How do you find the average value of the function #f(t)=4te^(-t^2)# on the interval [0, 5]?

Question

Isaiah Allen · Accepted Answer

It is #1/5(2-2/e^25) ~~ 2/5#

For a function that is non-negative on an interval #[a,b]# -- like this function is on #[0,5]# -- the average value has a nice geometric explanation: It is the height that a rectangle on the same base (the interval) would need to have is the area of the rectangle is to be equal to the area under the curve on the interval.

Here is a picture of the area under the graph of #f(t)=4te^(-t^2)# on the interval [0, 5]

y <=0 [-1.874, 6.893, -1.357, 3.028]} graph{(y - 4xe^(-x^2)(sqrt(6.25-(x-2.5)^2))/(sqrt(6.25-(x-2.5)^2)))

We integrate to determine the area under the curve:

#int_0^5 4te^(-t^2) dt#

Let #u = -t^2# to get:

#-2int_0^-25 e^u du = -2e^u]_0^-25#

# = 2(1-e^-25)#

#Area = 2(1-e^-25)#

So if a rectangle on base #[0,5]# is to have that area, its height must be

#"height" = "Area"/"base" = "Area"/5 = (2(1-e^-25))/5# Which can be rewritten:

#"Average Value" = 1/5(2-2/e^25) # or #2/5(1-1/e^25)#

The general formula is: The average value of an function, #f# on interval #[a,b]# is:

#"Average Value" = 1/(b-a) int_a^b f(x) dx#

Emma Aldridge · Answer

undefined To find the average value of the function $f(t) = 4te^{-t^2}$ on the interval [0, 5], you integrate the function over the interval and then divide by the length of the interval. The formula for average value is:

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

where $a$ and $b$ are the lower and upper bounds of the interval.

So, for $f(t) = 4te^{-t^2}$ on the interval [0, 5]:

$$ \text{Average value} = \frac{1}{5-0} \int_{0}^{5} 4te^{-t^2} \, dt $$