f(t)= -2te^-t^2, [0,5] Find average value.

Attempt: I know that the average value is the integration of f(t) from a to b divided by b-a. I used u substitution to get u= -t^2 and du= -2xdx. I
changed the limits of integration to [0,25] and I got integral of e^u du.
I am not sure if this is right, but then would the integral of that be
just e^u and I would just plug in 25, because I did that and the answer
was wrong.

I don't see anything wrong with your approach. Perhaps there was an arithmetic error lurking somewhere.

What did you get for the answer?
Did you divide the integral by 5 or 25?

You are almost right but since u=-t^(2) the limits are [0,-25] then put them into e^(u) and you get,

e^(-25)-e^(0)=
e^(-25)-1

Then divide that answer by 5 thus the answer is,

(e^(-25)-1)/5

To find the average value of a function f(t) over the interval [a, b], you are correct that you need to evaluate the integral of f(t) from a to b and divide it by b - a.

In this case, f(t) = -2te^(-t^2), and you want to find the average value over the interval [0, 5].

First, let's evaluate the integral:
∫(0 to 5) -2te^(-t^2) dt

To integrate this expression, you made a substitution:
u = -t^2
du = -2t dt

When changing the limits of integration, you need to consider that u depends on t. So, let's evaluate u at the limits [0, 5]:
u(0) = 0
u(5) = -(5^2) = -25

Now, substitute these limits into the integral:
∫(0 to -25) e^u du

To integrate e^u with respect to u, you can simply integrate it as you suggested:
∫ e^u du = e^u

Now, plug in the limits of integration:
e^u ∣ (0 to -25) = e^(-25) - e^0

Finally, you can calculate the average value by dividing this result by the length of the interval:
Average value = (e^(-25) - e^0) / (5 - 0) = (e^(-25) - 1) / 5

Remember to use a calculator to evaluate e^(-25) to get the numerical value of the average value.