We have an answer to "Find c such that fave =f(c),f(x)=(x−3)2,[2,5]"

Joan Thompson

Answered question

2021-12-20

Find c such that fave

= f (c), f (x) = {(x - 3)}^{2}, [2, 5]

Answer & Explanation

alexandrebaud43

Beginner2021-12-21Added 36 answers

Average value of the function f(x) over the interval $[a, b]$ is given by
$f_{a v g} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$
Average value of $f (x) = {(x - 3)}^{2}$ ovthe interval [2,5] is given by
$f_{a v g} = \frac{1}{5 - 2} \int_{2}^{5} {(x - 3)}^{2} d x$
Substitute $x - 3 = u$ and $d x = d u$
Limits of Integration will change from $\int_{-} 2^{5} \to \int_{-} 2 - 3^{5 - 3} = \int_{- 1}^{2}$
$= \frac{1}{3} \int_{- 1}^{2} u^{2} d u$
$= \frac{1}{3} {[\begin{matrix} \frac{u^{3}}{3} \end{matrix}]}_{- 1}^{2}$
$= \frac{1}{3} [\begin{matrix} \frac{8}{3} + \frac{1}{3} \end{matrix}] = 1$

Orlando Paz

Beginner2021-12-22Added 42 answers

Average value of the function f(x) over the interval [a,b] is given by
$f_{a v g} = \frac{1}{b - a} \int_{a}^{b} f_{(x)} d x$
Average value of $f (x) = {(x - 3)}^{2}$ over the interval $(2, 5]$ is given by
$f_{a v g} = \frac{1}{5 - 2} \int_{2}^{5} (x - 3)^{2} d x$
$= \frac{1}{3} {[\begin{matrix} \frac{(x - 3)^{3}}{3} \end{matrix}]}_{2}^{5}$
$= \frac{1}{3} [\begin{matrix} \frac{8}{3} + \frac{1}{3} \end{matrix}] = \frac{1}{3} \times \frac{9}{3} = 1$

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