Problem 50 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int\left(x^{3 / 5}+x^{5 / 3}\right) d x $$

Step-by-Step Solution

Answer

\( \frac{5}{8}x^{8/5} + \frac{3}{8}x^{8/3} + C \)

Step 1: Break Down the Integral

The given integral can be broken down into two separate integrals. Thus, the expression \[\int \left( x^{3/5} + x^{5/3} \right) \, dx\]can be rewritten as\[\int x^{3/5} \, dx + \int x^{5/3} \, dx.\] This allows us to handle each term separately.

Step 2: Apply Power Rule to First Term

For the first term, \[\int x^{3/5} \, dx,\]use the power rule for integration, which states that \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C,\] where \(n eq -1\). Here, \(n = 3/5\). Applying the power rule, we get:\[\frac{x^{3/5 + 1}}{3/5 + 1} = \frac{x^{8/5}}{8/5}.\] Simplifying, this becomes \[\frac{5}{8}x^{8/5}.\]

Step 3: Apply Power Rule to Second Term

For the second term, \[\int x^{5/3} \, dx,\]note again that we can use the power rule for integration. Here, \(n = 5/3\). Applying the power rule, \[\frac{x^{5/3 + 1}}{5/3 + 1} = \frac{x^{8/3}}{8/3}.\] On simplification, this becomes \[\frac{3}{8}x^{8/3}.\]

Step 4: Combine Results

Now that each term has been integrated separately, combine the results. The original integral \[\int \left( x^{3/5} + x^{5/3} \right) \, dx\] becomes\[\frac{5}{8}x^{8/5} + \frac{3}{8}x^{8/3} + C.\]Here, \(C\) is the constant of integration.

Key Concepts

Power Rule for Integration

The power rule for integration is one of the fundamental tools in calculus, widely used for computing indefinite integrals. It simplifies the process of integrating polynomial expressions. To apply it, consider the formula: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Here, \(n eq -1\) ensures the expression \(n+1\) doesn't become zero, which would otherwise render the formula invalid.

When you apply this rule, you are essentially increasing the power of the variable \(x\) by one and then dividing by the new power. Don't forget to add \(C\), the constant of integration. This practice honors the fact that indefinite integrals represent a family of functions, all differing by a constant.

In our exercise, the terms \(x^{3/5}\) and \(x^{5/3}\) were individually integrated using this rule, resulting in \(\frac{5}{8}x^{8/5}\) and \(\frac{3}{8}x^{8/3}\) respectively.

Integration Techniques

Integration techniques go beyond the power rule, equipping you with methods to tackle more complicated integrals. While the power rule works perfectly for basic polynomial expressions, other methods are necessary for integrals that involve products of functions, trigonometric functions, or more complex expressions.

Here are some additional techniques you should be aware of:

• Substitution: Similar to the reverse chain rule, useful for integrals involving composite functions.
• Integration by Parts: Useful for integrals of products of functions, using the formula \(\int u \, dv = uv - \int v \, du\).
• Partial Fractions: Handy when dealing with rational functions, breaking them into simpler fractions.

In this exercise, since the given expression is already in a polynomial form, the power rule was sufficient. But as you continue your calculus journey, mastering all these techniques will greatly enhance your integration skills.

Calculus Education

Understanding calculus is critical for many fields in science, engineering, and beyond. As a student, integrating these concepts into your thinking can take time, but they're foundational to understanding change and motion in the world around us.

Here are some strategies to deepen your calculus understanding:

• Practice regularly: The more you work through problems, the more intuitive calculus will become.
• Visualize problems: Graphs and diagrams can provide insights into the behavior of functions and integrals.
• Connect concepts: See how integration relates to other areas of calculus like derivatives and limits.

This particular exercise on indefinite integrals is an excellent opportunity to solidify your understanding of the power rule and recognize how it fits into the broader context of calculus. As you grow more comfortable with these ideas, you'll find them invaluable tools in your academic and professional toolkit.