Problem 17 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Verify Convergence

To determine whether the given integral converges or diverges, compare it with a known convergent integral. We know that, in general, the exponential function decays faster than any power of \(x\). Thus, the integral \(\int_{0}^{\infty} x^{2} e^{-x} dx\) converges because the \(e^{-x}\) term, as \(x\) approaches infinity, reduces faster than \(x^{2}\) can increase.

Step 2: Integrate Using the IBP Method (First Time)

To evaluate the integral, use integration by parts (IBP). Set \(u = x^2\) and \(dv = e^{-x} dx\). So, \(du = 2x dx\) and \(v = -e^{-x}\). Using the formula for IBP (\(\int u dv = uv - \int v du\)), the integral transforms into:\[\int_{0}^{\infty} x^{2} e^{-x} dx = -x^{2} e^{-x} \Biggr|_{0}^{\infty} + \int_{0}^{\infty} 2x e^{-x} dx.\]

Step 3: Apply Limits to the First Term

We first evaluate the term \(-x^2e^{-x}\) from 0 to infinity, which gives us 0. Hence, the integral reduces to: \[\int_{0}^{\infty} 2x e^{-x} dx.\]

Step 4: Integrate Using the IBP Method (Second Time)

We still have an integral we need to evaluate. We again use IBP. This time we set \(u = x\) and \(dv = 2e^{-x} dx\). We then find \(du = dx\) and \(v = -2e^{-x}\). Using the IBP formula again, the integral transforms into:\[\int_{0}^{\infty} 2x e^{-x} dx = -2x e^{-x} \Biggr|_{0}^{\infty} + 2 \int_{0}^{\infty} e^{-x} dx.\]

Step 5: Evaluate the Remaining Integral and Apply Limits

We now need to evaluate the integral \(\int_{0}^{\infty} e^{-x} dx\), which is simply \(-e^{-x}\) from 0 to infinity. Evaluating the limits and summing all terms, we find that the original improper integral, \(\int_{0}^{\infty} x^{2} e^{-x} dx\), is equal to 2.