What is the average value of the function #f(t)=te^(-t^2 )# on the interval #[0,5]#?

Question

Charles Anctil · Accepted Answer

It is #1/10(1-e^-25)# #1/(5-0) int_0^5 te^(-t^2)dt = -1/10 int_0^5 e^(-t^2)(-2t) dt# # = -1/10[e^(-t^2)]_0^5# # = -1/10(e^-25 - e^0)# # = 1/10(1-e^-25)#

Axel Alvarez · Answer

undefined To find the average value of the function $ f(t) = te^{-t^2} $ on the interval $[0,5]$, you can use the formula:

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

In this case, $ a = 0 $ and $ b = 5 $. Substituting these values into the formula gives:

$$ \text{Average value} = \frac{1}{5-0} \int_{0}^{5} t e^{-t^2} \, dt $$

Now, integrate $ te^{-t^2} $ with respect to $ t $ from 0 to 5:

$$ \int_{0}^{5} t e^{-t^2} \, dt $$

Let $ u = -t^2 $, then $ du = -2t \, dt $. This leads to:

$$ -\frac{1}{2} \int e^u \, du $$

Integrating $ e^u $ gives $ e^u $, so:

$$ -\frac{1}{2} e^{-t^2} \bigg|_{0}^{5} $$

Plugging in the upper and lower limits:

$$ -\frac{1}{2} (e^{-25} - e^{0}) $$

Simplifying further:

$$ \frac{1}{2} (1 - e^{-25}) $$

Therefore, the average value of the function $ f(t) = te^{-t^2} $ on the interval $[0,5]$ is approximately 0.4999.

Aurora Alvarado · Answer

undefined To find the average value of the function $ f(t) = t e^{-t^2} $ on the interval $[0,5]$, you can use the formula:

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

In this case, $a = 0$ and $b = 5$. So, the average value of the function on the interval $[0,5]$ is:

$$ \text{Average value} = \frac{1}{5-0} \int_{0}^{5} t e^{-t^2} \, dt $$