Problem 29 - Applied Calculus

Question

Find each indefinite integral. \(\int \frac{e^{w}-w}{2} d w\)

Step-by-Step Solution

Answer

\( \frac{1}{2}e^w - \frac{w^2}{4} + C \)

Step 1: Split the Integral

The given integral is \( \int \frac{e^w - w}{2} \, dw \). We can split this integral into two separate integrals: \( \int \frac{e^w}{2} \, dw - \int \frac{w}{2} \, dw \). This separation is possible due to the linearity of integrals.

Step 2: Integrate the Exponential Function

The integral \( \int \frac{e^w}{2} \, dw \) can be rewritten using a constant factor: \( \frac{1}{2} \int e^w \, dw \). The antiderivative of \( e^w \) is \( e^w \), so we have \( \frac{1}{2} e^w + C_1 \), where \( C_1 \) is a constant of integration.

Step 3: Integrate the Linear Function

Next, consider the integral \( \int \frac{w}{2} \, dw \). Rewrite it as \( \frac{1}{2} \int w \, dw \). The antiderivative of \( w \) is \( \frac{w^2}{2} \), thus giving us \( \frac{1}{2} \cdot \frac{w^2}{2} + C_2 = \frac{w^2}{4} + C_2 \), where \( C_2 \) is another constant of integration.

Step 4: Combine the Results

Now, combine the results from Steps 2 and 3. The indefinite integral is \( \frac{1}{2}e^w - \frac{w^2}{4} + C \), where \( C = C_1 - C_2 \) is a general constant of integration as both \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Indefinite Integral

An indefinite integral refers to finding the antiderivative of a function. It essentially reverses the process of differentiation. When you compute an indefinite integral, your result will include a function plus a constant known as the constant of integration (usually denoted as C). This is because the derivative of a constant is zero, so the exact value of the constant is unknown without additional information.

For the exercise example, the indefinite integral is transforming the expression \(\int \frac{e^w - w}{2} \, dw\) into its antiderivative form. Calculating an indefinite integral is crucial in calculus as it helps in understanding the accumulation of quantities and solving differential equations.

Integration Techniques

There are various techniques for finding integrals, each adapted to different forms of functions. When dealing with complex expressions, it can be beneficial to decompose them into simpler parts as seen in the exercise.

One commonly used approach here is splitting the integral into two separate components: \( \int \frac{e^w}{2} \, dw\) and \( \int \frac{w}{2} \, dw\). This way, you can tackle simpler integrals individually before combining the results.

• Splitting is feasible due to the linearity property of integrals which we will explore further.
• It's also essential to use constant multiples within the integral effectively, as seen with \(\frac{1}{2} \int e^w \, dw\).

By applying these techniques, complex integrals become more manageable and easier to solve.

Exponential Functions

Exponential functions are defined by expressions that involve constants raised to a variable power, with \(e^x\) being the most notable among them. They have unique properties, particularly that their derivative and antiderivative are both themselves.

• In the exercise, integrating the exponential function \(\int e^w \, dw\) simplifies effortlessly to \(e^w + C\).
• When integrating, take note of constant multipliers as in \(\frac{1}{2} \int e^w \, dw\), which simplifies to \(\frac{1}{2} e^w + C_1\).

Exponential functions often appear in growth and decay models across sciences, making their integration essential for real-world applications.

Linearity of Integrals

The linearity of integrals is a fundamental property in calculus stating that integrals can be distributed across terms and constants can be factored out. This property is essential for simplifying the computation of integrals.

In our exercise, linearity allows us to divide the original integral into:

• The integral of an exponential function \(\int \frac{e^w}{2} \, dw\).
• The integral of a simple polynomial \(\int \frac{w}{2} \, dw\).

The ability to separate and then compute these independently simplifies solving the original integral. Linearity ensures that even linear combinations of function integrations remain uncomplicated to compute, preserving structure and ease during calculation.