img3.imageshack.us/img3/5082/mathfp5.jpg
• May 6th 2010, 01:49 AM
tonio
Quote:

Originally Posted by JollyJolly15
Can someone please help me out with this problem? It is the last one I have for my calculus assignment due tomorrow! We haven't done anything like this in class, so I have NO idea what to do.

Here is a link to a picture of the problem because I couldn't type it on here.

http://img3.imageshack.us/img3/5082/mathfp5.jpg

The average value of a Riemann-integrable functuon $f(x)$ over an interval $(a,b)\,,\,a is given by $\frac{1}{b-a}\int\limits^b_af(x)\,dx$

In your case, note that $\sec^2x=(\tan x)'$ , so the integral of that function is pretty easy...(Wink)

Tonio
• May 6th 2010, 07:39 AM
grgrsanjay
simple
we should do differentiation or integration??
if its deferenciation then
use the product principle rule (Cool)
e^3tanx+1 * d/dx sec^2x + sec^2x * d/dx e^3tanx+1
= e^3tanx+1 * 2secx(secxtanx) + sec^2x * e^3tanx+1(3sec^2x)
then simplyfy it and substitute x as 0 and x as 45 and then add the values and divide them by 2,
• May 6th 2010, 08:05 AM
HallsofIvy
Quote:

Originally Posted by grgrsanjay
we should do differentiation or integration??
if its deferenciation then
use the product principle rule (Cool)
e^3tanx+1 * d/dx sec^2x + sec^2x * d/dx e^3tanx+1
= e^3tanx+1 * 2secx(secxtanx) + sec^2x * e^3tanx+1(3sec^2x)
then simplyfy it and substitute x as 0 and x as 45 and then add the values and divide them by 2,