Solved on Mar 27, 2024

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

STEP 1

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

STEP 2

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

Simplify the fraction.
$-16$

STEP 15

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

STEP 16

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

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

STEP 1

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

STEP 2

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

Simplify the fraction.
$-16$

STEP 15

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

STEP 16

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

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Find the average value of x3−9xx^3 - 9xx3−9x on the interval −1≤x≤3-1 \leq x \leq 3−1≤x≤3.

STEP 1

STEP 2

1. Set up the integral for the average value of f(x)f(x)f(x) on the interval [−1,3][-1, 3][−1,3].2. Compute the integral of f(x)f(x)f(x) from −1-1−1 to 333.3. 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)=x3−9xf(x) = x^3 - 9xf(x)=x3−9x on the interval [−1,3][-1, 3][−1,3].Average value=13−(−1)∫−13(x3−9x) dx \text{Average value} = \frac{1}{3 - (-1)} \int_{-1}^{3} (x^3 - 9x) \, dx Average value=3−(−1)1​∫−13​(x3−9x)dx

STEP 4

STEP 5

Compute the integral of f(x)=x3−9xf(x) = x^3 - 9xf(x)=x3−9x from −1-1−1 to 333.∫−13(x3−9x) dx \int_{-1}^{3} (x^3 - 9x) \, dx ∫−13​(x3−9x)dx

STEP 6

Find the antiderivative of f(x)f(x)f(x).∫(x3−9x) dx=x44−9x22+C \int (x^3 - 9x) \, dx = \frac{x^4}{4} - \frac{9x^2}{2} + C ∫(x3−9x)dx=4x4​−29x2​+C

STEP 7

Evaluate the antiderivative from −1-1−1 to 333.(x44−9x22)∣−13 \left. \left( \frac{x^4}{4} - \frac{9x^2}{2} \right) \right|_{-1}^{3} (4x4​−29x2​)∣∣​−13​

STEP 8

STEP 9

Simplify the expression by calculating the values.(814−812)−(14−92) \left( \frac{81}{4} - \frac{81}{2} \right) - \left( \frac{1}{4} - \frac{9}{2} \right) (481​−281​)−(41​−29​)

STEP 10

Combine the terms within each parenthesis.(814−1624)−(14−184) \left( \frac{81}{4} - \frac{162}{4} \right) - \left( \frac{1}{4} - \frac{18}{4} \right) (481​−4162​)−(41​−418​)

STEP 11

Further simplify the expression.(−814)−(−174) \left( -\frac{81}{4} \right) - \left( -\frac{17}{4} \right) (−481​)−(−417​)

STEP 12

Subtract the two terms.−814+174 -\frac{81}{4} + \frac{17}{4} −481​+417​

STEP 13

Combine the fractions.−644 -\frac{64}{4} −464​

STEP 14

Simplify the fraction.−16 -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) Average value=41​⋅(−16)

STEP 16

Multiply the fraction by the result of the integral.Average value=−4 \text{Average value} = -4 Average value=−4The average value of x3−9xx^{3}-9 xx3−9x on the interval −1≤x≤3-1 \leq x \leq 3−1≤x≤3 is −4-4−4.

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

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

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

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

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

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

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

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

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

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

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

Simplify the fraction.
$-16$

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

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