Question

At what value of \(\mathrm{x}\) does the function \(\mathrm{I}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{xe}^{-\mathrm{x}^{2}} \mathrm{dx}\) have an extremum? What is it equal to?

Short answer

Answer: The function \(I(x)\) has a minimum extremum at \(x = 0\), and the value of the extremum is 0.

Step-by-step solution

Evaluate the integral

We want to evaluate the integral \(\mathrm{I}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{xe}^{-\mathrm{x}^{2}}\mathrm{dx}\). We will use integration by substitution. Let \(u = -x^2\), so \(\frac{du}{dx} = -2x\). Therefore, \(dx = \frac{-du}{2x}\). Now, we will rewrite the integral in terms of u: \(I(x) = \int_{0}^{x} xe^{-x^2} dx = -\frac{1}{2}\int_{0}^{-x^2} e^u du\) Now, integrating e^u with respect to u will give us: \(I(x) = -\frac{1}{2}[e^u |_0^{-x^2}] = -\frac{1}{2}(e^{-x^2} - 1)\)

Find the derivative

Now we have the function \(I(x) = -\frac{1}{2}(e^{-x^2} - 1)\), and we want to find its derivative, which is denoted as \(I'(x)\). \(I'(x) = -\frac{1}{2}\frac{d}{dx}(e^{-x^2} - 1) = -\frac{1}{2}(-2xe^{-x^2})\) Simplifying, we get: \(I'(x) = xe^{-x^2}\)

Find critical points

To find critical points, we need to set \(I'(x)\) equal to zero and solve for x: \(0 = xe^{-x^2}\) \(x = 0\) or \(e^{-x^2} = 0\) Since \(e^{-x^2}\) is never equal to 0, our only critical point is \(x = 0\).

Confirm extremum

To determine if this critical point is an extremum, we can use the second derivative (concavity) test. First, find the second derivative, \(I''(x)\): \(I''(x) = \frac{d}{dx}(xe^{-x^2}) = e^{-x^2} - 2x^2e^{-x^2}\) Now, we will evaluate the second derivative at \(x = 0\): \(I''(0) = e^0 - 2(0)^2e^0 = 1 - 0 = 1\) Since \(I''(0) > 0\), the critical point \(x=0\) corresponds to a minimum.

Find the value of the extremum

Finally, we can determine the value of the function I(x) at the minimum: \(I(0) = -\frac{1}{2}(e^{-0^2} - 1) = -\frac{1}{2}(1 - 1) = 0\) Hence, the function \(\mathrm{I}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{xe}^{-\mathrm{x}^{2}}\mathrm{dx}\) has a minimum extremum at \(x = 0\), and the value of the extremum is 0.

Key concepts

Integration by Substitution

When faced with the challenge of integrating functions that are not straightforward, integration by substitution offers a powerful technique much like the chain rule in differentiation. In the given exercise, the integration by substitution method is used to simplify the integral

Using the substitution method, we can transform the integrand into a simpler form, where \(u = -x^2\) replaces the \(x^2\) component, and \(du\) represents the differential of \(u\), which replaces \(dx\). It requires expressing \(dx\) in terms of \(du\) and involves a bit of algebraic manipulation. This strategy reduces complex functions to more elementary forms, making the integration process more manageable. It often simplifies an otherwise complicated integrand into a basic form, which is especially helpful with exponential functions as seen in this exercise.

Practicing this technique is essential as it comes up often in calculus. Understanding how to choose the right substitution can save time and dramatically ease the process of finding antiderivatives.

Finding Critical Points

After integrating, the next step in identifying extremums is finding critical points. A function's critical points occur where its derivative equals zero or does not exist. In this case, setting the derivative \(I'(x)\) equal to zero and solving for \(x\) leads us to discover these critical points.

Understanding this concept is vital for analyzing the behavior of functions since it provides the potential locations for local extremums. The derivative \(I'(x)\) simplifies to \(xe^{-x^2}\), which, when set to zero, highlights \(x = 0\) as a critical point. Remember that exponential functions, like \(e^{-x^2}\), are never zero; hence, it's another way to reinforce that only \(x = 0\) is a valid critical point in this context. Recognizing and interpreting critical points lay the groundwork for determining where functions attain their maximum and minimum values, a pivotal element in calculus and real-world problem solving.

Second Derivative Test

Once we've found potential critical points, we employ the second derivative test to determine if each point corresponds to a local minimum, local maximum, or neither. This test involves evaluating the second derivative of the function at each critical point. If the second derivative \(I''(x)\) is greater than zero at a critical point, the function has a local minimum there. If the second derivative is less than zero, it indicates a local maximum.

In our exercise, the calculation of \(I''(x)\) and its subsequent evaluation at \(x = 0\) shows that \(I''(0) > 0\). Hence, the critical point corresponds to a local minimum. The beauty of the second derivative test lies in its simplicity and how it provides a quick and reliable way to classify critical points without the need for extensive graphing or further computations. It's a powerful tool for students and professionals alike, giving them the ability to fast-track their analysis of a function's concavity and behavior at given points.