What is the average value of the function #f(t) = t (sqrt (1 + t^2) )# on the interval #[0,5]#?

Question

4573 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T14:15:12

#1/15(26sqrt26-1)#

The average value of the function #f(t)# on #t in [a.b]# is given by:

#1/(b-a)int_a^bf(t)dt#

So here, the average value of #f(t)=tsqrt(1+t^2)# on #t in[0,5]# is given by:

#1/(5-0)int_0^5tsqrt(1+t^2)dt#

Use the substitution #u=1+t^2#. This implies that #du=2tdt#. When we change the bounds, #t=0# becomes #u=0^2+1=1# and #t=5 # becomes #u=5^2+1=26#:

#=1/(5xx2)int_0^5sqrt(1+t^2)(2tdt)=1/10int_1^26sqrtudu=1/10int_1^26u^(1/2)du#

Using #intu^ndu=u^(n+1)/(n+1)#:

#=1/10[u^(3/2)/(3/2)]_1^26=1/10(2/3)[u^(3/2)]_1^26=1/15(26^(3/2)-1^(3/2))=1/15(26sqrt26-1)#