How do you find the average value of the function for $f (x) = 2 - \frac{1}{2} x, 0 \leq x \leq 4$ $f(x)=2-1/2x, 0<=x<=4$ ?

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How do you find the average value of the function for $f (x) = 2 - \frac{1}{2} x, 0 \leq x \leq 4$ $f(x)=2-1/2x, 0<=x<=4$ ?

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Answer 1 · 2018-03-22T15:55:03

The average value of the function on the given interval is $1$ $1$ .

Explanation:

Average value of a function is given by

$A = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ $A = 1/(b - a) int_a^b f(x) dx$

$A = \frac{1}{4} \int_{0}^{4} 2 - \frac{1}{2} x d x$ $A = 1/4 int_0^4 2 - 1/2x dx$

$A = \frac{1}{4} {[2 x - \frac{1}{4} x^{2}]}_{0}^{2}$ $A = 1/4[2x - 1/4x^2]_0^4$

$A = \frac{1}{4} (2 (4) - \frac{1}{4} {(4)}^{2})$ $A = 1/4(2(4) - 1/4(4)^2)$

$A = \frac{1}{4} (8 - 4)$ $A = 1/4(8 - 4)$

$A = \frac{1}{4} (4)$ $A = 1/4(4)$

$A = 1$ $A = 1$

Hopefully this helps!

Answer 2 · 2018-03-22T15:57:39

The average value of $f$ $f$ over $[0, 4]$ $[0,4]$ is 1.

Explanation:

The average value of a function over an interval is its (definite) integral over that interval divided by the length of the interval.

$\int_{0}^{4} (2 - \frac{x}{2}) d x = {[2 x - \frac{x^{2}}{4}]}_{0}^{4}$ $int_0^4 (2 - x/2) dx = [2x-x^2/4]_0^4$

$= 4 - 0$ $= 4 - 0$

$= 4$ $= 4$

$\frac{\int_{0}^{4} (2 - \frac{x}{2}) d x}{4 - 0} = \frac{4}{4} = 1$ $(int_0^4 (2 - x/2) dx)/(4-0) = 4/4 = 1$

The average value of $f$ $f$ over $[0, 4]$ $[0,4]$ is 1.

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How do you find the average value of the function for $f (x) = 2 - \frac{1}{2} x, 0 \leq x \leq 4$ $f(x)=2-1/2x, 0<=x<=4$ ?

2 Answers

Explanation:

Explanation:

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How do you find the average value of the function for f(x)=2-1/2x, 0<=x<=4f(x)=2−12x,0≤x≤4f(x)=2-1/2x, 0<=x<=4?

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How do you find the average value of the function for $f (x) = 2 - \frac{1}{2} x, 0 \leq x \leq 4$ $f(x)=2-1/2x, 0<=x<=4$ ?