Problem 23 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(\frac{4}{z^{3}}+\frac{1}{\sqrt{z}}\right) d z\)

Step-by-Step Solution

Answer

\(-\frac{2}{z^2} + 2\sqrt{z} + C\).

Step 1: Rewrite the Integral

The first step is to rewrite the integral so that each term is in a form that is easier to integrate. Both terms involve fractions with powers of \(z\). Rewrite \(\frac{4}{z^{3}}\) as \(4z^{-3}\) and \(\frac{1}{\sqrt{z}}\) as \(z^{-1/2}\). So, the integral becomes:\[ \int \left( 4z^{-3} + z^{-1/2} \right) \, dz \]

Step 2: Integrate Each Term Separately

Next, integrate each term separately using the power rule for integration. The power rule states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n eq -1\).- Integrate \(4z^{-3}\): \[ \int 4z^{-3} \, dz = 4 \cdot \frac{z^{-2}}{-2} = -2z^{-2} \]- Integrate \(z^{-1/2}\): \[ \int z^{-1/2} \, dz = \frac{z^{1/2}}{1/2} = 2z^{1/2} \]

Step 3: Combine the Results and Simplify

Add the results of the integrals from the previous step and include the constant of integration, \(C\).The combined integral is:\[ -2z^{-2} + 2z^{1/2} + C \]This can also be rewritten, if needed, as:\[ -\frac{2}{z^2} + 2\sqrt{z} + C \]

Step 4: Verify the Integration

Optionally, differentiate the result to check your work.Differentiating \(-2z^{-2} + 2z^{1/2} + C\) should yield the original function:- Differentiating \(-2z^{-2}\) gives \(4z^{-3}\).- Differentiating \(2z^{1/2}\) gives \(z^{-1/2}\).The derivative matches the original integrand: \(\frac{4}{z^3} + \frac{1}{\sqrt{z}}\).

Key Concepts

Power Rule for Integration

The power rule is a fundamental technique used in calculus for finding indefinite integrals, also known as antiderivatives. This rule helps to simplify the integration process by providing a straightforward method to integrate functions of the form \(x^n\). The power rule states:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for any real number \(n\) except \(-1\).

It's important to remember the "plus one to the exponent, divide by the new exponent" approach. "C" is included to represent the constant of integration since indefinite integrals yield a family of functions.

In our problem, the expressions \(4z^{-3}\) and \(z^{-1/2}\) need to be integrated. Using the power rule:

• Integrate \(4z^{-3}\) by adding one to the exponent (\(-3+1 = -2\)), and dividing by the new exponent to get \(-2z^{-2}\).
• Integrate \(z^{-1/2}\) similarly by handling the fractional exponent. The result is \(2z^{1/2}\).

Integration Techniques

Integration techniques are methods used to solve integrals that are not immediately solvable through basic rules alone. For many problems, rewriting or simplifying functions before integrating makes them more manageable.

In our case, the integral \(\int \left( \frac{4}{z^3} + \frac{1}{\sqrt{z}} \right) \, dz\) was not convenient with the original format. By rewriting terms using negative exponents, we converted the integrand to \( \int (4z^{-3} + z^{-1/2}) \, dz \). This step transformed the expressions to a form suitable for applying the power rule directly.

Sometimes you may encounter more complex integrals requiring techniques like substitution or integration by parts. With practice, determining which technique to use becomes more intuitive.

Fractional Exponents

Fractional exponents can sometimes appear complex, but they follow the same basic principles as other powers. A fractional exponent like \(z^{-1/2}\) signifies both a power and a root. Here, \(-1/2\) implies the reciprocal (negative sign) of the square root (Roots are indicated by denominators in the fraction).

When dealing with fractional exponents in integration,

• First, rewrite them in terms of powers. For instance, \(\frac{1}{\sqrt{z}}\) becomes \(z^{-1/2}\).
• Next, apply the power rule to integrate: add 1 to the fractional exponent and divide by the new exponent.

In the exercise given, \(z^{-1/2}\) integrates to \(2z^{1/2}\). Understanding fractional exponents is essential in not only integration but also in simplifying and rewriting expressions across various areas of mathematics.