1. The problem asks to find the derivative of the function $f(x) = x$ at $x=5$. 2. The derivative of a function $f(x)$, denoted $f'(x)$, represents the rate of change or slope of the function at any point $x$. 3. For the function $f(x) = x$, the derivative is constant because the function is a straight line with slope 1. 4. Using the power rule, $\frac{d}{dx} x = 1$, so $f'(x) = 1$ for all $x$. 5. Therefore, $f'(5) = 1$. 6. The second problem asks to evaluate the integral $\int_{-1}^0 e^{-x^2} dx$ given that $\int_{-1}^1 e^{-x^2} dx = k$. 7. The function $e^{-x^2}$ is an even function because $e^{-(-x)^2} = e^{-x^2}$. 8. For even functions, the integral over symmetric limits satisfies $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$. 9. Given $\int_{-1}^1 e^{-x^2} dx = k$, it follows that $k = 2 \int_0^1 e^{-x^2} dx$. 10. We want $\int_{-1}^0 e^{-x^2} dx$, which by symmetry equals $\int_0^1 e^{-x^2} dx$. 11. Therefore, $\int_{-1}^0 e^{-x^2} dx = \frac{k}{2}$. 12. Since the integral is positive, the answer is $\frac{k}{2}$, not $-\frac{k}{2}$. Final answers: - $f'(5) = 1$ - $\int_{-1}^0 e^{-x^2} dx = \frac{k}{2}$