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

Question

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

Samantha Adkison · Accepted Answer

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#

Evelyn Agard · Answer

undefined To find the average value of the function $ f(x) = \sqrt{2x - 1} $ over the interval $ 1 \leq x \leq 5 $, you use the formula for the average value of a function over an interval:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Substitute $ a = 1 $ and $ b = 5 $ into the formula and evaluate the integral:

$$ \text{Average value} = \frac{1}{5 - 1} \int_{1}^{5} \sqrt{2x - 1} \, dx $$

$$ = \frac{1}{4} \int_{1}^{5} \sqrt{2x - 1} \, dx $$

This integral can be evaluated using techniques of integration. After integration, divide the result by $ 4 $ to find the average value of the function over the interval $ 1 \leq x \leq 5 $.

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

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