Problem 15 - Essential Calculus

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{0} x e^{-2 x} d x $$

Step-by-Step Solution

Answer

The improper integral \(\int_{-\infty}^{0} x e^{-2x} dx\) converges and its value is \(\frac{1}{2}\).

Step 1: Swap the Limits

Swap the limits of integration to get rid of the negative infinity. This step results in a change of sign during integration. Therefore, our integral becomes: \(-\int_{0}^{\infty} x e^{-2x} dx\)

Step 2: Integration by Parts

Utilize integration by parts theorem, let \(u=x\) and \(dv=e^{-2x} dx\). Calculating for \(du\) and \(v\), we get \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\). The integration by parts theorem gives us that the integral of \(u dv\) is equal to \(uv - \int v du\). Substituting for \(u\), \(v\), \(du\) and \(dv\) we get the equation \(-(-x\frac{1}{2}e^{-2x}-\frac{1}{2}\int_{0}^{\infty} e^{-2x} dx)\). This simplifies to \(x\frac{1}{2}e^{-2x}+\frac{1}{2}\int_{0}^{\infty} e^{-2x} dx\).

Step 3: Solve the remaining integral

Solve the remaining integral, that is, \(\int_{0}^{\infty} e^{-2x} dx\). This is a standard integral and its solution is \(-\frac{1}{2}e^{-2x}\) evaluated from 0 to infinity which gives us \(\frac{1}{2}\).

Step 4: Substitute values

Substitute the values of \(x\) and the evaluated integral solution into the equation from Step 2, giving us the following expression: \[\lim_{{x-> \infty}} [x\frac{1}{2}e^{-2x}]+\frac{1}{2}.\] The limit of \(x\frac{1}{2}e^{-2x}\) as \(x\) approaches infinity is 0. Therefore, our final answer becomes \(\frac{1}{2}\).