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Chapter 8: Problem 23

Exercises \(17-32,\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$\int_{0}^{\infty} x^{2} e^{-x} d x$$

Short Answer

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

Step by step solution

01

Apply the integration by parts

The first step in this exercise is to identify \(u\) and \(dv\), as we want to apply the integration by parts rule. The best choice here is to set \(u = x^2\) and \(dv = e^{-x} dx\). The differentials \(du = 2x dx\) and \(v = -e^{-x}\) are computed. According to the formula for integration by parts, we now evaluate the integral: \(\int_{0}^{\infty} x^{2} e^{-x} dx = -x^{2} e^{-x} \Big|_{0}^{\infty} - \int_{0}^{\infty} -2x e^{-x} dx\) or \(- \int_{0}^{\infty} 2x e^{-x} dx\).
02

Again apply the integration by parts

Now we have another integral that we can solve by applying the integration by parts. We set \(u = x\) and \(dv = 2e^{-x} dx\), compute the differentials \(du = dx\) and \(v = -2e^{-x}\). The result of the integration by parts is: \(- 2x e^{-x} \Big|_{0}^{\infty} - \int_{0}^{\infty} -2 e^{-x} dx\) or \(2 \int_{0}^{\infty} e^{-x} dx\). The final result is then: \(\int_{0}^{\infty} x^{2} e^{-x} dx = -x^{2} e^{-x} \Big|_{0}^{\infty} + 2x e^{-x} \Big|_{0}^{\infty} - 2 \int_{0}^{\infty} e^{-x} dx\)
03

Evaluate the integral and verify convergence/divergence

Now we proceed with the evaluation of the remaining integral and the limits. Using the limits of integration \(0\) and \(\infty\), and simplifying the expression, we have: \[ -\lim_{x \rightarrow \infty}(x^{2} e^{-x}) + \lim_{x \rightarrow \infty} (2x e^{-x}) + 2( - e^{-x} \Big|_{0}^{\infty} ) \] All of these limits turn out to be zero, hence this improper integral is convergent and its value is zero.

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