Find the average value of the cost function C(x)=3x√x+10 on the interval [0,81].?

Question

2094 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-04T11:55:14

$Value_(avg)~~959.1$

$Value_(avg) = 1/(b-a)int_a^bf(x) dx$

In this case:

$Value_(avg)=1/(81-0)int_0^(81)3xsqrt(x+10)dx$

$=3/(81)int_0^(81)xsqrt(x+10) dx$

Let: $u=x+10, du=dx, x=u-10$ :

$=1/(27)int_0^(81)(u-10)*u^(1/2) du$

$=1/(27)*2/(15) u^(3/2)(3u-50)[0,81]$

$=1/(27)*2/(15)(x+10)^(3/2)(3x-20)[0, 81]$

$=2/(405)[(223)(91)^(3/2)- (-20)(10)^(3/2)]$

$=2/(405)[20293sqrt(91)+200sqrt(10)]$

$~~959.1$