What is the average value of e^(x) on the interval [3,9] ? Choose 1 answer: (A) (e^(9)+e^(3))/(6) (B) (e^(9)+e^(3))/(2) (C) (e^(9)-e^(3))/(6) (D) (e^(9)-e^(3))/(2)

Question

What is the average value of  e^(x) on the interval  [3,9] ? Choose 1 answer: (A)  (e^(9)+e^(3))/(6) (B)  (e^(9)+e^(3))/(2) (C)  (e^(9)-e^(3))/(6) (D)  (e^(9)-e^(3))/(2)

Bytelearn · Accepted Answer

Formula for Average Value: The average value of a continuous function f(x) on the interval [a, b] is given by the formula: Average value = (1/(b-a)) * ∫ from a to b of f(x) dx For the function e^(x) on the interval [3,9], we have: Average value = (1/(9-3)) * ∫ from 3 to 9 of e^(x) dxCalculate Integral of e^(x): Calculate the integral of e^(x) from 3 to 9. The integral of e^(x) with respect to x is e^(x), so we evaluate it at the bounds 9 and 3: ∫ from 3 to 9 of e^(x) dx = e^(x) | from 3 to 9 = e^(9) - e^(3)Substitute Integral Result: Substitute the integral result into the average value formula: Average value = (1/(9-3)) * (e^(9) - e^(3)) Average value = (1/6) * (e^(9) - e^(3))Simplify to Find Average Value: Simplify the expression to find the average value: Average value = (e^(9) - e^(3)) / 6 This corresponds to answer choice (C).

What is the average value of $e^{x}$ on the interval $[3,9]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{e^{9}+e^{3}}{6}$ $\newline$ (B) $\frac{e^{9}+e^{3}}{2}$ $\newline$ (C) $\frac{e^{9}-e^{3}}{6}$ $\newline$ (D) $\frac{e^{9}-e^{3}}{2}$

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