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

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Answer 1 · 2017-10-31T08:55:08

The average value is #=1/2#

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#