How do you find the average value of the function f(x)=122x−1 on the interval of [1,5]?

Average value =f(x2)−f(x1)x2−x1wherex1=1x2=5f(x2)=122⋅5−1=123=4f(x1)=122⋅1−1=12Average value =4−125−1=−2

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gabolzm6d

Answered question

2022-04-13

How do you find the average value of the function

f (x) = \frac{12}{\sqrt{2 x - 1}}

on the interval of [1,5]?

Answer & Explanation

blessgansgxei

Beginner2022-04-14Added 10 answers

Average value

= \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}}

where

x_{1} = 1

x_{2} = 5

f (x_{2}) = \frac{12}{\sqrt{2 \cdot 5 - 1}} = \frac{12}{3} = 4

f (x_{1}) = \frac{12}{\sqrt{2 \cdot 1 - 1}} = 12

Average value

= \frac{4 - 12}{5 - 1} = - 2

kinoNoidae1wj

Beginner2022-04-15Added 7 answers

The idea for positive functions:
The average value is the height of a horizontal line that gives a rectangle on the interval whose area is equal to the area under the curve.
Average value of f(x) on [a,b] is

\frac{1}{b - a} \int_{a}^{b} f (x) d x

So find

\int_{1}^{5} \frac{12}{\sqrt{2 x - 1}} d x

(By substitution.)

= 12 \sqrt{2 x - 1} ∣_{1}^{5}

=12(3)-12(1)=24
Now divide by the length of the interval:

\frac{24}{5 - 1} = \frac{24}{4} = 6

The average value is 6

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