1. **Problem statement:** Given a function $f$ defined on $\mathbb{R}$ such that $f(-x) = f(x)$ (meaning $f$ is an even function), and the integrals $\int_{-5}^1 f(x)\,dx = 24$ and $\int_1^5 f(x)\,dx = 16$, find $\int_0^1 f(x)\,dx$. 2. **Key property of even functions:** For an even function, $f(-x) = f(x)$, so the integral over symmetric intervals satisfies: $$\int_{-a}^a f(x)\,dx = 2 \int_0^a f(x)\,dx$$ 3. **Use the given integrals:** We can write: $$\int_{-5}^1 f(x)\,dx = \int_{-5}^0 f(x)\,dx + \int_0^1 f(x)\,dx$$ Since $f$ is even, $$\int_{-5}^0 f(x)\,dx = \int_0^5 f(x)\,dx$$ 4. **Express $\int_0^5 f(x)\,dx$ using the given integral from 1 to 5:** $$\int_0^5 f(x)\,dx = \int_0^1 f(x)\,dx + \int_1^5 f(x)\,dx$$ 5. **Substitute back into the equation for $\int_{-5}^1 f(x)\,dx$:** $$24 = \int_0^5 f(x)\,dx + \int_0^1 f(x)\,dx = \left(\int_0^1 f(x)\,dx + 16\right) + \int_0^1 f(x)\,dx = 16 + 2 \int_0^1 f(x)\,dx$$ 6. **Solve for $\int_0^1 f(x)\,dx$:** $$24 - 16 = 2 \int_0^1 f(x)\,dx$$ $$8 = 2 \int_0^1 f(x)\,dx$$ $$\int_0^1 f(x)\,dx = 4$$ **Final answer:** $\boxed{4}$, which corresponds to option (b).