Problem 2 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 20 , find the indicated indefinite integral. \(\int\left(x^{2 / 3}-\frac{1}{x}+5+\sqrt{x}\right) d x\)

Step-by-Step Solution

Answer

\(\frac{3}{5} x^{5/3} - \text{ln}|x| + 5x + \frac{2}{3} x^{3/2} + C\)

Step 1: Write the integral expression

The given integral is \(\bigintss (x^{2/3} - \frac{1}{x} + 5 + \frac{x^{1/2}}{2}) dx \).

Step 2: Integrate term-by-term

Integrate each term separately within the integral. For the expression inside the integral \(\bigintss (x^{2/3} dx \), use the power rule, \(\bigintss x^n dx = \frac{x^{n+1}}{n+1} + C \).

Step 3: Integrate the first term

The first term is \(\bigintss x^{2/3} dx \). Using the power rule: \(\frac{x^{(2/3) + 1}}{(2/3) + 1} = \frac{x^{5/3}}{5/3} = \frac{3}{5} x^{5/3}\).

Step 4: Integrate the second term

The second term is \(\bigintss -\frac{1}{x} dx \). This can be rewritten as \(\bigintss -x^{-1} dx \). Using the rule for integrating \(\frac{d}{dx}[\text{ln}|x|] = \frac{1}{x} \), the integral becomes \(- \text{ln}|x| \).

Step 5: Integrate the third term

The third term is \(\bigintss 5 dx \). Integrating a constant gives: \(\bigintss 5 dx = 5x \).

Step 6: Integrate the fourth term

The fourth term is \(\bigintss \frac{x^{1/2}}{2} dx \). Using the power rule: \(\bigintss x^{1/2} dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}\).

Step 7: Combine all terms and add the constant of integration

Combine all the integrated terms: \(\frac{3}{5} x^{5/3} - \text{ln}|x| + 5x + \frac{2}{3} x^{3/2} + C \).

Key Concepts

Power Rule in Calculus

The power rule is a foundational concept in calculus for finding antiderivatives, or indefinite integrals. It states that if you have an integrand of the form \(x^n\), where \(n\) is any real number except -1, the integral is given by:
\(\bigintss x^n dx = \frac{x^{n+1}}{n+1} + C\)
Here's how it works:

• Increase the exponent by one.
• Divide by the new exponent.

For example, if integrating \(x^{2/3}\), you increase \(2/3\) by 1 to get \(5/3\), then divide by \(5/3\):

\(\bigintss x^{2/3} dx = \frac{x^{5/3}}{5/3} = \frac{3}{5} x^{5/3}\)
The power rule makes integration straightforward and is highly useful when dealing with polynomials and power functions.

Logarithmic Integration

Logarithmic integration is used when integrating terms involving \(\frac{1}{x}\). The integral of \(\frac{1}{x}\) is a special case:
\(\bigintss \frac{1}{x} dx = \text{ln}|x| + C\)
This is because the derivative of \(\text{ln}|x|\) is \(\frac{1}{x}\). Let’s apply this to \(-\frac{1}{x}\):

• Rewrite \(-\frac{1}{x}\) as \(-x^{-1}\).
• Integrate using the logarithmic rule.

Conclusion: the integral is:
\(\bigintss -\frac{1}{x} dx = -\text{ln}|x| + C\).
Logarithmic integration is very useful for handling terms where the variable is in the denominator.

Constant of Integration

In indefinite integration, we always add a constant of integration \(C\). This constant accounts for all possible vertical shifts of the antiderivative function.
Consider the integral:
\(\bigintss f(x) dx = F(x) + C\)
Why is this necessary?

• Integration gives a family of functions.
• The constant \(C\) ensures generality.

For instance, if you integrate the function 5:
\(\bigintss 5 dx = 5x + C\).
Without the \(C\), you might miss some valid solutions.
In the exercise, the final answer includes \(C\) to represent all possible antiderivatives of the given function.
It ensures the solution is as general as possible, covering every potential solution.