Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
