How do you find the average value of the function for f(x)=sqrt(2x-1), 1<=x<=5?

1 Answer
Oct 31, 2017

The average value is =1/2

Explanation:

We need

intx^ndx=x^(n+1)/(n+1)+C(n!=-1)

The average value is

bary=1/(5-1)int_1^5(1/sqrt(2x-1))dx

Perform first the integral by substitution

Let u=2x-1, =>, du=2dx

Therefore,

int(1/sqrt(2x-1))dx=1/2int(du)/sqrtu=1/2*2sqrtu=sqrtu=sqrt(2x-1)

So,

bary=1/4[sqrt(2x-1)]_1^5=1/4((sqrt9)-(sqrt1))=1/4*2=1/2