The function is tsin(t^2) on the interval [0,10]

So it'd end up being 1/10 the integral from 0 to 10 of tsin(t^2), but that's where I'm stuck. How in the world would I take the antiderivative of that function?

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- Mar 5th 2009, 10:14 AMfattydqFinding the average value of this function
The function is tsin(t^2) on the interval [0,10]

So it'd end up being 1/10 the integral from 0 to 10 of tsin(t^2), but that's where I'm stuck. How in the world would I take the antiderivative of that function? - Mar 5th 2009, 10:17 AMfattydq
Could somebody confirm or deny that I'm heading in the right direction by saying my next step would be 1/10((-1/2)cos(t^2) evaluated at 10, minus that evaluated at 0?

- Mar 5th 2009, 01:11 PMfattydq
I'm sorry I wasn't very clear, I'm just wondering if the antiderivative of tsin(t^2) is just -1/2cos(t^2) yes or no?

- Mar 5th 2009, 02:09 PMMentia
Are you sure you arent finding the average value of sin(t^2) on the interval [0,10] ?

Then the average value is:

$\displaystyle \int_{0 }^{10 } tSin( { t}^{2 }) $

then substitute u = t^2 and solve.

You should get around .0688.

If you are finding the average value of tsin(t^2), the average value is:

$\displaystyle \int_{0 }^{10 } { t}^{ 2} Sin( { t}^{2 }) $

and this integral is now very difficult and involves the Fresnel integral. The value here is about -4.01. - Mar 5th 2009, 02:22 PMfattydq
- Mar 5th 2009, 02:25 PMMentia
Wow you are totally right. I was thinking about probability density functions, whoops! Yeah so just do the substitution then divide by 10.