f(x)dx =F(b)- F(a) Please explein in detar L, . To evaluate the integral, first find F(s).) How to get from A to B Let f(s) = 1+s F(s)=s-2s 1/2+C lue of E/h) Ela) Thus there is no need to

Calculus: Early Transcendentals
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please explain in detail, how to get from A to B.
**Evaluate the Integral**

Evaluate the integral \(\int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds\).

Begin by simplifying the integrand of \(\int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds\) and expressing it in exponential form.

\[
\frac{s^2 + \sqrt{5}}{s^2} = 1 + s^{-3/2}
\]

According to the second part of the fundamental theorem of calculus, if \( f \) is continuous at every point of \([a,b]\) and \( F \) is any antiderivative of \( f \) on \([a,b]\), then the definite integral can be evaluated using the following formula.

\[
\int_{a}^{b} f(x) \, dx = F(b) - F(a)
\]

**Explanation of the Method:**

1. **Simplify the Function:**
   - Let \( f(s) = 1 + s^{-3/2} \).
   - To evaluate the integral, first find \( F(s) \).

2. **Find \( F(s) \):**
   - \( F(s) = s - 2s^{-1/2} + C \)

   Because of the subtraction, a constant in \( F(x) \) will not affect the value of \( F(b) - F(a) \). Thus, there is no need to include the constant \( C \).

3. **Integral Evaluation:**
   \[
   \int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds = F(\sqrt{6}) - F(1)
   \]
   \[
   = [s - 2s^{-1/2}]_{1}^{\sqrt{6}}
   \]

4. **Simplify \( F(\sqrt{6}) \):**
   \[
   F(\sqrt{6}) = (\sqrt{6}) - 2 \cdot (\sqrt{6})^{-1/2}
   \]
   \[
   = (\sqrt{6}) - 2 \cdot (6^{1/2})^{-1/2}
   \]
   \[
   = (\
Transcribed Image Text:**Evaluate the Integral** Evaluate the integral \(\int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds\). Begin by simplifying the integrand of \(\int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds\) and expressing it in exponential form. \[ \frac{s^2 + \sqrt{5}}{s^2} = 1 + s^{-3/2} \] According to the second part of the fundamental theorem of calculus, if \( f \) is continuous at every point of \([a,b]\) and \( F \) is any antiderivative of \( f \) on \([a,b]\), then the definite integral can be evaluated using the following formula. \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] **Explanation of the Method:** 1. **Simplify the Function:** - Let \( f(s) = 1 + s^{-3/2} \). - To evaluate the integral, first find \( F(s) \). 2. **Find \( F(s) \):** - \( F(s) = s - 2s^{-1/2} + C \) Because of the subtraction, a constant in \( F(x) \) will not affect the value of \( F(b) - F(a) \). Thus, there is no need to include the constant \( C \). 3. **Integral Evaluation:** \[ \int_{1}^{\sqrt{6}} \frac{s^2 + \sqrt{5}}{s^2} \, ds = F(\sqrt{6}) - F(1) \] \[ = [s - 2s^{-1/2}]_{1}^{\sqrt{6}} \] 4. **Simplify \( F(\sqrt{6}) \):** \[ F(\sqrt{6}) = (\sqrt{6}) - 2 \cdot (\sqrt{6})^{-1/2} \] \[ = (\sqrt{6}) - 2 \cdot (6^{1/2})^{-1/2} \] \[ = (\
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