Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int(z+\sqrt{2} z)^{2} d z $$

Step-by-Step Solution

Answer

\(\left(1 + \frac{2\sqrt{2}}{3}\right)z^3 + C\) is the integral.

Step 1: Simplify the Expression Inside the Integral

First, simplify the expression inside the integral. Given \[z + \sqrt{2}z = z(1+\sqrt{2}).\]Now, square this expression to get:\[(z + \sqrt{2}z)^2 = (z(1+\sqrt{2}))^2 = z^2 (1+\sqrt{2})^2.\]We now have:\[(1+\sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}.\]Thus, the expression becomes:\[z^2 (3 + 2\sqrt{2}).\]

Step 2: Express the Integral in Terms of Simplified Expression

Write the integral using the simplified expression:\[\int (z^2 (3 + 2\sqrt{2})) \, dz.\]Factor out the constant term:\[(3 + 2\sqrt{2}) \int z^2 \, dz.\]

Step 3: Integrate Using Power Rule

Now, apply the power rule for integration which states \(\int z^n \, dz = \frac{z^{n+1}}{n+1} + C\) for \(n eq -1\).Substitute \(n = 2\):\[\int z^2 \, dz = \frac{z^{3}}{3} + C\]So the expression becomes:\[(3 + 2\sqrt{2}) \frac{z^3}{3} + C\]

Step 4: Simplify the Final Expression

Multiply the terms:\[\frac{3 + 2\sqrt{2}}{3} z^3 + C.\]Simplify the constant:\[\left(1 + \frac{2\sqrt{2}}{3}\right)z^3 + C.\]Thus, the integral evaluates to:\[\left(1 + \frac{2\sqrt{2}}{3}\right)z^3 + C.\]

Key Concepts

Simplifying Expressions

Before we can find the definite or indefinite integral, it's often essential to simplify expressions. Think of it like cleaning your workspace before starting a project. By simplifying, we make integration steps more straightforward.

In our problem, the original expression inside the integral is \((z + \sqrt{2}z)^2\). At first glance, this might look a bit daunting, but there's a neat trick. Notice that both terms inside the bracket have a common factor, which is \(z\). So, you can factor \(z\) out to get \(z(1 + \sqrt{2})\).

Next, we need to square the entire expression. This involves squaring \(z\) and \((1 + \sqrt{2})\). The square of \(z\) is simple: it's just \(z^2\). For \((1+\sqrt{2})^2\), expand it as \(1^2 + 2 \cdot 1 \cdot \sqrt{2} + (\sqrt{2})^2\), which simplifies to \(3 + 2\sqrt{2}\).

• Multiply \(z^2\) by the simplified constant, \(3 + 2\sqrt{2}\).
• This gives a new, much neater expression under the integral: \(z^2 (3 + 2\sqrt{2})\).

A clean expression will make the integration much easier and less prone to errors.

Integration Techniques

Once you've simplified the expression, it's time to integrate. Various integration techniques can be applied depending on the function. Here, we'll use the technique of factoring out constants.

The given simplified integral expression is \(\int z^2 (3 + 2\sqrt{2}) \, dz\). You might notice there is a constant multiplier \((3 + 2\sqrt{2})\) multiplied by \(z^2\). According to the rules of integration, constants can be factored out of the integral sign.

• Factor \((3 + 2\sqrt{2})\) out to simplify the integration process.
• This leaves the integral \((3 + 2\sqrt{2}) \int z^2 \, dz\).

Factoring allows us to focus on integrating \(z^2\) only, which is significantly simpler. This separation of constant and variable parts helps in easily applying further integration techniques without unnecessary complication.

Power Rule for Integration

The power rule is one of the most fundamental and widely used techniques in integration. It is a straightforward process to integrate powers of a single variable.

In this problem, after factoring out \((3 + 2\sqrt{2})\), you're left with \(\int z^2 \, dz\). This is a perfect candidate for the power rule.

The power rule states that \(\int z^n \, dz = \frac{z^{n+1}}{n+1} + C\), where \(n\) is not equal to \(-1\). In our problem, \(n\) is \(2\). Using the rule:

• Plug \(n=2\) into the formula, which gives \(\frac{z^{3}}{3} + C\).

Once you have this result, it's time to multiply it by the factored constant \((3 + 2\sqrt{2})\):

• This gives \((3 + 2\sqrt{2}) \cdot \frac{z^3}{3} + C\).
• Finally, you simplify it to \(\left(1 + \frac{2\sqrt{2}}{3}\right)z^3 + C\).

The power rule helps break down integrals of polynomial functions easily, making it a vital tool in calculus.