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

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

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

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