Problem 21 - College Algebra and Calculus: An Applied Approach

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{\infty} 2 x e^{-3 x^{2}} d x $$

Step-by-Step Solution

Answer

The improper integral \(\int_{-\infty}^{\infty} 2x e^{-3x^2}\) dx converges and its value is 0.

Step 1: Split the integral into two parts

We can rewrite the original integral as the sum of two integrals: \(\int_{-\infty}^{0} 2x e^{-3x^2} dx + \int_{0}^{\infty} 2x e^{-3x^2} dx\). This allows us to deal with each part of the integral separately and also allows us to use different techniques of integration for each integral if necessary.

Step 2: Perform the integration

For both integrals, we can perform a substitution \(u = -3x^2\), \(du = -6x dx\), and \(x dx = - du / 6\). We then re-rewrite and integrate: \(\int_{-\infty}^{0} 2x e^{-3x^2} dx = -\int_{\infty}^{0} e^u du / 3\), \(\int_{0}^{\infty} 2x e^{-3x^2} dx = -\int_{0}^{\infty} e^u du / 3\). Both integrals will then evaluate to \(-[-e^u / 3]_{\infty}^{0}\) and \(-[-e^u / 3]_{0}^{\infty}\) respectively.

Step 3: Evaluate the limits and check convergence

Evaluating the upper and lower limits of each integral gives us \(-[0+e^0 / 3] = -1/3\) for the first integral and \(-[-e^0 / 3 + 0] = 1/3\) for the second integral. Adding these two values gives us \(-1/3 + 1/3 = 0\). So the original improper integral converges to a value of 0.