9. Determine if the Mean Value Theorem for Integrals applies to the function f(x)=√x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem. No, the theorem does not apply. Yes, x=- Yes, x= 16 Yes, x= 16
9. Determine if the Mean Value Theorem for Integrals applies to the function f(x)=√x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem. No, the theorem does not apply. Yes, x=- Yes, x= 16 Yes, x= 16
Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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![### Applying the Mean Value Theorem for Integrals
#### Problem Statement:
Determine if the Mean Value Theorem for Integrals applies to the function \( f(x) = \sqrt{x} \) on the interval \([0, 4]\). If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
#### Options:
- ⃝ No, the theorem does not apply.
- ⃝ Yes, \( x = \frac{16}{9} \).
- ⃝ Yes, \( x = \frac{1}{4} \).
- ⃝ Yes, \( x = \frac{1}{16} \).
#### Explanation:
The Mean Value Theorem for Integrals states that if \( f \) is continuous on the closed interval \([a, b]\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[
f(c) \cdot (b - a) = \int_{a}^{b} f(x) \, dx
\]
### Detailed Solution:
First, verify that \( f(x) = \sqrt{x} \) is continuous on \([0, 4]\). Since \( \sqrt{x} \) is continuous on \([0, 4]\), the Mean Value Theorem for Integrals applies. We then need to:
1. Calculate the definite integral of \( f(x) \):
\[
\int_{0}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \left( 4^{3/2} - 0 \right) = \frac{2}{3} \cdot 8 = \frac{16}{3}
\]
2. Apply the theorem:
There exists \( c \) in \((0, 4)\) such that:
\[
\sqrt{c} \cdot 4 = \frac{16}{3}
\]
\[
\sqrt{c} = \frac{4}{3}
\]
\[
c = \left( \frac{4}{3} \right)^2 = \frac{16}{9}
\]
Thus, the correct answer is:
⃝ Yes, \( x = \frac{16}{9} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bf8d9f8-9f3c-4f11-9f6a-4feeeac00eb3%2Fd3a614b7-e252-4653-ae95-78dbc76009a3%2Fdtd0i6p_processed.png&w=3840&q=75)
Transcribed Image Text:### Applying the Mean Value Theorem for Integrals
#### Problem Statement:
Determine if the Mean Value Theorem for Integrals applies to the function \( f(x) = \sqrt{x} \) on the interval \([0, 4]\). If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
#### Options:
- ⃝ No, the theorem does not apply.
- ⃝ Yes, \( x = \frac{16}{9} \).
- ⃝ Yes, \( x = \frac{1}{4} \).
- ⃝ Yes, \( x = \frac{1}{16} \).
#### Explanation:
The Mean Value Theorem for Integrals states that if \( f \) is continuous on the closed interval \([a, b]\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[
f(c) \cdot (b - a) = \int_{a}^{b} f(x) \, dx
\]
### Detailed Solution:
First, verify that \( f(x) = \sqrt{x} \) is continuous on \([0, 4]\). Since \( \sqrt{x} \) is continuous on \([0, 4]\), the Mean Value Theorem for Integrals applies. We then need to:
1. Calculate the definite integral of \( f(x) \):
\[
\int_{0}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \left( 4^{3/2} - 0 \right) = \frac{2}{3} \cdot 8 = \frac{16}{3}
\]
2. Apply the theorem:
There exists \( c \) in \((0, 4)\) such that:
\[
\sqrt{c} \cdot 4 = \frac{16}{3}
\]
\[
\sqrt{c} = \frac{4}{3}
\]
\[
c = \left( \frac{4}{3} \right)^2 = \frac{16}{9}
\]
Thus, the correct answer is:
⃝ Yes, \( x = \frac{16}{9} \).
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