Solved on Mar 27, 2024

Find the average value of x39xx^3 - 9x on the interval 1x3-1 \leq x \leq 3.

STEP 1

1. The average value of a function f(x)f(x) on the interval [a,b][a, b] is given by the formula 1baabf(x)dx\frac{1}{b-a} \int_{a}^{b} f(x) \, dx.
2. The function to be averaged is f(x)=x39xf(x) = x^3 - 9x.
3. The interval over which the average is to be taken is [1,3][-1, 3].

STEP 2

1. Set up the integral for the average value of f(x)f(x) on the interval [1,3][-1, 3].
2. Compute the integral of f(x)f(x) from 1-1 to 33.
3. Divide the result of the integral by the length of the interval to find the average value.

STEP 3

Set up the integral for the average value of f(x)=x39xf(x) = x^3 - 9x on the interval [1,3][-1, 3].
Average value=13(1)13(x39x)dx \text{Average value} = \frac{1}{3 - (-1)} \int_{-1}^{3} (x^3 - 9x) \, dx

STEP 4

Simplify the denominator of the fraction which is the length of the interval [a,b][a, b] where a=1a = -1 and b=3b = 3.
Average value=13(1)13(x39x)dx=1413(x39x)dx \text{Average value} = \frac{1}{3 - (-1)} \int_{-1}^{3} (x^3 - 9x) \, dx = \frac{1}{4} \int_{-1}^{3} (x^3 - 9x) \, dx

STEP 5

Compute the integral of f(x)=x39xf(x) = x^3 - 9x from 1-1 to 33.
13(x39x)dx \int_{-1}^{3} (x^3 - 9x) \, dx

STEP 6

Find the antiderivative of f(x)f(x).
(x39x)dx=x449x22+C \int (x^3 - 9x) \, dx = \frac{x^4}{4} - \frac{9x^2}{2} + C

STEP 7

Evaluate the antiderivative from 1-1 to 33.
(x449x22)13 \left. \left( \frac{x^4}{4} - \frac{9x^2}{2} \right) \right|_{-1}^{3}

STEP 8

Plug in the upper limit of the integral and then the lower limit, and subtract the two.
(3449322)((1)449(1)22) \left( \frac{3^4}{4} - \frac{9 \cdot 3^2}{2} \right) - \left( \frac{(-1)^4}{4} - \frac{9 \cdot (-1)^2}{2} \right)

STEP 9

Simplify the expression by calculating the values.
(814812)(1492) \left( \frac{81}{4} - \frac{81}{2} \right) - \left( \frac{1}{4} - \frac{9}{2} \right)

STEP 10

Combine the terms within each parenthesis.
(8141624)(14184) \left( \frac{81}{4} - \frac{162}{4} \right) - \left( \frac{1}{4} - \frac{18}{4} \right)

STEP 11

Further simplify the expression.
(814)(174) \left( -\frac{81}{4} \right) - \left( -\frac{17}{4} \right)

STEP 12

Subtract the two terms.
814+174 -\frac{81}{4} + \frac{17}{4}

STEP 13

Combine the fractions.
644 -\frac{64}{4}

STEP 14

Simplify the fraction.
16 -16

STEP 15

Divide the result of the integral by the length of the interval to find the average value.
Average value=14(16) \text{Average value} = \frac{1}{4} \cdot (-16)

STEP 16

Multiply the fraction by the result of the integral.
Average value=4 \text{Average value} = -4
The average value of x39xx^{3}-9 x on the interval 1x3-1 \leq x \leq 3 is 4-4.

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