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Chapter 5: Problem 23

Evaluate the integrals in Exercises \(1-28\). $$\int_{1}^{\sqrt{2}} \frac{s^{2}+\sqrt{s}}{s^{2}} d s$$

Short Answer

Expert verified
The integral evaluates to \( \sqrt{2} + 1 - \frac{2}{\sqrt{\sqrt{2}}} \).

Step by step solution

01

Simplify the Integrand

Start by simplifying the fraction inside the integral:\[ \frac{s^2 + \sqrt{s}}{s^2} = \frac{s^2}{s^2} + \frac{\sqrt{s}}{s^2} = 1 + \frac{s^{1/2}}{s^2} = 1 + s^{-3/2} \]Now, you have an easier integrand: \(1 + s^{-3/2}\).
02

Break Down the Integral

The integral can be broken down into two separate integrals:\[ \int_{1}^{\sqrt{2}} \left(1 + s^{-3/2}\right) ds = \int_{1}^{\sqrt{2}} 1 \, ds + \int_{1}^{\sqrt{2}} s^{-3/2} \, ds \]
03

Integrate the Constant

The integral of a constant, 1, over the interval from 1 to \(\sqrt{2}\) is:\[ \int_{1}^{\sqrt{2}} 1 \, ds = s \bigg|_{1}^{\sqrt{2}} = \sqrt{2} - 1 \]
04

Integrate the Power Function

To integrate \(s^{-3/2}\), use the power rule for integration:\[ \int s^{-3/2} \, ds = \frac{s^{-3/2 + 1}}{-3/2 + 1} = \frac{s^{-1/2}}{-1/2} = -2s^{-1/2} \]Evaluate this from \(1\) to \(\sqrt{2}\):\[ -2s^{-1/2} \bigg|_{1}^{\sqrt{2}} = -2(\sqrt{2})^{-1/2} + 2(1)^{-1/2} = -2 \cdot \frac{1}{\sqrt{\sqrt{2}}} + 2 \cdot 1 \]\[ = -2 \cdot \frac{1}{2^{1/4}} + 2 \]\[ = -\frac{2}{\sqrt{\sqrt{2}}} + 2 \]
05

Combine the Results

Combine the results from Step 3 and Step 4:\[ (\sqrt{2} - 1) + [-2(\sqrt{2})^{-1/2} + 2] \]Calculate:\[ = \sqrt{2} - 1 - \frac{2}{\sqrt{\sqrt{2}}} + 2 \]\[= \sqrt{2} + 1 - \frac{2}{\sqrt{\sqrt{2}}} \]Simplify the expression as needed to find the final result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrand Simplification
In calculus, integrating a complex expression can often be made easier by simplifying the integrand. An integrand is the function you are integrating. To simplify, break down complex fractions or manipulate algebraic expressions.

For example, consider the expression \( \frac{s^2 + \sqrt{s}}{s^2} \). Here, you can simplify by separating the terms in the numerator:
  • First, \( \frac{s^2}{s^2} = 1 \).
  • Then, \( \frac{\sqrt{s}}{s^2} = s^{-3/2} \).
Putting it together, we get the simplified integrand \( 1 + s^{-3/2} \).This simplification allows us to more easily apply integration rules. It transforms a fraction into a sum which is simpler to handle.

Remember, simplifying is a key step before diving into solving the integral itself.
Power Rule for Integration
One of the fundamental rules of integration is the power rule. This rule helps to find the integral of functions in the form of powers of \( s \). The rule states:
  • If \( f(s) = s^n \), then the integral is \( \int s^n \, ds = \frac{s^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) is the constant of integration.
In our problem, we need to integrate \( s^{-3/2} \). Utilizing the power rule:
  • Add 1 to the exponent: \(-3/2 + 1 = -1/2 \).
  • Divide by the new exponent: \( \frac{s^{-1/2}}{-1/2} = -2s^{-1/2} \).
After integration, don't forget to evaluate it over the given limits to get a numerical answer.

This rule transforms differentiating powers of \( s \) into integrating them, making calculations manageable.
Combining Results of Integrals
Once you have calculated separate integrals, the last step is to combine them to get the final result. This is the process of putting together the solutions from each part you integrated separately.

For our exercise, the integral expression \( \int (1 + s^{-3/2}) ds \) was split into two simpler integrals:
  • \(\int 1 \, ds = s |_{1}^{\sqrt{2}} = \sqrt{2} - 1 \).
  • \(-2s^{-1/2} |_{1}^{\sqrt{2}} = -2(\sqrt{2})^{-1/2} + 2(1)^{-1/2} = -\frac{2}{\sqrt{\sqrt{2}}} + 2 \).
The sum of these results gives:
  • \( (\sqrt{2} - 1) + [-\frac{2}{\sqrt{\sqrt{2}}} + 2] = \sqrt{2} + 1 - \frac{2}{\sqrt{\sqrt{2}}} \).
Combining ensures that all parts of the expression are accounted for in the final answer.

This step requires attention to detail as it's easy to make mistakes when putting various parts together. Always double-check your work to ensure accuracy.

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