the average value of (sec^2)x over the interval 0<x<pi/4?

Question

on the (sec^2)x the sec is squared and the x is NOT an exponent. and the less than signs are less than or equal to (i just didn't know how to type it :) thanks)

Answer 1

When taking the average value, the formula is the integral from a to b divided by (a - b)

So, the integral of (sec^2)x is tanx. Taking the integral from 0 to pi/4 gives us 1. Then, divide by (pi/4-0), or pi/4. So we've got (1/(pi/4)). This gives us a final answer of 4/pi.

Answer 2

Average value = 1/(pi/4)Int (pi/4,0) sec^2 xdx
= 4/pi(tan x) between (0,pi/4)= 4/pi as tan(pi/4) =1