Problem 26 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Either evaluate the given improper integral or show that it diverges. $$ \int_{0}^{+\infty} 2 x^{2} e^{-x^{3}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \frac{2}{3}\.

Step 1: Identify the type of integral

Recognize that the integral in question is an improper integral because its upper limit is infinity. The integral is from 0 to \(+\infty\) of \(2x^{2}e^{-x^{3}}\).

Step 2: Use a substitution

Use the substitution \(u = x^{3}\). Then, \(du = 3x^{2}dx\), or equivalently, \(dx = \frac{du}{3x^{2}}\). Also, note that when \(x = 0\), \(u = 0\) and when \(x = \infty\), \(u = \infty\).

Step 3: Simplify the integral

Substitute \(u = x^{3}\) and \(dx = \frac{du}{3x^{2}}\) into the integral: \(\begin{aligned} \int_{0}^{+\infty} 2x^{2}e^{-x^{3}}dx &= \int_{0}^{+\infty} 2x^{2}e^{-u} \frac{du}{3x^{2}} \end{aligned}\).

Step 4: Integrate

Simplify the integral to get \(\begin{aligned} \int_{0}^{+\infty} \frac{2}{3}e^{-u}du = \frac{2}{3} \int_{0}^{+\infty} e^{-u}du. \end{aligned}\)

Step 5: Evaluate the integral

Now integrate \(\begin{aligned} \int_{0}^{+\infty} e^{-u}du \end{aligned}\). Since \(e^{-u}\) is the standard exponential decay function, its integral from \(0\) to \(+fty\) is \(1\). Thus, \(\begin{aligned} \int_{0}^{+\infty} \frac{2}{3}e^{-u}du = \frac{2}{3} [1] = \frac{2}{3} \end{aligned}\).

Key Concepts

Exponential Functions

Exponential functions play a vital role in many fields of mathematics and science. They are characterized by a constant base raised to a variable exponent. Generally, an exponential function has the form \( f(x) = a^{x} \), where \(a\) is a constant and \(x\) is a variable.
One of the most commonly used exponential functions is the natural exponential function \( e^{x} \), where \( e \) (approximately 2.718) is a mathematical constant known as Euler's number.
Exponential functions can model growth and decay processes. For instance:

• Population growth
• Radioactive decay
• Interest calculations

In the given integral, \( e^{-x^{3}}\), we deal with an exponential function representing a decay process. Understanding this function's behavior is key to solving various integration problems involving exponentials. The function decays to zero as \( x \) approaches infinity, which helps determine the convergence of the improper integral.

Substitution Method

The substitution method is a powerful technique for evaluating integrals, especially when dealing with complicated functions. It involves changing the variable of integration to simplify the integral.
Here's how it works:

• Choose a substitution \( u = g(x) \). The goal is to transform the original function into a simpler form.
• Find the derivative \( du = g'(x)dx \) and solve for \( dx \).
• Substitute \( u \) and \( dx \) in the original integral. Also, update the limits of integration if necessary.

In our example, we use the substitution \( u = x^{3} \). Then, \( du = 3x^{2}dx\), which implies \( dx = \frac{du}{3x^{2}} \). This substitution greatly simplifies the integral:
\ \int_{0}^{+\infty} 2x^{2}e^{-x^{3}}dx = \int_{0}^{+\infty} 2x^{2}e^{-u} \frac{du}{3x^{2}} = \int_{0}^{+\infty} \frac{2}{3}e^{-u}du\ By simplifying the integral, it becomes much easier to solve.

Integration Techniques

Integration techniques are methods used to find the integral of a function, which is the reverse process of differentiation. Different techniques are suited for different types of functions.
Some common techniques include:

• Substitution Method
• Integration by Parts
• Partial Fraction Decomposition
• Trigonometric Substitution

In our problem, we applied the substitution method to evaluate the improper integral. Once the substitution \( u = x^{3} \) was made, the integral was transformed to \( \frac{2}{3} \int_{0}^{+\infty} e^{-u}du \). This integral is simpler to evaluate because \( e^{-u} \) is a standard exponential decay function.
The integral of \( e^{-u} \) from 0 to \( +\infty \) is well-known and equals 1, leading to the final result:
\ \frac{2}{3} \int_{0}^{+\infty} e^{-u}du = \frac{2}{3} * 1 = \frac{2}{3}\ Understanding and mastering these integration techniques allow students to solve a wide range of problems efficiently.