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

Question

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

1 Answer

Related questions

The profit (in dollars) from the sale of x lawn mowers is ...
What is the average value of the function $f (x) = {(x - 1)}^{2}$ $f(x)=(x-1)^2$ on the interval from x=1 to x=5?
What is the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[0,sqrt pi]$ ?
What is the average value of the function $f (x) = cos (\frac{x}{2})$ $f(x)=cos(x/2)$ on the interval $[- 4, 0]$ $[-4,0]$ ?
What is the average value of the function $f (x) = x^{2}$ $f(x) = x^2$ on the interval $[0, 3]$ $[0,3]$ ?
What is the average value of the function $f (t) = t e^{- t^{2}}$ $f(t)=te^(-t^2 )$ on the interval $[0, 5]$ $[0,5]$ ?
What is the average value of the function $f (x) = x - (x^{2})$ $f(x) = x - (x^2)$ on the interval $[0, 2]$ $[0,2]$ ?
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 $[0, 5]$ $[0,5]$ ?
What is the average value of the function $f (x) = sec x tan x$ $f(x) = sec x tan x$ on the interval $[0, \frac{π}{4}]$ $[0,pi/4]$ ?
What is the average value of the function $f (x) = 2 x^{3} {(1 + x^{2})}^{4}$ $f(x) = 2x^3(1+x^2)^4$ on the interval $[0, 2]$ $[0,2]$ ?

Impact of this question

3459 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-31T08:55:08

The average value is $= \frac{1}{2}$ $=1/2$

Explanation:

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

The average value is

$¯ y = \frac{1}{5 - 1} \int_{1}^{5} (\frac{1}{\sqrt{2 x - 1}}) d x$ $bary=1/(5-1)int_1^5(1/sqrt(2x-1))dx$

Perform first the integral by substitution

Let $u = 2 x - 1$ $u=2x-1$ , $\Rightarrow$ $=>$ , $d u = 2 d x$ $du=2dx$

Therefore,

$\int (\frac{1}{\sqrt{2 x - 1}}) d x = \frac{1}{2} \int \frac{d u}{\sqrt{u}} = \frac{1}{2} \cdot 2 \sqrt{u} = \sqrt{u} = \sqrt{2 x - 1}$ $int(1/sqrt(2x-1))dx=1/2int(du)/sqrtu=1/2*2sqrtu=sqrtu=sqrt(2x-1)$

So,

$¯ y = \frac{1}{4} {[\sqrt{2 x - 1}]}_{1}^{5} = \frac{1}{4} ((\sqrt{9}) - (\sqrt{1})) = \frac{1}{4} \cdot 2 = \frac{1}{2}$ $bary=1/4[sqrt(2x-1)]_1^5=1/4((sqrt9)-(sqrt1))=1/4*2=1/2$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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