1. **Problem statement:** Given that $f$ is an odd function, and that $$\int_0^4 f(|x|) \, dx = 1.6,$$ we need to find: (i) $$\int_{-4}^0 f(x) \, dx$$ (ii) $$\int_{-4}^4 (f(|x|) + f(x)) \, dx$$ 2. **Recall properties of odd functions:** - A function $f$ is odd if $$f(-x) = -f(x)$$ for all $x$. - For odd functions, $$\int_{-a}^a f(x) \, dx = 0$$ for any $a$. 3. **Evaluate (i) $$\int_{-4}^0 f(x) \, dx$$:** Since $f$ is odd, $$\int_{-4}^0 f(x) \, dx = -\int_0^4 f(x) \, dx$$ But note that $$f(|x|) = f(x)$$ for $x \geq 0$ because $|x|=x$ when $x \geq 0$. Therefore, $$\int_0^4 f(|x|) \, dx = \int_0^4 f(x) \, dx = 1.6$$ Hence, $$\int_{-4}^0 f(x) \, dx = -1.6$$ 4. **Evaluate (ii) $$\int_{-4}^4 (f(|x|) + f(x)) \, dx$$:** Split the integral: $$\int_{-4}^4 f(|x|) \, dx + \int_{-4}^4 f(x) \, dx$$ - For the first integral, since $f(|x|)$ is an even function (because $|x|$ is even), $$\int_{-4}^4 f(|x|) \, dx = 2 \int_0^4 f(|x|) \, dx = 2 \times 1.6 = 3.2$$ - For the second integral, since $f$ is odd, $$\int_{-4}^4 f(x) \, dx = 0$$ Therefore, $$\int_{-4}^4 (f(|x|) + f(x)) \, dx = 3.2 + 0 = 3.2$$ **Final answers:** (i) $$\int_{-4}^0 f(x) \, dx = -1.6$$ (ii) $$\int_{-4}^4 (f(|x|) + f(x)) \, dx = 3.2$$