What is the average value of the function #F(x) = xe ^ (x^2) # on the interval #[0,2]#?

1 Answer
Dec 19, 2015

#1/4 e^{4}-1/4 approx 13.3995#

Explanation:

The average value of a continuous function #F# over an interval #[a,b]# is #1/(b-a) int_{a}^[b}\ F(x)\ dx#.

For #F(x)=x e^{x^{2}}# over the interval #[0,2]#, this becomes

#1/2 int_{0}^{2}x e^{x^{2}}\ dx#

The integral can be done using the substitution #u=x^{2}#, making #du=2x\ dx#. Also, when #x=0#, #u=0# and when #x=2#, #u=2^{2}=4#. Therefore, the quantity above can be written as

#1/2 int_{0}^{4} 1/2 e^{u}\ du=[1/4 e^{u}]_{0}^{4}=1/4 e^{4}-1/4 e^{0}=1/4 e^{4}-1/4 approx 13.3995#