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Chapter 5: Problem 33

$$ \int x^{2} e^{-2 x^{3}} d x $$

Short Answer

Expert verified
The integral is \(-\frac{1}{6} e^{-2x^3} + C\).

Step by step solution

01

Identify the Substitution

Notice that the exponential function has a composite argument. We can try substitution by setting the inner function in the exponent as a new variable. Let's set \( u = -2x^3 \). This substitution will help simplify the integration process.
02

Differentiate for Substitution

Differentiate the expression for \( u \) with respect to \( x \). When \( u = -2x^3 \), the differentiation gives \( \frac{du}{dx} = -6x^2 \). Therefore, \( du = -6x^2 \, dx \). We will need this to change the variables in the integral.
03

Solve for dx

Next, solve for \( dx \) to substitute into the integral. From \( du = -6x^2 \, dx \), we can express \( dx \) as \( dx = \frac{du}{-6x^2} \). This will allow us to substitute into the integral.
04

Substitute into the Integral

Substitute \( x^2 \cdot e^{-2x^3} \) from the given integral. Using the substitution \( u = -2x^3 \), we have \( x^2 \, dx = -\frac{1}{6} \, du \). The original integral \( \int x^2 e^{-2x^3} \, dx \) becomes \( -\frac{1}{6} \int e^u \, du \).
05

Integrate with new variable

Integrate \( -\frac{1}{6} \int e^u \, du \). The integral of \( e^u \) with respect to \( u \) is \( e^u \). Therefore, we obtain \( -\frac{1}{6} e^u + C \), where \( C \) is the integration constant.
06

Back-Substitute the Original Variable

Substitute back the original variable. We had \( u = -2x^3 \). Replace \( u \) in \( -\frac{1}{6} e^u + C \) with \( -2x^3 \) to get \( -\frac{1}{6} e^{-2x^3} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a way to calculate the net area under a curve from one point to another on the x-axis. Unlike indefinite integrals, which include a constant of integration, definite integrals result in a specific numeric value. They are useful for finding
  • Areas under curves
  • Displacements in physics
  • Total accumulated quantities
To evaluate a definite integral, you need to find the antiderivative of the function and then apply the limits. While our original exercise focused on finding the indefinite integral, understanding definite integrals is essential as they frequently arise in applications. Remember, to transition from indefinite to definite, you simply substitute the upper and lower limits into the antiderivative and subtract.
Exponential Function
Exponential functions are mathematical functions of the form \( f(x) = a^x \), where \( a \) is a constant and \( x \) is the exponent. This type of function grows very quickly and is used to model exponential growth or decay, such as
  • Population growth
  • Radioactive decay
  • Compound interest
In the exercise, we deal with an exponential function, specifically \( e^{-2x^3} \). The base \( e \) is approximately equal to 2.71828 and is an important constant in mathematics. Its unique property is that the derivative and integral of an exponential function with base \( e \) are proportional to its value. This makes it particularly easy to integrate with respect to natural base \( e \), as seen in converting the variable for integration.
Integration Techniques
Integration techniques are strategies used to find integrals that aren't easily deducible at first sight. They include substitution, by-parts, partial fractions, among others. The integration by substitution is a method similar to solving a chain of functions. It's akin to using a change of variables to simplify an expression.Here's how it works:
  • Identify a part of the integrand that looks like the derivative of another part.
  • Define a new variable for this part (the substitution step).
  • Derive the differential of this new variable.
  • Rearrange to substitute into the original integral.
  • Integrate with respect to the new variable and back-substitute the original variable.
In our original exercise, we set \( u = -2x^3 \) and found that \( du = -6x^2 dx \). After substituting, the complicated-looking function simplified, making the integral much easier to solve. Mastery of substitution gives flexibility and reveals the hidden simplicity in complex integrals.

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Most popular questions from this chapter

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