What is the average value of a function # f(t)= -2te^(-t^2)# on the interval #[0, 8]#?

Question

Ezekiel Alexander · Accepted Answer

The average value is #1/(8e^64) - 1/8 ~~ -0.125#

The average value of a function on the continuous interval #[a, b]# is given by

#A = 1/(b - a) int_a^b f(x) dx#

In this case we would have

#A = 1/8 int_0^8 -2te^(-t^2)dt#

This expression can be integrated using the substitution #u = -t^2#. Then #du = -2tdt# and #dt = (du)/(-2t)#. We change the bounds of integration accordingly.

#A = 1/8int_0^-64 e^u du#

#A = 1/8[e^u]_0^-64#

#A = 1/8[e^(-t^2)])_0^8#

#A = 1/8e^(-64) - 1/8#

If you want an approximation, use a calculator to get

#A = -0.125#

Hopefully this helps!

Emilia Aguilar · Answer

undefined To find the average value of the function $ f(t) = -2te^{-t^2} $ on the interval $[0, 8]$, we use the formula for the average value of a function on a closed interval $[a, b]$:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(t) \, dt $$

For the given function $ f(t) = -2te^{-t^2} $ and the interval $[0, 8]$, we have:

$$ \text{Average value} = \frac{1}{8 - 0} \int_{0}^{8} (-2t)e^{-t^2} \, dt $$

Now, we integrate $ f(t) $ from $ 0 $ to $ 8 $:

$$ \int_{0}^{8} (-2t)e^{-t^2} \, dt = \left[ -e^{-t^2} \right]_{0}^{8} = -e^{-64} + e^0 $$

So, the average value of $ f(t) $ on the interval $[0, 8]$ is:

$$ \frac{1}{8} \left( -e^{-64} + 1 \right) $$