# Average value

## Homework Statement

find the average value of cos^2(x) [0,(pi/4)]

## Homework Equations

I don't think i'm doing it correctly I know the formula,(I won't put my attempt because I don't have math symbols on my computer) but I get 1 as the answer every time which doesn't seem correct

## The Attempt at a Solution

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robphy
Homework Helper
Gold Member
You should try to give your attempt.
or "Quote" an earlier post, you'll see a small $$\Sigma$$ symbol, which will help you compose the mathematical symbols.

ok so my attempt is
(4/pi)$$\int\overline{}4/pi$$$$\underline{}0$$(1/2)(1+cos2x)

(2/pi)[x+sin2x/2]$$\overline{}pi/4$$$$\underline{}0$$

(2/pi)[(pi/4+(1/2)=1

hmm im sorry im new i don't think i can get it right the 4/pi is the upper bound the 0 is the lower bound

Dick
Homework Helper
ok so my attempt is
(4/pi)$$\int\overline{}4/pi$$$$\underline{}0$$(1/2)(1+cos2x)

(2/pi)[x+sin2x/2]$$\overline{}pi/4$$$$\underline{}0$$

(2/pi)[(pi/4+(1/2)=1
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.

Last edited:
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.
Soo that would be equal to
(1/pi)+(1/2)?

Dick