Evaluate $\int_{-1}^{1} \left(x^{4} + x^{2}\right)\,dx$.

Explanation

1. 01

   $x^4 + x^2$ is even: $f(-x) = f(x)$, so $\int_{-1}^{1} f = 2\int_0^1 f$.

2. 02

   $\int_0^1 (x^4 + x^2)\,dx = \frac{1}{5} + \frac{1}{3} = \frac{8}{15}$, so the integral is $\frac{16}{15}$.