Problem 29 - Applied Calculus

Question

Find each indefinite integral. \(\int(1+10 w) \sqrt{w} d w\)

Step-by-Step Solution

Answer

\(\frac{2}{3} w^{3/2} + 4w^{5/2} + C\)

Step 1: Understand the Integral

The integral we want to solve is \(\int(1+10w) \sqrt{w} \, dw\). This is an indefinite integral, meaning we are looking for a general function whose derivative gives us the integrand \( (1 + 10w) \sqrt{w} \).

Step 2: Rewrite the Integrand

First, let's rewrite the integrand by expressing it in a form that is easier to integrate. Notice that \( \sqrt{w} = w^{1/2} \). Thus, we have:\[(1 + 10w) \sqrt{w} = 1 \cdot w^{1/2} + 10w \cdot w^{1/2} = w^{1/2} + 10w^{3/2}\].

Step 3: Separate the Integral

We can now separate the integral into more manageable parts:\[\int w^{1/2} \, dw + 10 \int w^{3/2} \, dw\].

Step 4: Integrate Each Term

Let's integrate each term individually using the power rule for integration:- The power rule states \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(n eq -1\).- Applying this to \(\int w^{1/2} \, dw\), we get: \[ \frac{w^{3/2}}{3/2} = \frac{2}{3} w^{3/2} \].- For \(10 \int w^{3/2} \, dw\): \[ 10 \cdot \frac{w^{5/2}}{5/2} = 10 \cdot \frac{2}{5} w^{5/2} = 4w^{5/2} \].

Step 5: Combine Results

Now we combine the results of the two integrals: \[ \frac{2}{3} w^{3/2} + 4w^{5/2} + C \].Here, \(C\) is the constant of integration.

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental tool that makes solving many types of integrals straightforward. Just like its cousin, the power rule for differentiation, it provides a simple formula to follow. When you're facing an integral of the form \( \int x^n \, dx \), the power rule specifies that the solution is \( \frac{x^{n+1}}{n+1} + C \) (where \( n eq -1 \)).
This rule is immensely helpful because it turns a complicated-looking integral into a simple calculation. Here's how it works in our exercise: the integral has been split into terms that look like \( w^{1/2} \) and \( w^{3/2} \). For each term, simply increase the power by 1 and then divide by this new power. For example, increasing \( 1/2 \) by 1 makes \( 3/2 \), and therefore, \( \int w^{1/2} \, dw = \frac{w^{3/2}}{3/2} \).
This makes integration, which might seem complex at first, much more manageable. Plus, don't forget the constant of integration \( C \), as integrating an indefinite integral yields a family of functions.

Integral Calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. The primary focus of integral calculus is on accumulation and the concept of "seeing the whole from its parts."
This branch of math deals with integrals, and the most common tasks include finding areas under curves and solving differential equations. It's a crucial tool in science and engineering as it helps to model and analyze real-world phenomena.
In our exercise, we are asked to find an indefinite integral. An indefinite integral is essentially the reverse of a derivative, meaning it's a function whose derivative is the original integrand you started with. Think of it as a way to "undo" differentiation. This means taking a variety of expressions and finding the function that originally gave rise to them when differentiated.

Integration Techniques

Integration techniques are various strategies used to tackle difficult integrals. While the power rule covers a wide range of problems, sometimes integrals require more complicated approaches.
For the given exercise, we used:

• Rewriting the integrand: This helps to simplify the expression. In our example, transforming \( \sqrt{w} \) into \( w^{1/2} \) made it easier to apply the power rule.
• Separating the integral: Breaking the original integral into smaller, simpler parts allows us to integrate each part individually, making the entire process more manageable.

Each technique is aimed at making the problem as easy as possible to solve, by transforming it into a more familiar format. Learning these techniques gives you a powerful set of tools for solving a broad array of problems in calculus.