1. **State the problem:** Given the function $f(x) = \cos x + x^2$, we know that $$\int_0^a f(x)\,dx = 16$$ We need to find $$\int_{-a}^a f(x)\,dx$$ 2. **Recall properties of integrals and functions:** - The integral of a sum is the sum of the integrals. - $\cos x$ is an even function, meaning $\cos(-x) = \cos x$. - $x^2$ is also an even function, meaning $(-x)^2 = x^2$. - The integral of an even function over $[-a,a]$ is twice the integral from $[0,a]$. 3. **Split the integral:** $$\int_{-a}^a f(x)\,dx = \int_{-a}^a \cos x\,dx + \int_{-a}^a x^2\,dx$$ 4. **Use even function property:** Since both $\cos x$ and $x^2$ are even, $$\int_{-a}^a \cos x\,dx = 2 \int_0^a \cos x\,dx$$ $$\int_{-a}^a x^2\,dx = 2 \int_0^a x^2\,dx$$ 5. **Combine:** $$\int_{-a}^a f(x)\,dx = 2 \int_0^a (\cos x + x^2)\,dx = 2 \int_0^a f(x)\,dx$$ 6. **Substitute given value:** $$\int_0^a f(x)\,dx = 16$$ Therefore, $$\int_{-a}^a f(x)\,dx = 2 \times 16 = 32$$ **Final answer:** $$\boxed{32}$$