What is the average value of the function f(x)=2x on the interval [1, 10]?

f(x)=2xBy definition:fave=∫abf(x)dxb−ahere we havefave=∫1102xdx10−1=[2ln⁡(x)]1109=29ln⁡(10)

The average rate of change of a function f(x) on an interval [a,b] is given byA(x)=f(b)−f(a)b−a=f(10)−f(1)10−1f(1)=2f(10)=15=15−210−115−210−1=−15Result:−15

f(x)=2xBy definition:fave=∫abf(x)dxb−ahere we havefave=∫1102xdx10−1=[2ln⁡(x)]1109=29ln⁡(10)

The average rate of change of a function f(x) on an interval [a,b] is given byA(x)=f(b)−f(a)b−a=f(10)−f(1)10−1f(1)=2f(10)=15=15−210−115−210−1=−15Result:−15

Best solutions to - "What is the average value of the function..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Joaquin Salas

Answered question

2022-04-18

What is the average value of the function

f (x) = \frac{2}{x}

on the interval [1, 10]?

Answer & Explanation

ysnlm8eut

Beginner2022-04-19Added 14 answers

f (x) = \frac{2}{x}

By definition:

f_{a v e} = \frac{\int_{a}^{b} f (x) d x}{b - a}

here we have

f_{a v e} = \frac{\int_{1}^{10} \frac{2}{x} d x}{10 - 1}

= \frac{{[2 \ln (x)]}_{1}^{10}}{9}

= \frac{2}{9} \ln (10)

Zaria Lamb

Beginner2022-04-20Added 8 answers

The average rate of change of a function f(x) on an interval [a,b] is given by

A (x) = \frac{f (b) - f (a)}{b - a}

= \frac{f (10) - f (1)}{10 - 1}

f(1)=2

f (10) = \frac{1}{5}

= \frac{\frac{1}{5} - 2}{10 - 1}

\frac{\frac{1}{5} - 2}{10 - 1} = - \frac{1}{5}

Result:

- \frac{1}{5}

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