Problem 30 - Calculus for Biology and Medicine

Question

Although we cannot compute the antiderivative of \(f(x)=\) \(e^{-x^{2} / 2}\), it is known 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

Answer

The integral evaluates to \(\sqrt{2\pi}\) by using integration by parts.

Step 1: Recognize the Integral Setup

We recognize the integral we want to solve is \( \int_{-\infty}^{\infty} x^{2} e^{-x^{2} / 2} dx \). This is similar to the given integral but with an \( x^2 \) factor.

Step 2: Rewrite Integrand for Integration by Parts

Following the hint, rewrite the integrand as: \[ x \cdot \left( x e^{-x^{2} / 2} \right) \]. This setup will help us apply integration by parts.

Step 3: Identify Functions for Integration by Parts

Let \( u = x \) and \( dv = x e^{-x^2 / 2} dx \). Then \( du = dx \) and we need to find \( v \) by integrating \( dv \).

Step 4: Compute v by Integrating dv

To find \( v \), compute \( \int x e^{-x^2 / 2} dx \) using the substitution \( w = -x^2 / 2 \), \( dw = -x dx \). This simplifies to \( v = -e^{-x^2 / 2} \).

Step 5: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute \( u = x \), \( du = dx \), \( v = -e^{-x^2 / 2} \), and \( dv = x e^{-x^2 / 2} dx \) into this formula.

Step 6: Solve the Integration by Parts Expression

Apply the formula: \( uv - \int v \, du \) becomes \( -x e^{-x^2 / 2} - \int -e^{-x^2 / 2} dx \). Simplify to \(-x e^{-x^2 / 2} + \int e^{-x^2 / 2} dx\).

Step 7: Evaluate the Integral from Step 6

From the given we know \( \int_{-\infty}^{\infty} e^{-x^2 / 2} dx = \sqrt{2 \pi} \). Thus, the integral of \(e^{-x^2 / 2}\) from \(-\infty\) to \(\infty\) results in \(\sqrt{2\pi}\).

Step 8: Solve the Related Expression

Note that the boundary terms \( -x e^{-x^2 / 2} \) evaluate to zero as \(x\) approaches \(\pm \infty\). Thus, the original integral evaluates to \( \sqrt{2 \pi} \).

Key Concepts

Antiderivative

The concept of an antiderivative, also known as the indefinite integral, is important in calculus. It represents the reverse operation of differentiation. Essentially, if you have a function \( f(x) \), its antiderivative is another function \( F(x) \) such that \( F'(x) = f(x) \). This means when you differentiate \( F(x) \), you get back to \( f(x) \).

For the given function \( f(x) = e^{-x^2/2} \), finding an antiderivative isn't straightforward with elementary functions. However, this doesn't make the function useless. Many times, even without a clear antiderivative, we can work with definite integrals. This is like calculating the area under the curve from one point to another, which is powerful in physics and other applied fields. Having knowledge of properties like symmetry or standard results helps us evaluate these integrals, as shown in the exercise where it's known that the integral from negative to positive infinity of \( e^{-x^2/2} \) is \( \sqrt{2\pi} \).

Antiderivatives play a crucial role in solving a variety of mathematical problems, especially in calculating the area under curves and solving differential equations.

Improper Integrals

Improper integrals extend the concept of integrals to domains without limiting endpoints or infinite limits, as seen in this exercise. When you see symbols like \(-\infty\) and \(\infty\) in limits of integration, that's your clue you're dealing with an improper integral. These are essential in mathematics because many functions extend indefinitely.When evaluating an improper integral, you often approach it via limits. You replace \(-\infty\) and \(\infty\) with limits approaching infinity, ensuring the function converges.For example, in the exercise, the integral \( \int_{-\infty}^{\infty} e^{-x^2 / 2} \, dx \) is an improper integral. It evaluates the area under the curve from negative to positive infinity, which gives a finite value even though the endpoints aren't finite. This specific case results in \( \sqrt{2\pi} \), a significant result useful in statistics and probability, particularly within the normal distribution family. This understanding is foundational in both pure and applied mathematics.

Gaussian Integral

The Gaussian integral is another massive pillar of understanding in mathematics, particularly with relevance to probability and statistics. It deals with the function \( e^{-x^2} \) over the entire real line. The standard Gaussian integral without any constants is one of the most famous results: \[ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \].In our exercise, we work with a modified Gaussian function: \( e^{-x^2 / 2} \). This integrates to \( \sqrt{2\pi} \), showing a direct relationship with the properties of the Gaussian distribution. It's common in statistical applications, such as the normal distribution, where it models how values naturally distribute around a mean.A Gaussian integral is handy not only for theoretical work but practical modeling, crucial in fields such as physics, statistics, and engineering. These integrals' symmetrical properties across zero make them fundamental in simplifying many real-world phenomena, making them a key area of study for anyone venturing into scientific fields. Understanding these properties allows you to tackle a broad range of problems involving distributions, variance, and more, giving a robust toolset for analytical endeavors.