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Average value

  • Thread starter Allie G
  • Start date
  • #1
7
0

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

 

Answers and Replies

  • #2
robphy
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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 [tex]\Sigma[/tex] symbol, which will help you compose the mathematical symbols.
 
  • #3
7
0
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
 
  • #4
7
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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
 
  • #5
Dick
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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
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:
  • #6
7
0
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)?
 
  • #7
Dick
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Soo that would be equal to
(1/pi)+(1/2)?
Yessss. It would. That's 0.8183... Not 1.
 
  • #8
7
0
thank you so much
 

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