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

It is

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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.

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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 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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