Problem 16 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The improper integral converges and its value is 4.

Step 1: Identify the equation format

The given equation is an example of an improper integral due to its upper limit of \( \infty \). It is in the format \( \int_{a}^{b} u dv \) which indicates it is suitable for the integration by parts method. Here, identify \( u = x \) and \( dv = e^{-x / 2} dx \). Then we differentiate \( u \) and integrate \( dv \) to find \( du \) and \( v \) respectively.

Step 2: Apply Integration by Parts

According to the integration by parts formula, \( \int u dv = uv - \int v du \). Substitute \( u \), \( v \), \( du \), and \( dv \) into the formula. We get \( uv - \int v du = x \cdot 2e^{-x/2} - \int 2e^{-x/2} dx \). This integral is easier to solve.

Step 3: Evaluate the new integral and solve

The integral \( \int 2e^{-x/2} dx \) can be solved by simple integration to get -4e^{-x/2}. Thus, the original integral simplifies to \( x \cdot 2e^{-x/2} + 4e^{-x/2} \). Apply the limits of integration from 0 to \( \infty \).

Step 4: Check for Convergence

Finally, to determine convergence or divergence, substitute the limits into the equation. If the result is a real number, the integral converges, otherwise it diverges. Here, the integral is found to converge and the evaluation gives a result of 4.