Average Value Function

  • Thread starter circa415
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  • #1
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Find the average value of

[tex]
\int_{x}^{1} cos(t^2)dt
[/tex]
on [0,1]

I have no idea where to even begin..
 

Answers and Replies

  • #2
shmoe
Science Advisor
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You understand that your expression is just a function of x? namely you can write:

[tex]f(x)=\int_{x}^{1} cos(t^2)dt[/tex]

Now what's the general expression for the average value of a function f(x) on the interval [0,1]? Substitute the above in and see what happens.
 
  • #3
20
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I just get
1/1 * (sin 1 - sin(x^2))

but that seems too simple? am I doing something wrong?
 
  • #4
shmoe
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How did you get that? Please provide some details..

The end asnwer should be a number, and will have no x's or other variables in it.
 
  • #5
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I used 1/b-a *[tex]
\int_{x}^{1} cos(t^2)dt
[/tex]

and I tried to evaluate the integral
 
  • #6
shmoe
Science Advisor
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sin(t^2) is not an antiderivative of cos(t^2), but that's beside the point here.

You want the average value of the function f(x), the thing you were given. Ignore for a moment that it's defined by an integral and just treat it like any old function. The average value is given by [tex]\frac{1}{1-0}\int_{0}^{1}f(x)dx[/tex]

Now substitute your integral equation for f(x):

[tex]\int_{0}^{1}\left(\int_{x}^{1} cos(t^2)dt\right)dx[/tex]
 

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