Chapter 6: Problem 35
$$ \text { Use integration by parts to find each integral. } $$ $$ \begin{array}{l} \int x^{3} e^{x^{2}} d x \quad\left[\text { Hint }: \text { Take } u=x^{2}, \quad d v=x e^{x^{2}},\right.\\\ \text { and use a substitution to find } v \text { from } d v \text { . } \end{array} $$
Short Answer
Expert verified
The integral is \( \frac{1}{2} e^{x^2} (x^2 - 1) + C \).
Step by step solution
01
Choose Functions for Integration by Parts
For integration by parts, choose \( u = x^2 \) and \( dv = x e^{x^2} \, dx \). This requires that we find \( du \) and \( v \) before applying the integration by parts formula.
02
Differentiate and Integrate Separately
Differentiate \( u \) to get \( du = 2x \, dx \). Next, integrate \( dv = x e^{x^2} \, dx \). Let \( w = x^2 \), then \( dw = 2x \, dx \). Thus, the integral becomes \( \int e^w \frac{dw}{2} = \frac{1}{2} e^w + C = \frac{1}{2} e^{x^2} + C \), so \( v = \frac{1}{2} e^{x^2} \).
03
Apply Integration by Parts Formula
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting the parts, we get \( \int x^3 e^{x^2} \, dx = x^2 \cdot \frac{1}{2} e^{x^2} - \int \frac{1}{2} e^{x^2} \cdot 2x \, dx \). Simplifying, \( \int x^3 e^{x^2} \, dx = \frac{x^2}{2} e^{x^2} - \int x e^{x^2} \, dx \).
04
Resolve the Remaining Integral
The remaining integral \( \int x e^{x^2} \, dx \) is straightforward because it matches the integrated form from Step 2. Substitute \( w = x^2 \), then \( dw = 2x \, dx \), and the integral becomes \( \frac{1}{2} \int e^w \frac{dw}{2} = \frac{1}{2} e^w + C = \frac{1}{2} e^{x^2} + C \).
05
Combine Results
Now, combine the parts: \( \frac{x^2}{2} e^{x^2} - (\frac{1}{2} e^{x^2} + C) = \frac{x^2}{2} e^{x^2} - \frac{1}{2} e^{x^2} - C \). This simplifies to \( \frac{1}{2} e^{x^2} (x^2 - 1) - C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Integrals
Integrals are a fundamental concept in calculus dealing with the accumulation of quantities. Essentially, an integral calculates the area under a curve, defined by a function, across a specific interval. There are two main types of integrals: definite and indefinite.
- **Indefinite Integrals:** These refer to integrals without specific limits of integration. The result is a general form plus a constant, represented mathematically as \( \int f(x) \, dx = F(x) + C \). The constant \( C \) is crucial because integration is the reverse process of differentiation, and for any function \( F(x) \), there can be many antiderivatives differing by a constant.
- **Definite Integrals:** These involve specific upper and lower limits and represent the total area under the curve between those limits. It's written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits. The result of a definite integral is a specific number, not a function.
Substitution Method in Integration
The substitution method is a powerful technique for simplifying the integration process, especially when faced with complex functions. This technique is akin to reversing the chain rule in differentiation. Here's how it works:
- Choose a new variable, typically denoted by \( u \), and express part of the original integral in terms of \( u \). The goal is to make the integral easier to handle.
- Determine \( du \) by differentiating \( u \) with respect to \( x \). This helps in replacing \( dx \) in the integral.
- Convert the entire integral from terms of \( x \) to terms of \( u \). At this stage, the integral should be simpler, allowing for straightforward integration.
- After integrating, revert back to the original variable \( x \).
Definite and Indefinite Integrals
Definite and indefinite integrals, as mentioned, are two sides of the calculus coin. Each has its unique characteristics and applications.
- **Indefinite Integrals** involve finding a general formula for the antiderivative of a function. They're specified without limits and always include a constant \( + C \). The integration by parts and substitution techniques often aim to solve indefinite integrals.
- **Definite Integrals** provide the evaluated result of an area between the curve and the x-axis from one point to another. The evaluation of a definite integral uses the Fundamental Theorem of Calculus, simplifying integration by using antiderivatives.