Problem 30 - Applied Calculus

Question

Find each indefinite integral. \(\int \frac{e^{w}+w^{2}}{3} d w\)

Step-by-Step Solution

Answer

The indefinite integral is \(\frac{1}{3} e^w + \frac{w^3}{9} + C\).

Step 1: Identify the Integral

We need to find the indefinite integral of \(\int \frac{e^w + w^2}{3} \, dw\). Notice that this is a sum of two functions divided by a constant, which can be separated out.

Step 2: Separate the Integral

Separate the integral into two simpler integrals: \(\int \frac{e^w}{3} \, dw + \int \frac{w^2}{3} \, dw\).

Step 3: Integrate Each Term Separately

Solve each integral separately: First, \(\int \frac{e^w}{3} \, dw = \frac{1}{3} \int e^w \, dw = \frac{1}{3} e^w + C_1\). Next, \(\int \frac{w^2}{3} \, dw = \frac{1}{3} \int w^2 \, dw\).

Step 4: Integrate Polynomial Term

Solve \(\int w^2 \, dw\) which is \(\frac{w^3}{3} + C_2\). Therefore, \(\int \frac{w^2}{3} \, dw = \frac{1}{3} \left( \frac{w^3}{3} \right) = \frac{w^3}{9} + C_2\).

Step 5: Combine Integrals

Combine the results of the integrals: \(\frac{1}{3} e^w + \frac{w^3}{9} + C\), where \(C\) is the constant of integration after combining \(C_1\) and \(C_2\).

Key Concepts

Integration

Integration is a central concept in calculus that involves finding the integral of a function. The integral is essentially the "reverse" of differentiation and is used to determine the area under a curve, among other things. Indefinite integrals, like the one we are dealing with in this exercise, do not have specified limits and thus include a constant of integration, represented by the letter \(C\).

One crucial aspect when dealing with integration is recognizing how to separate and simplify functions, as showcased in our exercise. By expressing functions in a way that is easier to integrate, such as breaking them down into sums or differences, students can tackle complex integrals step by step. This process enables the application of simpler integration rules to each part.

In our example, the integral was separated into parts containing the exponential function and the polynomial function, then integrated individually. This step-by-step simplification is often key in solving integration problems efficiently and correctly.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to the power of a variable. They are denoted as \(e^x\), where \(e\) is Euler's number, approximately equal to 2.71828, a constant fundamental to the natural logarithm.

In integration, exponential functions have a straightforward rule: the integral of \(e^w\) is \(e^w\) itself. This is due to the unique property of exponential functions where their rate of growth (derivative) and accumulation (integral) are the same. Thus, integrating \(e^w\) is direct, where you simply write \(e^w + C\), with \(C\) being the constant of integration.

In the given exercise, the term \(\frac{e^w}{3}\) was simplified by factoring out constant coefficients, making it conducive for easy integration. This showed how constants can be managed effectively by focusing on the primary function form.

Polynomial Functions

Polynomial functions consist of variables raised to whole number powers, each term having a coefficient. In the form \(ax^n\), polynomial functions are ubiquitous in mathematical computations due to their straightforward form and the simplicity in handling their derivatives and integrals.

When integrating polynomial functions, the power rule comes into play. For a term like \(w^n\), you integrate by adding 1 to the power and dividing by the new power: \(\int w^n \, dw = \frac{w^{n+1}}{n+1} + C\). It's a remarkably simple way to find the antiderivative of polynomials, as seen in our exercise with \(w^2\).

To solve \(\int w^2 \, dw\), you increase the power from 2 to 3, and divide by the new power, 3. The result \(\frac{w^3}{3} + C\), showcases the elegance of the power rule. Understanding these fundamental integration techniques assists students in tackling a wide array of polynomial functions they might encounter.