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To find the value(s) of [tex]\( c \)[/tex] as guaranteed by the Mean Value Theorem for Integrals for the function [tex]\( f(x) = 7 \sqrt{x} \)[/tex] over the interval [tex]\([4, 9]\)[/tex], we'll go through the following steps:
### 1. Define the Function and Interval
We have the function [tex]\( f(x) = 7 \sqrt{x} \)[/tex] and the interval [tex]\([4, 9]\)[/tex].
### 2. Calculate the Definite Integral of [tex]\( f(x) \)[/tex] over [tex]\([4, 9]\)[/tex]
We need to find the definite integral of [tex]\( f(x) = 7 \sqrt{x} \)[/tex] from 4 to 9. This integral represents the total area under the curve of [tex]\( f(x) \)[/tex] from [tex]\( x = 4 \)[/tex] to [tex]\( x = 9 \)[/tex].
Result:
[tex]\[ \int_{4}^{9} 7 \sqrt{x} \, dx = \frac{266}{3} \][/tex]
### 3. Calculate the Average Value of [tex]\( f(x) \)[/tex] over [tex]\([4, 9]\)[/tex]
The average value of the function over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average Value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \][/tex]
For our specific function:
[tex]\[ \text{Average Value} = \frac{1}{9 - 4} \int_{4}^{9} 7 \sqrt{x} \, dx = \frac{1}{5} \cdot \frac{266}{3} = \frac{266}{15} \][/tex]
### 4. Identify the Value(s) of [tex]\( c \)[/tex] such that [tex]\( f(c) = \)[/tex] Average Value
The Mean Value Theorem for Integrals states that there exists at least one [tex]\( c \)[/tex] in the interval [tex]\([4, 9]\)[/tex] such that [tex]\( f(c) \)[/tex] is equal to this average value.
We set up the equation:
[tex]\[ f(c) = \frac{266}{15} \][/tex]
Substituting [tex]\( f(x) = 7 \sqrt{x} \)[/tex]:
[tex]\[ 7 \sqrt{c} = \frac{266}{15} \][/tex]
Now solve for [tex]\( c \)[/tex]:
[tex]\[ \sqrt{c} = \frac{266}{105} \Rightarrow c = \left(\frac{266}{105}\right)^2 \][/tex]
### 5. Evaluate [tex]\( c \)[/tex]
Calculating the value of [tex]\( c \)[/tex]:
[tex]\[ c = \left(\frac{266}{105}\right)^2 \approx 6.4178 \][/tex]
### 6. Verification and Rounding
Lastly, we verify that this [tex]\( c \)[/tex] is within the interval [tex]\([4, 9]\)[/tex] and is correctly rounded to four decimal places.
Therefore, the value of [tex]\( c \)[/tex] is:
[tex]\[ c \approx 6.4178 \][/tex]
### Summary
The value(s) of [tex]\( c \)[/tex] guaranteed by the Mean Value Theorem for Integrals for the function [tex]\( f(x) = 7 \sqrt{x} \)[/tex] over the interval [tex]\([4, 9]\)[/tex] is:
[tex]\[ c = 6.4178 \][/tex]
### 1. Define the Function and Interval
We have the function [tex]\( f(x) = 7 \sqrt{x} \)[/tex] and the interval [tex]\([4, 9]\)[/tex].
### 2. Calculate the Definite Integral of [tex]\( f(x) \)[/tex] over [tex]\([4, 9]\)[/tex]
We need to find the definite integral of [tex]\( f(x) = 7 \sqrt{x} \)[/tex] from 4 to 9. This integral represents the total area under the curve of [tex]\( f(x) \)[/tex] from [tex]\( x = 4 \)[/tex] to [tex]\( x = 9 \)[/tex].
Result:
[tex]\[ \int_{4}^{9} 7 \sqrt{x} \, dx = \frac{266}{3} \][/tex]
### 3. Calculate the Average Value of [tex]\( f(x) \)[/tex] over [tex]\([4, 9]\)[/tex]
The average value of the function over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average Value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \][/tex]
For our specific function:
[tex]\[ \text{Average Value} = \frac{1}{9 - 4} \int_{4}^{9} 7 \sqrt{x} \, dx = \frac{1}{5} \cdot \frac{266}{3} = \frac{266}{15} \][/tex]
### 4. Identify the Value(s) of [tex]\( c \)[/tex] such that [tex]\( f(c) = \)[/tex] Average Value
The Mean Value Theorem for Integrals states that there exists at least one [tex]\( c \)[/tex] in the interval [tex]\([4, 9]\)[/tex] such that [tex]\( f(c) \)[/tex] is equal to this average value.
We set up the equation:
[tex]\[ f(c) = \frac{266}{15} \][/tex]
Substituting [tex]\( f(x) = 7 \sqrt{x} \)[/tex]:
[tex]\[ 7 \sqrt{c} = \frac{266}{15} \][/tex]
Now solve for [tex]\( c \)[/tex]:
[tex]\[ \sqrt{c} = \frac{266}{105} \Rightarrow c = \left(\frac{266}{105}\right)^2 \][/tex]
### 5. Evaluate [tex]\( c \)[/tex]
Calculating the value of [tex]\( c \)[/tex]:
[tex]\[ c = \left(\frac{266}{105}\right)^2 \approx 6.4178 \][/tex]
### 6. Verification and Rounding
Lastly, we verify that this [tex]\( c \)[/tex] is within the interval [tex]\([4, 9]\)[/tex] and is correctly rounded to four decimal places.
Therefore, the value of [tex]\( c \)[/tex] is:
[tex]\[ c \approx 6.4178 \][/tex]
### Summary
The value(s) of [tex]\( c \)[/tex] guaranteed by the Mean Value Theorem for Integrals for the function [tex]\( f(x) = 7 \sqrt{x} \)[/tex] over the interval [tex]\([4, 9]\)[/tex] is:
[tex]\[ c = 6.4178 \][/tex]
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