Sure! I'll solve **Problem 1 (a)** as an example, following your requested format: **"Evaluate each integral or state that it diverges."** --- ## 1(a) \(\int_{}^{\infty} xe^{-x^2} dx\) ### **Step 1: Substitution** Let’s use substitution to evaluate the integral. Let \( u = x^2 \implies du = 2x dx \implies x dx = \frac{du}{2} \). **Limits:** When \( x = \), \( u = \) When \( x \to \infty \), \( u \to \infty \) So the integral becomes: \[ \int_{}^{\infty} x e^{-x^2} dx = \int_{u=}^{u=\infty} e^{-u} \frac{du}{2} = \frac{1}{2} \int_{}^{\infty} e^{-u} du \] **Explanation:** Substitution simplifies the integral to a standard exponential integral. --- ### **Step 2: Evaluate the Integral** \[ \int_{}^{\infty} e^{-u} du = \left[ -e^{-u} \right]_{}^{\infty} = - (-1) = 1 \] So, \[ \int_{}^{\infty} x e^{-x^2} dx = \frac{1}{2} \times 1 = \boxed{\frac{1}{2}} \] **Explanation:** The exponential integral from \(\) to \(\infty\) is \(1\), and the factor \(\frac{1}{2}\) comes from the substitution. --- ## **Final Answer** \[ \boxed{\int_{}^{\infty} x e^{-x^2} dx = \frac{1}{2}} \] --- If you want the next part, just let me know which one!

Sure! Let's solve all the given problems step-by-step in the same structured format. --- ## 1(a) \(\int_0^{\infty} xe^{-x^2} dx\) ### **Step 1: Substitution** Let \( u = x^2 \), then \( du = 2x dx \) or \( x dx = \frac{du}{2} \). **Limits:** - When \( x = 0 \), \( u = 0 \) - When \( x \to \infty \), \( u \to \infty \) The integral becomes: \[ \int_0^{\infty} x e^{-x^2} dx = \frac{1}{2} \int_0^{\infty} e^{-u} du \] **Explanation:** Substituting \(u\) simplifies the integral to a standard exponential form. --- ### **Step 2: Evaluate the Integral** \[ \int_0^{\infty} e^{-u} du = \left[ -e^{-u} \right]_0^{\infty} = 0 - (-1) = 1 \] Thus, \[ \int_0^{\infty} x e^{-x^2} dx = \frac{1}{2} \times 1 = \boxed{\frac{1}{2}} \] **Explanation:** The integral of \(e^{-u}\) over \([0, \infty)\) evaluates to \(1\), and we multiply by \(\frac{1}{2}\) from the substitution. --- ## 1(b) \(\int_0^{\infty} e^{-x^2} dx\) ### **Step 1: Recognize the Gaussian Integral** The integral \(\int_0^{\infty} e^{-x^2} dx\) is known as the Gaussian integral. **Explanation:** This integral evaluates to \(\frac{\sqrt{\pi}}{2}\) without needing further calculations. --- ### **Step 2: State the Result** \[ \int_0^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \] **Explanation:** This result is derived from the full Gaussian integral over \((- \infty, \infty)\), which is \(\sqrt{\pi}\). --- ## **Final Answers** 1. **Problem 1(a):** \(\boxed{\frac{1}{2}}\) 2. **Problem 1(b):** \(\boxed{\frac{\sqrt{\pi}}{2}}\) --- If you have more problems to solve or need further explanations, feel free to ask!