# Finding the average value of this function

• Mar 5th 2009, 10:14 AM
fattydq
Finding 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 AM
fattydq
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 PM
fattydq
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 PM
Mentia
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 PM
fattydq
Quote:

Originally Posted by Mentia
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.

Umm I thought the formula for fave is 1/b-a times the integral you gave...at least that's what it says in my notes and in my book.
• Mar 5th 2009, 02:25 PM
Mentia
Wow you are totally right. I was thinking about probability density functions, whoops! Yeah so just do the substitution then divide by 10.