Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int \frac{s(s+1)^{2}}{\sqrt{s}} d s $$

Step-by-Step Solution

Answer

Start by simplifying the integrand, then distribute and integrate each term separately.

Step 1: Rewrite and expand

\(\frac{s(s+1)^2}{\sqrt{s}} = s^{1/2}(s+1)^2 = s^{1/2}(s^2+2s+1) = s^{5/2}+2s^{3/2}+s^{1/2}\)

Step 2: Integrate term by term

\(\int (s^{5/2}+2s^{3/2}+s^{1/2})\,ds = \frac{s^{7/2}}{7/2}+\frac{2s^{5/2}}{5/2}+\frac{s^{3/2}}{3/2} = \frac{2}{7}s^{7/2}+\frac{4}{5}s^{5/2}+\frac{2}{3}s^{3/2}+C\)

Key Concepts

Integrand Simplification

The first important concept to grasp when solving an integral is simplifying the integrand. In our case, the integrand is \( \frac{s(s+1)^{2}}{\sqrt{s}} \). At first glance, this expression can seem intimidating, but simplifying it makes the integral much easier to solve.

A crucial step in simplification is identifying terms that can be rewritten in a more manageable form. For instance, we know \( s \) is \( s^{1} \) and the square root of \( s \), \( \sqrt{s} \), is \( s^{1/2} \). Recognizing these allows us to write the fraction as a product:

• Replace \( \sqrt{s} \) with \( s^{1/2} \).
• Write the division as a multiplication with an exponent: \( \frac{s}{\sqrt{s}} = s^{1} \cdot s^{-1/2} = s^{1/2} \).

Rewriting in this way, we simplify the original integrand to \( s^{1/2}(s+1)^2 \), which is ready for further manipulation through other processes like distribution of exponents.

Distribution of Exponents

The process of distributing exponents involves applying the exponent to each term within a parenthesis. After simplification, our expression from the integrand is \( s^{1/2}(s+1)^2 \). Here, it's essential to understand how exponents work.

For the expression \((s+1)^2\), distributing exponents means:

• Recognize that \((s+1)^2\) is \((s+1)(s+1)\).
• Apply the distributive property: \( s \cdot s = s^2 \), \(s \cdot 1 = s\), \(1 \cdot s = s\), and \(1 \cdot 1 = 1\).
• Combine like terms to simplify: \[s^2 + 2s + 1\] .

So now the expression \( s^{1/2}(s^2 + 2s + 1) \) becomes the focus. Performing this distribution helps unpack more complex expressions, making them suitable for integration.

Integration Techniques

Now that the integrand is simplified to \( s^{1/2}(s^2 + 2s + 1) \), we proceed with integration techniques.We can use one of the fundamental techniques called integration by term:

• First, distribute \( s^{1/2} \) across the expression \( s^2 + 2s + 1 \) to get separate terms: \( s^{1/2}s^2, s^{1/2}2s, s^{1/2} \).
• These become \( s^{2+1/2}, 2s^{1+1/2}, s^{1/2} \) or \( s^{5/2}, 2s^{3/2}, s^{1/2} \) respectively.
• To integrate, utilize the power rule: \( \int s^n \, ds = \frac{s^{n+1}}{n+1} + C \).
• Apply this rule to each term:
  • \( \int s^{5/2} \, ds = \frac{s^{7/2}}{7/2} \).
  • \( \int 2s^{3/2} \, ds = 2 \frac{s^{5/2}}{5/2} \).
  • \( \int s^{1/2} \, ds = \frac{s^{3/2}}{3/2} \).

Combining these results, integrating term by term gives a complete solution to the integral. Each of these steps ensures the operation respects the rules of integration, leading to a correct final answer.