What is the average value of this function on the interval [0,1]?

[circa415]

Find the average value of

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

I have no idea where to even begin..

 

[shmoe]

You understand that your expression is just a function of x? namely you can write:

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

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.

 

[circa415]

I just get
1/1 * (sin 1 - sin(x^2))

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

 

[shmoe]

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.

 

[circa415]

I used 1/b-a *$$\int_{x}^{1} cos(t^2)dt$$

and I tried to evaluate the integral

 

[shmoe]

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 $$\frac{1}{1-0}\int_{0}^{1}f(x)dx$$

Now substitute your integral equation for f(x):

$$\int_{0}^{1}\left(\int_{x}^{1} cos(t^2)dt\right)dx$$