Chapter 7: Problem 50
Use either substitution or integration by parts to evaluate each integral. $$ \int x e^{-2 x^{2}} d x $$
Short Answer
Expert verified
The integral evaluates to \(-\frac{1}{4} e^{-2x^2} + C\).
Step by step solution
01
Identify the Type of Integration Technique to Use
The integral \( \int x e^{-2x^2} \, dx \) has an exponential function with a quadratic expression \(-2x^2\) in the exponent. This suggests that substitution might be a suitable method to simplify the integration.
02
Define the Substitution
To simplify the integral, set \( u = -2x^2 \). This choice is made because it aligns with the exponent of the exponential function.
03
Find the Differential \( du \)
Differentiate \( u = -2x^2 \ \) to find \( du \): \[ \frac{du}{dx} = -4x \] \ \[ du = -4x \, dx \]
04
Solve for \( x \, dx \)
Rearrange \( du = -4x \, dx \) to express \( x \, dx \) in terms of \( du \): \[ x \, dx = -\frac{1}{4} \, du \]
05
Substitute in the Integral
Substitute \( u = -2x^2 \) and \( x \, dx = -\frac{1}{4} \, du \) into the integral: \[ \int x e^{-2x^2} \, dx = \int -\frac{1}{4} e^u \, du \] The integral simplifies to: \[ -\frac{1}{4} \int e^u \, du \]
06
Integrate
Integrate \( e^u \) with respect to \( u \): \[ -\frac{1}{4} \int e^u \, du = -\frac{1}{4} e^u + C \] \where \( C \) is the constant of integration.
07
Substitute Back to Original Variable
Replace \( u \) with the original expression in terms of \( x \) to return to the variable \( x \):\[ -\frac{1}{4} e^u + C = -\frac{1}{4} e^{-2x^2} + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used in integration to simplify complex equations. By cleverly choosing a new variable, we can turn a daunting integral into a more manageable one, especially when dealing with compositions of functions. For example, in the original exercise with the integral \( \int x e^{-2x^2} \, dx \), substitution greatly eases the process. Here's how it works:
- We identify a part of the integrand that, when substituted, simplifies the function. In our example, the exponent \(-2x^2\) was chosen.
- We set \( u = -2x^2 \) and then find its derivative, \( du = -4x \, dx \), to substitute for \( x \, dx \).
Integration by Parts
Integration by parts is another essential tool in tackling integrals, especially when dealing with products of functions. It's based on the product rule for differentiation and provides a way to find the integral of a product of two functions. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]Here's how you can apply it:
- Choose \( u \) and \( dv \) from the integrand such that differentiating \( u \) and integrating \( dv \) simplifies the problem.
- Compute \( du = \frac{du}{dx} \, dx \) and integrate \( dv \) to find \( v \).
Exponential Functions
Exponential functions are crucial in calculus, frequently appearing in integration problems due to their unique properties. The function \( e^x \) is its own derivative and integral, making it easier to work with in both differentiating and integrating. In the context of the given exercise \( \int x e^{-2x^2} \, dx \), the exponential function \( e^{-2x^2} \) plays a central role. Here's what makes exponential functions interesting in integration:
- The exponential function grows rapidly but remains easily integrable due to its derivative property.
- When combined with other functions, like polynomials, special techniques such as substitution or integration by parts are often used.