Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int \frac{\left(z^{2}+1\right)^{2}}{\sqrt{z}} d z $$

Step-by-Step Solution

Answer

\(\frac{2}{9}z^{9/2} + \frac{4}{5}z^{5/2} + 2z^{1/2} + C\)

Step 1: Rewrite the Integral

The given integral is \( \int \frac{(z^2 + 1)^2}{\sqrt{z}} \, dz \). First, let's rewrite the expression in a more manageable form by expressing \(\sqrt{z}\) as a power of , which means can be written as \(z^{1/2}\). Thus, the integral becomes: \[\int (z^2 + 1)^2 \cdot z^{-1/2} \, dz\]

Step 2: Expand the Polynomial in the Numerator

Next, expand the expression \((z^2 + 1)^2\). This can be done using the formula \((a+b)^2 = a^2 + 2ab + b^2\). Here, \(a = z^2\) and \(b = 1\). Therefore, \[(z^2 + 1)^2 = (z^2)^2 + 2(z^2)(1) + 1^2 = z^4 + 2z^2 + 1\]Substitute this into the integral to get:\[\int (z^4 + 2z^2 + 1) z^{-1/2} \, dz\]

Step 3: Distribute the Power of z

Now, distribute \(z^{-1/2}\) to each term inside the integral:\[\int (z^4 \cdot z^{-1/2} + 2z^2 \cdot z^{-1/2} + 1 \cdot z^{-1/2}) \, dz\]This simplifies to:\[\int (z^{7/2} + 2z^{3/2} + z^{-1/2}) \cdot dz\]

Step 4: Integrate Each Term Separately

Integrate each term in the expression separately using the power rule for integrals, which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\).For the term \(z^{7/2}\):\[\int z^{7/2} \, dz = \frac{z^{9/2}}{9/2} = \frac{2}{9}z^{9/2}\]For the term \(2z^{3/2}\):\[2 \int z^{3/2} \, dz = 2 \left( \frac{z^{5/2}}{5/2} \right) = \frac{4}{5}z^{5/2}\]For the term \(z^{-1/2}\):\[\int z^{-1/2} \, dz = \frac{z^{1/2}}{1/2} = 2z^{1/2}\]

Step 5: Combine the Results

Combine the results from Step 4 and add the constant of integration \(C\):\[\frac{2}{9}z^{9/2} + \frac{4}{5}z^{5/2} + 2z^{1/2} + C\]

Key Concepts

Definite Integrals

In calculus, definite integrals play a vital role in calculating the area under a curve, which can represent a wide variety of physical quantities such as distance, probability, or volume. Unlike indefinite integrals, which include a constant of integration, definite integrals calculate a specific value between two limits. Here, you focus only on a given interval and find the accumulated value inside that range.
To understand definite integrals, imagine slicing the area under the curve into small strips and summing their areas. This process involves limits and is formalized using the integral sign \( \int \) with specified upper and lower bounds.
Since the problem at hand concerns itself with integrating a function without specified limits, it represents an indefinite integral, but the concept of definite integrals aids in understanding the larger application of integration.

Power Rule

The power rule is fundamental for integration and is similar to its counterpart in differentiation. This rule simplifies finding antiderivatives of power functions and states:

• If \( n eq -1 \), then \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).

This rule makes it easy to integrate monomials by increasing the exponent by one and dividing by the new exponent. In the given exercise, the power rule allows us to find the antiderivatives of terms like \( z^{7/2} \), \( 2z^{3/2} \), and \( z^{-1/2} \), ultimately leading to solving the integral. Keep in mind this rule only directly applies when the function is a power of \( x \), meaning it must only involve \( x \) raised to a constant power, simplifying complex calculus problems.

Polynomial Expansion

Polynomial expansion enables simplification of complex expressions, making integration more manageable. In the given exercise, the expression \((z^2 + 1)^2\) is expanded using the formula \((a+b)^2 = a^2 + 2ab + b^2\).
This process results in a sum of polynomials that is much easier to work with during integration. Expanding polynomial expressions helps reveal individual terms that can be separately integrated, employing rules like the power rule.
The expanded form, \( z^4 + 2z^2 + 1 \), allows each part combined with \( z^{-1/2} \) to be integrated individually, breaking down a complex integral into simpler, solvable parts, which is crucial in solving calculus exercises effectively.

Antiderivatives

Antiderivatives, or indefinite integrals, represent the inverse process of differentiation. They play a crucial role in calculus, offering solutions to integrate functions to their original form (or close to it), before differentiation.
An antiderivative of a function \( f(x) \) is a function \( F(x) \), such that \( F'(x) = f(x) \). This property allows integration to restore functions back into their primitive forms after having been differentiated.

• The problem requires finding the antiderivative of \( \frac{(z^2 + 1)^2}{\sqrt{z}} \).
• This involves polynomial expansion, applications of the power rule, and meticulous manipulation.

Once integrated, all terms are combined, often with a constant \( C \) representing family of solutions. Solving for antiderivatives requires careful steps and understanding each term's transformation through integration.