What is the average value of root(3)(x) on the interval -5 <= x <= 9 ? Choose 1 answer: (A) (3)/(56)*(root(4)(9^(3))-root(4)(5^(3))) (B) (3)/(56)*(root(3)(9^(4))-root(3)(5^(4))) (C) (1)/(2)*(root(3)(9)-root(3)(-5)) (D) (1)/(2)*(root(3)(9)+root(3)(-5))

Set up integral: To find the average value of a function f(x) on the interval [a, b], we use the formula: Average value = (1/(b-a)) * ∫ from a to b of f(x) dx Here, f(x) = root(3)(x) = x^(1/3), a = -5, and b = 9.Calculate interval length: First, we need to set up the integral to find the average value: Average value = (1/(9 - (-5))) * ∫ from -5 to 9 of x^(1/3) dxApply power rule: Calculate the denominator of the fraction, which is the length of the interval: 9 - (-5) = 9 + 5 = 14Evaluate antiderivative: Now, the average value formula becomes: Average value = (1/14) * ∫ from -5 to 9 of x^(1/3) dxSubstitute values: To integrate x^(1/3), we use the power rule for integration, which states that ∫ x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Here, n = 1/3.Find average value: Applying the power rule, we get: ∫ x^(1/3) dx = (x^(1/3 + 1))/(1/3 + 1) + C = (x^(4/3))/(4/3) + C = (3/4) * x^(4/3) + CMatch with options: Now we need to evaluate this antiderivative from -5 to 9: (3/4) * [x^(4/3)] from -5 to 9 = (3/4) * [9^(4/3) - (-5)^(4/3)]Since we are dealing with real numbers, the cube root of a negative number is negative, and raising it to the fourth power will give us a positive result. So, (-5)^(4/3) is the same as (5^(1/3))^4, which is 5^(4/3).Now we substitute the values into the expression: (3/4) * [9^(4/3) - 5^(4/3)]Finally, we multiply this by the reciprocal of the interval length to find the average value: Average value = (1/14) * (3/4) * [9^(4/3) - 5^(4/3)] = (3/56) * [9^(4/3) - 5^(4/3)]We can now match our result with the given options. The correct answer is: (B) (3)/(56)*(root(3)(9^(4))-root(3)(5^(4)))

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What is the average value of
root(3)(x) on the interval
-5 <= x <= 9 ?
Choose 1 answer:
(A)
(3)/(56)*(root(4)(9^(3))-root(4)(5^(3)))
(B)
(3)/(56)*(root(3)(9^(4))-root(3)(5^(4)))
(C)
(1)/(2)*(root(3)(9)-root(3)(-5))
(D)
(1)/(2)*(root(3)(9)+root(3)(-5))

What is the average value of $\sqrt[3]{x}$ on the interval $-5 \leq x \leq 9$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{3}{56} \cdot\left(\sqrt[4]{9^{3}}-\sqrt[4]{5^{3}}\right)$ $\newline$ (B) $\frac{3}{56} \cdot\left(\sqrt[3]{9^{4}}-\sqrt[3]{5^{4}}\right)$ $\newline$ (C) $\frac{1}{2} \cdot(\sqrt[3]{9}-\sqrt[3]{-5})$ $\newline$ (D) $\frac{1}{2} \cdot(\sqrt[3]{9}+\sqrt[3]{-5})$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. What is the average value of

\sqrt[3]{x}

on the interval

-5 \leq x \leq 9

\newline

Choose

1

answer:

\newline

(A)

\frac{3}{56} \cdot\left(\sqrt[4]{9^{3}}-\sqrt[4]{5^{3}}\right)

\newline

(B)

\frac{3}{56} \cdot\left(\sqrt[3]{9^{4}}-\sqrt[3]{5^{4}}\right)

\newline

(C)

\frac{1}{2} \cdot(\sqrt[3]{9}-\sqrt[3]{-5})

\newline

(D)

\frac{1}{2} \cdot(\sqrt[3]{9}+\sqrt[3]{-5})

Set up integral: To find the average value of a function $f(x)$ on the interval $[a, b]$ , we use the formula: $\newline$ Average value = $\frac{1}{(b-a)} \cdot \int_{a}^{b} f(x) \, dx$ $\newline$ Here, $f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}$ , $a = -5$ , and $b = 9$ .
Calculate interval length: First, we need to set up the integral to find the average value: $\newline$ Average value = $(1/(9 - (-5))) \times \int_{-5}^{9} x^{(1/3)} \, dx$
Apply power rule: Calculate the denominator of the fraction, which is the length of the interval: $9 - (-5) = 9 + 5 = 14$
Evaluate antiderivative: Now, the average value formula becomes: $\newline$ Average value = $(1/14) \times \int_{-5}^{9} x^{(1/3)} \, dx$
Substitute values: To integrate $x^{\frac{1}{3}}$ , we use the power rule for integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ , where $C$ is the constant of integration. Here, $n = \frac{1}{3}$ .
Find average value: Applying the power rule, we get: $\newline$ $\int x^{\frac{1}{3}} dx = \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} + C$ $\newline$ = $\frac{x^{\frac{4}{3}}}{\frac{4}{3}} + C$ $\newline$ = $\frac{3}{4} \cdot x^{\frac{4}{3}} + C$
Match with options: Now we need to evaluate this antiderivative from $-5$ to $9$ : $\newline$ $(\frac{3}{4}) \cdot [x^{(\frac{4}{3})}]$ from $-5$ to $9$ $\newline$ = $(\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - (-5)^{(\frac{4}{3})}]$
Match with options: Now we need to evaluate this antiderivative from $-5$ to $9$ : $(\frac{3}{4}) \cdot [x^{(\frac{4}{3})}]$ from $-5$ to $9$ $= (\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - (-5)^{(\frac{4}{3})}]$ Since we are dealing with real numbers, the cube root of a negative number is negative, and raising it to the fourth power will give us a positive result. So, $(-5)^{(\frac{4}{3})}$ is the same as $(5^{(\frac{1}{3})})^4$ , which is $5^{(\frac{4}{3})}$ .
Match with options: Now we need to evaluate this antiderivative from $-5$ to $9$ :
$(\frac{3}{4}) \cdot [x^{(\frac{4}{3})}]$ from $-5$ to $9$
= $(\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - (-5)^{(\frac{4}{3})}]$ Since we are dealing with real numbers, the cube root of a negative number is negative, and raising it to the fourth power will give us a positive result. So, $(-5)^{(\frac{4}{3})}$ is the same as $(5^{(\frac{1}{3})})^4$ , which is $5^{(\frac{4}{3})}$ .Now we substitute the values into the expression:
$(\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - 5^{(\frac{4}{3})}]$
Match with options: Now we need to evaluate this antiderivative from $-5$ to $9$ :
$(\frac{3}{4}) \cdot [x^{(\frac{4}{3})}]$ from $-5$ to $9$
= $(\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - (-5)^{(\frac{4}{3})}]$ Since we are dealing with real numbers, the cube root of a negative number is negative, and raising it to the fourth power will give us a positive result. So, $(-5)^{(\frac{4}{3})}$ is the same as $(5^{(\frac{1}{3})})^4$ , which is $5^{(\frac{4}{3})}$ .Now we substitute the values into the expression:
$(\frac{3}{4}) \cdot [9^{(\frac{4}{3})} - 5^{(\frac{4}{3})}]$ Finally, we multiply this by the reciprocal of the interval length to find the average value:
Average value = $9$ $0$
= $9$ $1$
Match with options: Now we need to evaluate this antiderivative from $-5$ to $9$ :
$(3/4) \cdot [x^{(4/3)}]$ from $-5$ to $9$
= $(3/4) \cdot [9^{(4/3)} - (-5)^{(4/3)}]$ Since we are dealing with real numbers, the cube root of a negative number is negative, and raising it to the fourth power will give us a positive result. So, $(-5)^{(4/3)}$ is the same as $(5^{(1/3)})^4$ , which is $5^{(4/3)}$ .Now we substitute the values into the expression:
$(3/4) \cdot [9^{(4/3)} - 5^{(4/3)}]$ Finally, we multiply this by the reciprocal of the interval length to find the average value:
Average value = $9$ $0$
= $9$ $1$ We can now match our result with the given options. The correct answer is:
(B) $9$ $2$