1. We are asked to evaluate three integrals and determine if they converge. 2. For part (a), evaluate the integral $$\int_{-\infty}^{\ln 5} e^x \, dx$$. - The integral of $e^x$ is $e^x$. - Evaluate the improper integral by taking the limit: $$\lim_{t \to -\infty} \int_t^{\ln 5} e^x \, dx = \lim_{t \to -\infty} [e^x]_t^{\ln 5} = \lim_{t \to -\infty} (e^{\ln 5} - e^t) = 5 - 0 = 5$$. - So, the integral converges and equals 5. 3. For part (b), evaluate $$\int_{-3}^{+\infty} x^n \, dx$$ with $n \in \mathbb{N} \setminus \{1\}$. - Since $n$ is a natural number except 1, consider cases: - If $n \geq 0$ and $n \neq 1$, the integral diverges because $x^n$ grows without bound as $x \to +\infty$. - If $n$ is a natural number, it cannot be negative, so no convergence at infinity. - Therefore, the integral diverges for all $n \in \mathbb{N} \setminus \{1\}$. 4. For part (c), evaluate $$\int_{-1}^1 \frac{\sqrt[3]{x}}{\sqrt{x}} \, dx$$. - Rewrite the integrand using exponents: $$\frac{\sqrt[3]{x}}{\sqrt{x}} = \frac{x^{1/3}}{x^{1/2}} = x^{1/3 - 1/2} = x^{-1/6}$$. - The integral becomes $$\int_{-1}^1 x^{-1/6} \, dx$$. - Since $x^{-1/6}$ is not defined for negative $x$ in the real numbers (due to fractional powers of negative numbers), the integral is not defined over $[-1,1]$ in the real domain. - Hence, the integral does not converge in the real numbers. Final answers: - (a) Converges, value = 5 - (b) Diverges for all $n \in \mathbb{N} \setminus \{1\}$ - (c) Does not converge (not defined over $[-1,1]$ in real numbers)