Let's analyze the image step by step. The problem involves:

(a) Using the $u$-substitution method to evaluate $\int (2x \cdot e^{-x^2}) \, dx$, with all steps included, including a checking step.

(b) Evaluating the improper integral $\int_{0}^\infty (2x \cdot e^{-x^2}) \, dx$.

(a) Solution: $u$-Substitution Method

1. Start with the integral: $\int (2x \cdot e^{-x^2}) \, dx$

2. Let $u = x^2$. Then, $du = 2x \, dx$.

   Substituting into the integral: $\int (2x \cdot e^{-x^2}) \, dx = \int e^{-u} \, du$

3. Evaluate $\int e^{-u} \, du$: $\int e^{-u} \, du = -e^{-u} + C$

4. Substitute back $u = x^2$: $-e^{-u} + C = -e^{-x^2} + C$

Answer for (a): $\int (2x \cdot e^{-x^2}) \, dx = -e^{-x^2} + C$

Verification: Differentiate $-e^{-x^2} + C$: $\frac{d}{dx} \big(-e^{-x^2} + C\big) = e^{-x^2} \cdot (2x) = 2x \cdot e^{-x^2}$ This matches the integrand, so the solution is correct.

(b) Evaluating the Improper Integral

Now, consider: $\int_{0}^\infty (2x \cdot e^{-x^2}) \, dx$

1. Use the result from part (a): $\int (2x \cdot e^{-x^2}) \, dx = -e^{-x^2}$

   The improper integral becomes: $\lim_{b \to \infty} \bigg[ -e^{-x^2} \bigg]_0^b$

2. Evaluate the limits:

   • At the upper limit $b \to \infty$: -e^{-b^2} \to 0 \quad \text{(since e^{-b^2} \to 0)}

   • At the lower limit $x = 0$: $-e^{-0^2} = -e^0 = -1$

3. Combine the results: $\int_{0}^\infty (2x \cdot e^{-x^2}) \, dx = 0 - (-1) = 1$

Answer for (b): $\int_{0}^\infty (2x \cdot e^{-x^2}) \, dx = 1$

Would you like further explanation or a breakdown of the steps? Here are some related questions to explore:

1. What is the intuition behind choosing $u = x^2$ in $u$-substitution?
2. How do improper integrals differ from standard definite integrals?
3. Why does $e^{-b^2} \to 0$ as $b \to \infty$?
4. Can we solve $\int_{0}^\infty e^{-x^2} \, dx$ using a similar technique?
5. How does the Gaussian integral relate to part (b)?

Tip: For $u$-substitution, always check if your chosen $u$ simplifies both the integrand and the differential!