- #1

- 20

- 0

[tex]

\int_{x}^{1} cos(t^2)dt

[/tex]

on [0,1]

I have no idea where to even begin..

- Thread starter circa415
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- #1

- 20

- 0

[tex]

\int_{x}^{1} cos(t^2)dt

[/tex]

on [0,1]

I have no idea where to even begin..

- #2

shmoe

Science Advisor

Homework Helper

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[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

- 0

I just get

1/1 * (sin 1 - sin(x^2))

but that seems too simple? am I doing something wrong?

1/1 * (sin 1 - sin(x^2))

but that seems too simple? am I doing something wrong?

- #4

shmoe

Science Advisor

Homework Helper

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The end asnwer should be a number, and will have no x's or other variables in it.

- #5

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- 0

I used 1/b-a *[tex]

\int_{x}^{1} cos(t^2)dt

[/tex]

and I tried to evaluate the integral

\int_{x}^{1} cos(t^2)dt

[/tex]

and I tried to evaluate the integral

- #6

shmoe

Science Advisor

Homework Helper

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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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