1. **State the problem:** We need to evaluate the improper integral $$\int_a^{+\infty} \frac{1}{15} e^{-\frac{x}{15}} \, dx$$ where $a$ is a constant. 2. **Recall the formula and rules:** The integral of an exponential function of the form $$\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C$$ for $k \neq 0$. 3. **Set up the integral:** $$\int_a^{+\infty} \frac{1}{15} e^{-\frac{x}{15}} \, dx = \frac{1}{15} \int_a^{+\infty} e^{-\frac{x}{15}} \, dx$$ 4. **Integrate the inner function:** Let $k = -\frac{1}{15}$, then $$\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C = -15 e^{-\frac{x}{15}} + C$$ 5. **Evaluate the definite integral:** $$\int_a^{+\infty} \frac{1}{15} e^{-\frac{x}{15}} \, dx = \frac{1}{15} \left[ -15 e^{-\frac{x}{15}} \right]_a^{+\infty} = \left[- e^{-\frac{x}{15}} \right]_a^{+\infty}$$ 6. **Calculate the limits:** As $x \to +\infty$, $e^{-\frac{x}{15}} \to 0$, so $$- e^{-\frac{+\infty}{15}} = 0$$ At $x = a$, $$- e^{-\frac{a}{15}}$$ 7. **Final result:** $$0 - \left(- e^{-\frac{a}{15}}\right) = e^{-\frac{a}{15}}$$ **Answer:** $$\int_a^{+\infty} \frac{1}{15} e^{-\frac{x}{15}} \, dx = e^{-\frac{a}{15}}$$