Question

Find the average value of $f(x) = 2x^2 - 3x$ on the interval $[2, 6]$. $f_{ave} = \frac{68}{3}$ From the Mean Value Theorem for Integrals, we know that there is a $c$ in $(2, 6)$ such that $f(c)$ equals this average value you just found. Find $c$. c =

          Find the average value of $f(x) = 2x^2 - 3x$ on the interval $[2, 6]$.
$f_{ave} = \frac{68}{3}$
From the Mean Value Theorem for Integrals, we know that there is a $c$ in $(2, 6)$ such that $f(c)$ equals this average value you just found. Find $c$.
c =

Find the average value of f(x) = 2x^2 - 3x on the interval [2, 6].
fave = (68)/(3)
From the Mean Value Theorem for Integrals, we know that there is a c in (2, 6) such that f(c) equals this average value you just found. Find c.
c =

Added by Peggy D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Keondre Parker

Step 1

The average value of f(x) on the interval [2,6] is given by the formula: Average value = (1/(b-a)) * ∫[a,b] f(x) dx In this case, a = 2, b = 6, and f(x) = 22 - 3x. So we have: Average value = (1/(6-2)) * ∫[2,6] (22 - 3x) dx Simplifying, we get: Average value Show more…

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Find the average value of f(x) = 22 - 3x on the interval [2,6]. The average value is 68/3. From the Mean Value Theorem for Integrals, we know that there is a c in [2,6] such that f'(c) equals this average value you just found. Find c.