Allie G said:ok so my attempt is
(4/pi)[tex]\int\overline{}4/pi[/tex][tex]\underline{}0[/tex](1/2)(1+cos2x)
(2/pi)[x+sin2x/2][tex]\overline{}pi/4[/tex][tex]\underline{}0[/tex]
(2/pi)[(pi/4+(1/2)=1
Dick said:Don't worry about the formatting. Better luck next time. But you are doing great! Except (2/pi)[pi/4+(1/2)] doesn't equal 1. Notice where I put the brackets.
Allie G said:Soo that would be equal to
(1/pi)+(1/2)?
The formula for calculating the average value of cos^2(x) [0,(pi/4)] is:
avg = (1/(b-a))*∫abcos^2(x) dx, where a = 0 and b = (pi/4).
To find the average value of cos^2(x) [0,(pi/4)], you need to integrate cos^2(x) from 0 to (pi/4) and then divide the result by the length of the interval, which is (pi/4 - 0) = (pi/4).
The range of cos^2(x) is [0,1]. This means that the values of cos^2(x) can range from 0 to 1, inclusive.
No, the average value of cos^2(x) [0,(pi/4)] cannot be negative. This is because cos^2(x) is always positive on the interval [0,(pi/4)] and the average value is calculated by taking the mean of all the values within that interval.
Calculating the average value of cos^2(x) [0,(pi/4)] is useful in various applications, such as in calculating the average power in an AC circuit or in finding the average brightness of a light source. It also helps in understanding the behavior of the function over a specific interval and can be used to compare different functions in terms of their average values.