Problem 40 - Calculus and its applications

Question

Find each integral. $$ \int\left(x^{4}+\frac{1}{8 \sqrt{x}}-\frac{4}{5} x^{-2 / 5}\right) d x $$

Step-by-Step Solution

Answer

\( \frac{x^5}{5} + \frac{x^{1/2}}{4} - \frac{4}{3}x^{3/5} + C \).

Step 1: Express the terms with exponents

First, let's rewrite the integral to clearly see each term's exponent form. The integral given is \( \int\left( x^{4} + \frac{1}{8\sqrt{x}} - \frac{4}{5} x^{-2/5} \right) dx \). We can rewrite \( \frac{1}{8\sqrt{x}} \) as \( \frac{1}{8}x^{-1/2} \) since \( \sqrt{x} = x^{1/2} \). Now the integral is \( \int\left( x^{4} + \frac{1}{8}x^{-1/2} - \frac{4}{5} x^{-2/5} \right) dx \).

Step 2: Integrate each term separately

We will integrate each term using the power rule for integration. The power rule states \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Apply this to each term:1. \( \int x^4 \, dx = \frac{x^{5}}{5} \).2. \( \int \frac{1}{8} x^{-1/2} \, dx = \frac{1}{8} \cdot \frac{x^{1/2}}{1/2} = \frac{x^{1/2}}{4} \).3. \( \int -\frac{4}{5} x^{-2/5} \, dx = -\frac{4}{5} \cdot \frac{x^{3/5}}{3/5} = -\frac{4}{5} \cdot \frac{5}{3} x^{3/5} = -\frac{4}{3}x^{3/5} \).

Step 3: Combine the integrated results

Now combine all the integrated results to construct the final integral solution. Including the constant of integration \( C \), the integrated expression becomes:\( \frac{x^5}{5} + \frac{x^{1/2}}{4} - \frac{4}{3}x^{3/5} + C \). This is the final result for the integral of the given function.

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental technique that every calculus student should master. This rule is highly useful for integrating polynomial expressions and functions with power terms. It simplifies the process of finding antiderivatives for functions of the form \( x^n \), where \( n \) is any real number except \( -1 \).

The power rule states that the integral of \( x^n \) with respect to \( x \) is given by:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, the \(+1\) in the exponent raises the power of \( x \) by one. The denominator, \( n+1 \), adjusts the coefficient to reflect this change in power. Let's take a closer look at how this rule applies to each part of a given function:

• For \( x^4 \), applying the power rule results in \( \frac{x^{5}}{5} \).
• For terms like \( x^{-1/2} \), the rule works the same; you raise the power to \( -1/2 + 1 = 1/2 \), resulting in \( \frac{x^{1/2}}{1/2} \), which simplifies further.
• Negative or fractional exponents like \( -2/5 \) are handled similarly, yielding \( \frac{x^{3/5}}{3/5} \).

This powerful technique allows us to integrate each term individually, which is particularly useful for polynomials and sums of power functions.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are a central concept in calculus. When you perform an indefinite integral, you find a family of functions whose derivative is the integrand. Generally, an indefinite integral has no limits of integration and is denoted by the integral sign with a function, followed by \( dx \).

The process involves reversing differentiation to find a function, or set of functions, that has a given derivative.

• This is done without specifying the upper and lower bounds of integration, leaving the result in a general form.
• The solution represents a family of functions rather than a single answer.

In the given exercise, the indefinite integral is expressed with power terms like \( x^4 \) and \( x^{-2/5} \). By finding the antiderivative of each term, you construct a general solution, which helps in understanding the behavior of the function described by the integrand.

When working through indefinite integrals, it's important to remember that the integral represents the accumulation of area under the curve, but in a general sense since there are no bounds specified.

Constant of Integration

The constant of integration, often denoted as \( C \), is a crucial component when dealing with indefinite integrals. This constant accounts for the fact that antiderivatives are not unique. Since differentiation of a constant results in zero, different antiderivatives of a function can differ by a constant.

When performing indefinite integration:

• You always add \( C \) at the end of the integration process to encompass all possible functions that could have led to the given derivative.
• Ignoring \( C \) can lead to incomplete solutions, especially when applying boundary conditions to determine specific functions.
• The constant allows for flexibility in solving real-world problems, where particular solutions are needed given additional information such as initial conditions.

In the example exercise, once each term is integrated separately, the constant \( C \) is included in the final expression. This indicates a family of functions that each serve as solutions, showing that without additional information, all of these functions are equally valid.