Problem 32
Question
Although we cannot compute the antiderivative of \(f(x)=e^{-x^{2} / 2}\), it can be shown that: $$\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}$$ Use this fact to show that $$\int_{-\infty}^{\infty} x^{2} e^{-x^{2} / 2} d x=\sqrt{2 \pi}$$ Hint: Write the integrand as \(x \cdot\left(x e^{-x^{2} / 2}\right)\) and use integration by parts.
Step-by-Step Solution
Verified Answer
The integral equals \( \sqrt{2 \pi} \).
1Step 1: Setup the Integration by Parts Formula
Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). We will choose parts of the integrand to fit this formula. Let \( u = x \) and \( dv = x e^{-x^2 / 2} \, dx \).
2Step 2: Find Derivatives and Integrals
Differentiate \( u = x \) to get \( du = dx \). For \( dv = x e^{-x^2 / 2} \, dx \), let \( v = -e^{-x^2 / 2} \). This choice simplifies because the derivative of \( v \) will yield terms matching those needed for a Gaussian integral.
3Step 3: Apply Integration by Parts
Substitute the values found into the integration by parts formula: \[ \int x \cdot x e^{-x^2/2} \, dx = x \cdot (-e^{-x^2/2}) \bigg|_{-\infty}^{\infty} + \int e^{-x^2/2} \, dx. \] Note that the first term evaluates to zero due to the exponential decay of \( e^{-x^2/2} \) at infinity. The integral \( \int e^{-x^2/2} \, dx \) evaluates to \( \sqrt{2 \pi} \).
4Step 4: Simplify the Result
Since the first term is zero and the integral evaluates to \( \sqrt{2 \pi} \), we find that \( \int_{-\infty}^{\infty} x^2 e^{-x^2 / 2} \, dx = \sqrt{2 \pi} \).
Key Concepts
Gaussian IntegralAntiderivativeExponential Decay
Gaussian Integral
The Gaussian integral is one of the most famous integrals in mathematics, a fundamental result capitalized on in probability theory and various fields within science. When you hear of the Gaussian integral, it generally refers to the integral \( \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \). This integral is so significant because it represents the total area under the normal distribution curve, which plays a pivotal role in statistics.
For the exercise, you need to deal with a specific variant of the Gaussian integral: \( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi} \). This formula is the backbone for solving the given problem, particularly when evaluating the integral involving \( x^2 e^{-x^2/2} \, dx \). The notion relies on knowing the property of the Gaussian function and how it decays exponentially as \( x \) moves away from zero.
Utilizing integration by parts, this integral can be simplified by choosing appropriate functions for \( u \) and \( dv \). It's the exponential decay property of the Gaussian function that helps simplify calculations involving these types of integrals and makes them approach zero as \( x \) tends to infinity.
For the exercise, you need to deal with a specific variant of the Gaussian integral: \( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi} \). This formula is the backbone for solving the given problem, particularly when evaluating the integral involving \( x^2 e^{-x^2/2} \, dx \). The notion relies on knowing the property of the Gaussian function and how it decays exponentially as \( x \) moves away from zero.
Utilizing integration by parts, this integral can be simplified by choosing appropriate functions for \( u \) and \( dv \). It's the exponential decay property of the Gaussian function that helps simplify calculations involving these types of integrals and makes them approach zero as \( x \) tends to infinity.
Antiderivative
In calculus, an antiderivative, or sometimes called an indefinite integral, is a function that reverses differentiation. Finding the antiderivative means determining which function, when differentiated, gives rise to the function we started with. However, not all functions have elementary antiderivatives, as is the case of \( f(x) = e^{-x^2/2} \).
Interestingly, while the antiderivative for this function cannot be expressed in terms of elementary functions, we could still solve integrals involving \( e^{-x^2/2} \) using strategic calculus techniques. One such method is integration by parts, a powerful tool in integration used specifically when integrands can be product-divided into two distinct parts.
In our specific problem, by setting \( u = x \) and \( dv = x e^{-x^2/2} \, dx \), we can transform and simplify the integral. The process involves integrating by parts to a level where the resulting terms can be evaluated using known results such as the Gaussian integral. This approach exemplifies that even when an explicit antiderivative is challenging to find, mathematical techniques can still facilitate solving integrals.
Interestingly, while the antiderivative for this function cannot be expressed in terms of elementary functions, we could still solve integrals involving \( e^{-x^2/2} \) using strategic calculus techniques. One such method is integration by parts, a powerful tool in integration used specifically when integrands can be product-divided into two distinct parts.
In our specific problem, by setting \( u = x \) and \( dv = x e^{-x^2/2} \, dx \), we can transform and simplify the integral. The process involves integrating by parts to a level where the resulting terms can be evaluated using known results such as the Gaussian integral. This approach exemplifies that even when an explicit antiderivative is challenging to find, mathematical techniques can still facilitate solving integrals.
Exponential Decay
Exponential decay is a mathematical concept describing quantity decreasing at a rate proportional to its current value. It's a stark quality observed in functions similar to \( e^{-x^2/2} \).
In our exercise, exponential decay becomes crucial when evaluating integrals over infinite intervals. As \( x \) moves toward positive or negative infinity, \( e^{-x^2/2} \) rapidly approaches zero. This happens because the exponential function shrinks quickly, reducing any further contributions from distant portions of the integral.
When applying integration by parts in this exercise, recognizing the property of exponential decay effectively allows us to ignore terms that evaluate to zero, making calculations more straightforward. The quicker elements decay to zero, the fewer terms remain to be worried about. Thus, exponential decay is not only an important concept in calculus but also in real-world applications such as radioactive decay, cooling laws, and more, wherever time-dependent decrease is observed.
In our exercise, exponential decay becomes crucial when evaluating integrals over infinite intervals. As \( x \) moves toward positive or negative infinity, \( e^{-x^2/2} \) rapidly approaches zero. This happens because the exponential function shrinks quickly, reducing any further contributions from distant portions of the integral.
When applying integration by parts in this exercise, recognizing the property of exponential decay effectively allows us to ignore terms that evaluate to zero, making calculations more straightforward. The quicker elements decay to zero, the fewer terms remain to be worried about. Thus, exponential decay is not only an important concept in calculus but also in real-world applications such as radioactive decay, cooling laws, and more, wherever time-dependent decrease is observed.
Other exercises in this chapter
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