How do you find the average value of f ( x ) = 1 x 2 + 1 as x varies between [-1,1]?

The average value of f on the interval [a,b] is 1 b − a ∫ a b f ( x ) d xWe also need: ∫ 1 x 2 + 1 d x = tan − 1 ⁡ x + Cor, you may prefer the notation ∫ 1 x 2 + 1 d x = arctan ⁡ x + CThe average value is 1 − ( − 1 ) ∫ − 1 1 1 x 2 + 1 d x = 1 2 [ tan − 1 ⁡ x ] − 1 1 = 1 2 [ π 4 − ( − π 4 ) ] = π 4

Answered: "How do you find the average value of..."

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Alissa Hancock

Answered question

2022-07-14

How do you find the average value of

f (x) = \frac{1}{x^{2} + 1}

as x varies between [-1,1]?

Answer & Explanation

SweallySnicles3

Beginner2022-07-15Added 21 answers

The average value of f on the interval [a,b] is

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

We also need:

\int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x + C

or, you may prefer the notation

\int \frac{1}{x^{2} + 1} d x = \arctan x + C

The average value is

\frac{1}{- (- 1)} \int_{- 1}^{1} \frac{1}{x^{2} + 1} d x = \frac{1}{2} [\tan^{- 1} x]_{- 1}^{1}

= \frac{1}{2} [\frac{π}{4} - (- \frac{π}{4})]

= \frac{π}{4}

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