Calculating the Average Value of cos^2(x) [0,(pi/4)]

[Allie G]

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

 

[robphy]

You should try to give your attempt.
If you select "Go Advanced" during a Quick Reply
or "Quote" an earlier post, you'll see a small $$\Sigma$$ symbol, which will help you compose the mathematical symbols.

 

[Allie G]

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

 

[Allie G]

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

 

[Dick]

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.

 

[Allie G]

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]

Soo that would be equal to
(1/pi)+(1/2)?

Yessss. It would. That's 0.8183... Not 1.

 

[Allie G]

thank you so much